These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 2 Polynomials MCQS. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

Mark the correct alternative in each of the following :
Question 1.
If α, β are the zeros of the polynomial f(x) = x² + x + 1, then \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta }\) =
(a) 1
(b) -1
(c) 0
(d) None of these
Solution:
(b)

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Question 2.
If α, β are the zeros of the polynomial p(x) = 4x² + 3x + 7, then \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta }\) is equal to
(a) \(\frac { 7 }{ 3 }\)
(b) – \(\frac { 7 }{ 3 }\)
(c) \(\frac { 3 }{ 7 }\)
(d) – \(\frac { 3 }{ 7 }\)
Solution:
(d)

Question 3.
If one zero of the polynomial f(x) = (k² + 4) x² + 13x + 4k is reciprocal of the other, then k =
(a) 2
(b) -2
(c) 1
(d) -1
Solution:
(a) f (x) = (k² + 4) x² + 13x + 4k
Here a = k² + 4, b = 13, c = 4k
One zero is reciprocal of the other
Let first zero = α

k = 2

Question 4.
If the sum of the zeros of the polynomial f(x) = 2x³ – 3kx² + 4x – 5 is 6, then value of k is
(a) 2
(b) 4
(c) -2
(d) -4
Solution:
(b)

Question 5.
If α and β are the zeros of the polynomial f(x) = x² + px + q, then a polynomial having α and β is its zeros is
(a) x² + qx + p
(b) x² – px + q
(c) qx² + px + 1
(d) px² + qx + 1
Solution:
(c)

Question 6.
If α, β are the zeros of polynomial f(x) = x² – p (x + 1) – c, then (α + 1) (β + 1) =
(a) c – 1
(b) 1 – c
(c) c
(d) 1 + c
Solution:
(b)

Question 7.
If α, β are the zeros of the polynomial f(x) = x² – p(x + 1) – c such that (α + 1) (β + 1) = 0, then c =
(a) 1
(b) 0
(c) -1
(d) 2
Solution:
(a)

Question 8.
If f(x) = ax² + bx + c has no real zeros and a + b + c < 0, then
(a) c = 0
(b) c > 0
(c) c < 0
(d) None of these
Solution:
(d) f(x) = ax² + bx + c
Zeros are not real
b² – 4ac < 0 ….(i)
but a + b + c < 0
b < – (a + c)
Squaring both sides b² < (a + c)²
=> (a + c)² – 4ac < 0 {From (i)}
=> (a – c)² < 0
=> a – c < 0
=> a < c

Question 9.
If the diagram in figure shows the graph of the polynomial f(x) = ax² + bx + c, then
(a) a > 0, b < 0 and c > 0
(b) a < 0, b < 0 and c < 0
(c) a < 0, b > 0 and c > 0
(d) a < 0, b > 0 and c < 0

Solution:
(a) Curve ax² + bx + c intersects x-axis at two points and curve is upward.
a > 0, b < 0 and c> 0

Question 10.
Figure shows the graph of the polynomial f(x) = ax² + bx + c for which
(a) a < 0, b > 0 and c > 0
(b) a < 0, b < 0 and c > 0
(c) a < 0, b < 0 and c < 0
(d) a > 0, b > 0 and c < 0

Solution:
(b) Curve ax² + bx + c intersects x-axis at two points and curve is downward.
a < 0, b < 0 and c > 0

Question 11.
If the product of zeros of the polynomial f(x) = ax³ – 6x² + 11x – 6 is 4, then a =
(a) \(\frac { 3 }{ 2 }\)
(b) – \(\frac { 3 }{ 2 }\)
(c) \(\frac { 2 }{ 3 }\)
(d) – \(\frac { 2 }{ 3 }\)
Solution:
(a) f(x) = ax³ – 6x² + 11x – 6

Question 12.
If zeros of the polynomial f(x) = x³ – 3px² + qx – r are in AP, then
(a) 2p³ = pq – r
(b) 2p³ = pq + r
(c) p³ = pq – r
(d) None of these
Solution:
(a) f(x) = x³ – 3px² + qx – r
Here a = 1, b = -3p, c = q, d= -r
Zeros are in AP
Let the zeros be α – d, α, α + d

Question 13.
If the product of two zeros of the polynomial f(x) = 2x³ + 6x² – 4x + 9 is 3, then its third zero is
(a) \(\frac { 3 }{ 2 }\)
(b) – \(\frac { 3 }{ 2 }\)
(c) \(\frac { 9 }{ 2 }\)
(d) – \(\frac { 9 }{ 2 }\)
Solution:
(b)

Question 14.
If the polynomial f(x) = ax² + bx – c is divisible by the polynomial g(x) = ax² + bx + c, then ab =
(a) 1
(b) \(\frac { 1 }{ c }\)
(c) – 1
(d) – \(\frac { 1 }{ c }\)
Solution:
(a)

Question 15.
In Q. No. 14, ac =
(a) b
(b) 2b
(c) 2b2
(d) -2b
Solution:
(b) In the previous questions
Remainder = 0
(b – ac + ab²) = 0
b + ab² = ac
=> ac = b (1 + ab) = b (1 + 1) = 2b

Question 16.
If one root of the polynomial f(x) = 5x² + 13x + k is reciprocal of the other, then the value of k is
(a) 0
(b) 5
(c) \(\frac { 1 }{ 6 }\)
(d) 6
Solution:
(b)

Question 17.
If α, β, γ are the zeros of the polynomial f(x) = ax³ + bx² + cx + d, then \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } +\frac { 1 }{ \gamma }\) =
(a) – \(\frac { b }{ d }\)
(b) \(\frac { c }{ d }\)
(c) – \(\frac { c }{ d }\)
(d) \(\frac { c }{ a }\)
Solution:
(c)

Question 18.
If α, β, γ are the zeros of the polynomial f(x) = ax³ + bx² + cx + d, then α² + β² + γ² =

Solution:
(d)

Question 19.
If α, β, γ are the zeros of the polynomial f(x) = x³ – px² + qx – r, then \(\frac { 1 }{ \alpha \beta } +\frac { 1 }{ \beta \gamma } +\frac { 1 }{ \gamma \alpha }\) =
(a) \(\frac { r }{ p }\)
(b) \(\frac { p }{ r }\)
(c) – \(\frac { p }{ r }\)
(d) – \(\frac { r }{ p }\)
Solution:
(b)

Question 20.
If α, β are the zeros of the polynomial f(x) = ax² + bx + c, then \(\frac { 1 }{ { \alpha }^{ 2 } } +\frac { 1 }{ { \beta }^{ 2 } }\) =


Solution:
(b)

Question 21.
If two of the zeros of the cubic polynomial ax³ + bx² + cx + d are each equal to zero, then the third zero is
(a) \(\frac { -d }{ a }\)
(b) \(\frac { c }{ a }\)
(c) \(\frac { -b }{ a }\)
(d) \(\frac { b }{ a }\)
Solution:
(c) Two of the zeros of the cubic polynomial ax³ + bx² + cx + d are each equal to zero
Let α, β and γ are its zeros, then

Third zero will be \(\frac { -b }{ a }\)

Question 22.
If two zeros of x³ + x² – 5x – 5 are √5 and – √5 then its third zero is
(a) 1
(b) -1
(c) 2
(d) -2
Solution:
(b)

Question 23.
The product of the zeros of x³ + 4x² + x – 6 is
(a) – 4
(b) 4
(c) 6
(d) – 6
Solution:
(c)

Question 24.
What should be added to the polynomial x² – 5x + 4, so that 3 is the zero of the resulting polynomial ?
(a) 1
(b) 2
(c) 4
(d) 5
Solution:
(b) 3 is the zero of the polynomial f(x) = x² – 5x + 4
x – 3 is a factor of f(x)
Now f(3) = (3)² – 5 x 3 + 4 = 9 – 15 + 4 = 13 – 15 = -2
-2 is to be subtracting or 2 is added

Question 25.
What should be subtracted to the polynomial x² – 16x + 30, so that 15 is the zero of the resulting polynomial ?
(a) 30
(b) 14
(b) 15
(d) 16
Solution:
(c) 15 is the zero of polynomial f(x) = x² – 16x + 30
Then f(15) = 0
f(15) = (15)² – 16 x 15 + 30 = 225 – 240 + 30 = 255 – 240 = 15
15 is to be subtracted

Question 26.
A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 3, is
(a) x² – 9
(b) x² + 9
(c) x² + 3
(d) x² – 3
Solution:
(a) In a quadratic polynomial
Let α and β be its zeros
and α + β = 0
and one zero = 3
3 + β = 0 ⇒ β = -3 .
Second zero = -3
Quadratic polynomial will be
(x – 3) (x + 3) ⇒ x² – 9

Question 27.
If two zeroes of the polynomial x³ + x² – 9x – 9 are 3 and -3, then its third zero is
(a) -1
(b) 1
(c) -9
(d) 9
Solution:
(a)

=> γ = -1
Third zero = -1

Question 28.
If √5 and – √5 are two zeroes of the polynomial x³ + 3x² – 5x – 15, then its third zero is
(a) 3
(b) – 3
(c) 5
(d) – 5
Solution:
(b)

Question 29.
If x + 2 is a factor x² + ax + 2b and a + b = 4, then
(a) a = 1, b = 3
(b) a = 3, b = 1
(c) a = -1, b = 5
(d) a = 5, b = -1
Solution:
(b) x + 2 is a factor of x² + ax + 2b and a + b = 4
x + 2 is one of the factor
x = – 2 is its one zero
f(-2) = 0
=> (-2)² + a (-2) + 2b = 0
=> 4 – 2a + 2b = 0
=> 2a – 2b = 4
=> a – b = 2
But a + b = 4
Adding we get, 2a = 6 => a = 3
and a + b = 4 => 3 + b = 4 => b = 4 – 3 = 1
a = 3, b = 1

Question 30.
The polynomial which when divided by – x² + x – 1 gives a quotient x – 2 and remainder 3, is
(a) x³ – 3x² + 3x – 5
(b) – x³ – 3x² – 3x – 5
(c) – x³ + 3x² – 3x + 5
(d) x³ – 3x² – 3x + 5
Solution:
(c) Divisor = – x² + x – 1, Quotient = x – 2 and
Remainder = 3, Therefore
Polynomial = Divisor x Quotient+Remainder
= (-x² + x – 1) (x – 2) + 3
= – x³ + x² – x + 2x² – 2x + 2 + 3
= – x³ + 3x² – 3x + 5

Question 31.
The number of polynomials having zeroes -2 and 5 is
(a) 1
(b) 2
(c) 3
(d) more than 3
Solution:
(d)

Hence, the required number of polynomials are infinite i.e., more than 3.

Question 32.
If one of the zeroes of the quadratic polynomial (k – 1)x² + kx + 1 is -3, then the value of k is
(a) \(\frac { 4 }{ 3 }\)
(b) – \(\frac { 4 }{ 3 }\)
(c) \(\frac { 2 }{ 3 }\)
(d) – \(\frac { 2 }{ 3 }\)
Solution:
(a)

Question 33.
The zeroes of the quadratic polynomial x² + 99x + 127 are
(a) both positive
(b) both negative
(c) both equal
(d) one positive and one negative
Solution:
(b)

Question 34.
If the zeroes of the quadratic polynomial x² + (a + 1) x + b are 2 and -3, then
(a) a = -7, b = -1
(b) a = 5, b = -1
(c) a = 2, b = -6
(d) a = 0, b = -6
Solution:
(d)

Question 35.
Given that one of the zeroes of the cubic polynomial ax³ + bx² + cx + d is zero, the product of the other two zeroes is
(a) – \(\frac { c }{ a }\)
(b) \(\frac { c }{ a }\)
(c) 0
(d) – \(\frac { b }{ a }\)
Solution:
(b)

Question 36.
The zeroes of the quadratic polynomial x² + ax + a, a ≠ 0,
(a) cannot both be positive
(b) cannot both be negative
(c) area always unequal
(d) are always equal
Solution:
(a) Let p(x) = x² + ax + a, a ≠ 0
On comparing p(x) with ax² + bx + c, we get
a = 1, b = a and c = a

So, both zeroes are negative.
Hence, in any case zeroes of the given quadratic polynomial cannot both the positive.

Question 37.
If one of the zeroes of the cubic polynomial x³ + ax² + bx + c is -1, then the product of other two zeroes is
(a) b – a + 1
(b) b – a – 1
(c) a – b + 1
(d) a – b – 1
Solution:
(a)



=> α β = -a + b + 1
Hence, the required product of other two roots is (-a + b + 1)

Question 38.
Given that two of the zeroes of the cubic polynomial ax³ + bx² + cx + d are 0, the third zero is
(a) – \(\frac { b }{ a }\)
(b) \(\frac { b }{ a }\)
(c) \(\frac { c }{ a }\)
(d) – \(\frac { d }{ a }\)
Solution:
(a) Two of the zeroes of the cubic polynomial
ax³ + bx² + cx + d = 0, 0
Let the third zero be d
Then, use the relation between zeroes and coefficient of polynomial, we have
d + 0 + 0 = – \(\frac { b }{ a }\)
⇒ d = – \(\frac { b }{ a }\)

Question 39.
If one zero of the quadratic polynomial x² + 3x + k is 2, then the value of k is
(a) 10
(b) -10
(c) 5
(d) -5
Solution:
(b) Let the given quadratic polynomial be P(x) = x² + 3x + k
It is given that one of its zeros is 2
P(2) = 0
=> (2)² + 3(2) + k = 0 => 4 + 6 + k = 0
=> k + 10 = 0 => k = -10

Question 40.
If the zeroes of the quadratic polynomial ax² + bx + c, c ≠ 0 are equal, then
(a) c and a have opposite signs
(b) c and b have opposite signs
(c) c and a have the same sign
(d) c and b have the same sign
Solution:
(c) The zeroes of the given quadratic polynomial ax² + bx + c, c ≠ 0 are equal. If coefficient of x² and constant term have the same sign
i.e., c and a have the same sign. While b i.e., coefficient of x can be positive/negative but not zero.

Question 41.
If one of the zeroes of a quadratic polynomial of the form x² + ax + b is the negative of the other, then it
(a) has no linear term and constant term is negative.
(b) has no linear term and the constant term is positive.
(c) can have a linear term but the constant term is negative.
(d) can have a linear term but the constant term is positive.
Solution:
(a)

Given that, one of the zeroes of a quadratic polynomial p(x) is negative of the other.
αβ < 0
So, b < 0 [from Eq. (i)]
Hence, b should be negative Put a = 0, then,
p(x) = x² + b = 0 => x² = – b
=> x = ± √-b [ b < 0]
Hence, if one of the zeroes of quadratic polynomial p(x) is the negative of the other, then it has no linear term i.e., a = 0 and the constant term is negative i.e., b < 0. Alternate Method Let f(x) = x² + ax + b and by given condition the zeroes are a and -a. Sum of the zeroes = α – α = a => a = 0
f(x) = x² + b, which cannot be linear and product of zeroes = α (-α) = b
=> – α² = b
which is possible when, b < 0.
Hence, it has no linear term and the constant tenn is negative.

Question 42.
Which of the following is not the graph of a quadratic polynomial?




Solution:
(d) For any quadratic polynomial ax²+ bx + c, a 0, the graph of the corresponding equation y = ax² + bx + c has one of the two shapes either open upwards like ∪ or open downwards like ∩ depending on whether a > 0 or a < 0. These curves are called parabolas. So, option (d) cannot be possible.
Also, the curve of a quadratic polynomial crosses the X-axis on at most two points but in option (d) the curve crosses the X-axis on the three points, so it does not represent the quadratic polynomial.

Hope given RD Sharma Class 10 Solutions Chapter 2 Polynomials MCQS are helpful to complete your math homework.

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These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.1
• RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.2
• RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS
• RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

Mark the correct alternative in each of the following :
Question 1.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q such that OQ = 12 cm. Length PQ is
(a) 12 cm
(b) 13 cm
(c) 8.5 cm
(d) √119 cm
Solution:
(d) Radius of a circle OP = 5 cm OQ = 12 cm, PQ is tangent

OP ⊥ PQ
In right ∆OPQ,
OQ² = OP² + PQ² (Pythagoras Theorem)
=> (12)² = (5)2 + PQ²
=> 144 = 25 + PQ²
PQ² = 144 – 25 = 119
PQ = √119

Question 2.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(a) 7 cm
(b) 12 cm
(c) 15 cm
(d) 24.5 cm
Solution:
(a) Let PQ be the tangent from Q to the circle with O as centre
PQ = 24 cm
OQ = 25 cm

Let Radius OQ = r
OQ ⊥ PQ
Now in right ∆OPQ,
OQ² = OP² + PQ² (Pythagoras Theorem)
=> (25)² = r² + (24)²
=> 625 = r² + 576
=> r² = 625 – 576 = 49 = (7)²
r = 7
Radius of the circle = 7 cm

Question 3.
The length of the tangent from a point A at a circle, of radius 3 cm, is 4 cm. The distance of A from the centre of the circle is
(a) √7 cm
(b) 7 cm
(c) 5 cm
(d) 25 cm
Solution:
(c) Let AB be the tangent from A to the circle of centre O, then
OB = 3 cm
BA = 4 cm

OB ⊥ BA
In right ∆OBA,
OA² = OB² + BA² (Pythagoras Theorem) = (3)² + (4)² = 9 + 16 = 25 = (5)²
OA = 5
Distance of A from the centre O = 5 cm

Question 4.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80° then ∠POA is equal to
(a) 50°
(b) 60°
(c) 70°
(d) 80°
Solution:
(a) PA and PB are the tangents to the circle from P and ∠APB = 80°

∠AOB = 180° – ∠APB = 180°- 80° = 100°
But OP is the bisector of ∠AOB
∠POA = ∠POB = \(\frac { 1 }{ 2 }\) ∠AOB
=> ∠POA = \(\frac { 1 }{ 2 }\) x 100° = 50°

Question 5.
If TP and TQ are two tangents to a circle with centre O so that ∠POQ = 110°, then, ∠PTQ is equal to
(a) 60°
(b) 70°
(c) 80°
(d) 90°
Solution:
(b) TP and TQ are the tangents from T to the circle with centre O and OP, OQ are joined and ∠POQ = 110°

But ∠POQ + ∠PTQ = 180°
=> 110° + ∠PTQ = 180°
=> ∠PTQ = 180° – 110° = 70°

Question 6.
PQ is a tangent to a circle with centre O at the point P. If ∆OPQ is an isosceles triangle, then ∠OQP is equal to
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
(b) In a circle with centre O, PQ is a tangent to the circle at P and ∆OPQ is an isosceles triangle such that OP = PQ

OP is radius of the circle
OP ⊥ PQ
OP = PQ
∠POQ = ∠OQP
But ∠POQ = ∠PQO = 90° (OP ⊥ PQ)
∠OQP = ∠POQ = 45°

Question 7.
Two equal circles touch each other externally at C and AB is a common tangent to the circles. Then, ∠ACB =
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Solution:
(d) Two circles with centres O and O’ touch each other at C externally
A common tangent is drawn which touches the circles at A and B respectively.
Join OA, O’B and O’O which passes through C

AO = BO’ (radii of the equal circle)
AB || OO’
=> AOO’B is a rectangle
Draw another common tangent through C which intersects AB at D, then DA = DC = DB
ADCO and BDCO’ are squares
AC and BC are the diagonals of equal square
AC = BC
∠DAC = ∠DBC = 45°
∠ACB = 90°

Question 8.
ABC is a right angled triangle, right angled at B such that BC = 6 cm and AB = 8 cm. A circle with centre O is inscribed in ∆ABC. The radius of the circle is
(a) 1 cm
(b) 2 cm
(c) 3 cm
(d) 4 cm
Solution:
(b) In a right ∆ABC, ∠B = 90°
BC = 6 cm, AB = 8 cm

AC² = AB² + BC² (Pythagoras Theorem) = (8)² + (6)² = 64 + 36 = 100 = (10)²
AC = 10 cm
An incircle is drawn with centre 0 which touches the sides of the triangle ABC at P, Q and R
OP, OQ and OR are radii and AB, BC and CA are the tangents to the circle
OP ⊥ AB, OQ ⊥ BC and OR ⊥ CA
OPBQ is a square
Let r be the radius of the incircle
PB = BQ = r
AR = AP = 8 – r,
CQ = CR = 6 – r
AC = AR + CR
=> 10 = 8 – r + 6 – r
10 = 14 – 2r
=> 2r = 14 – 10 = 4
=> r = 2
Radius of the incircle = 2 cm

Question 9.
PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that ∠POR = 120°, then ∠OPQ is
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Solution:
(c) PQ is a tangent to the circle with centre O, from P, QOR is the diameter and ∠POR = 120°

OQ is radius and PQ is tangent to the circle
OQ ⊥ QP or ∠OQP = 90°
But ∠QOP + ∠POR = 180° (Linear pair)
=> ∠QOP + 120° = 180°
∠QOP = 180° – 120° = 60°
Now in ∆POQ
∠QOP + ∠OQP + ∠OPQ = 180° (Angles of a triangle)
=> 60° + 90° + ∠OPQ = 180°
=> 150° + ∠OPQ = 180°
=> ∠OPQ = 180° – 150° = 30°

Question 10.
If four sides of a quadrilateral ABCD are tangential to a circle, then
(a) AC + AD = BD + CD
(b) AB + CD = BC + AD
(c) AB + CD = AC + BC
(d) AC + AD = BC + DB
Solution:
(b) A circle is inscribed in a quadrilateral ABCD which touches the sides AB, BC, CD and DA at P, Q, R and S respectively then the sum of two opposite sides is equal to the sum of other two opposite sides
AB + CD = BC + AD

Question 11.
The length of the tangent drawn from a point 8 cm away from the centre of a circle of radius 6 cm is
(a) √7 cm
(b) 2√7cm
(c) 10 cm
(d) 5 cm
Solution:
(b) Radius of the circle = 6 cm
and distance of the external point from the centre = 8 cm
Length of tangent = √{(8)² – (6)²}
= √(64 – 36) = √28
= √(4 x 7) = 2√7 cm

Question 12.
AB and CD are two common tangents to circles which touch each other at C. If D lies on AB such that CD = 4 cm, then AB is equal to
(a) 4 cm
(b) 6 cm
(c) 8 cm
(d) 12 cm
Solution:
(c) AB and CD are two common tangents to the two circles which touch each other externally at C and intersect AB in D

CD = 4 cm
DA and DC are tangents to the first circle from D
CD = AD = 4 cm
Similarly DC and DB are tangents to the second circle from D
CD = DB = 4 cm
AB = AD + DB = 4 + 4 = 8 cm

Question 13.
In the adjoining figure, if AD, AE and BC are tangents to the circle at D, E and F respectively. Then,

(a) AD = AB + BC + CA
(b) 2AD = AB + BC + CA
(c) 3AD = AB + BC + CA
(d) 4AD = AB + BC + CA
Solution:
(b) AD, AE and BC are the tangents to the circle at D, E and F respectively
D and AE are tangents to the circle from A
AD = AE ……(i)
Similarly, CD = CF and BE = BF ….(ii)
Now AB + AC + BC = AE – BE + AD – CD + CF + BF
= AD – BE + AD – CD + BE + BE
= 2AD [From (i) and (ii)]
or 2 AD = AB + BC + CA

Question 14.
In the figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and QR = 4 cm, then OR =

(a) 8 cm
(b) 3 cm
(c) 2.5 cm
(d) 5 cm
Solution:
(d) In the figure, 0 is the centre of the circle
QR is tangent to the circle and QOS is a diameter SQ = 6 cm, QR = 4 cm

OQ = \(\frac { 1 }{ 2 }\) QS = \(\frac { 1 }{ 2 }\) x 6 = 3 cm
OQ is radius
OQ ⊥ QR
Now in right ∆OQR
OR² = QR² + QO² = (3)² + (4)² = 9 + 16 = 25 = (5)²
OR = 5 cm

Question 15.
In the figure, the perimeter of ∆ABC is
(a) 30 cm
(b) 60 cm
(c) 45 cm
(d) 15 cm

Solution:
(a) ∆ABC is circumscribed of circle with centre O
AQ = 4 cm, CP = 5 cm and BR = 6 cm
AQ and AR the tangents to the circle AQ = AR = 4 cm
Similarly BP and BR are tangents,
BP = BR = 6 cm
and CP and CQ are the tangents
CQ = CP = 5 cm

AB = AR + BR = 4 + 6 = 10 cm
BC = BP + CP = 6 + 5 = 11 cm
AC = AQ + CQ = 4 + 5 = 9 cm
Perimeter of ∆ABC = AB + BC + AC = 10 + 11 + 9 = 30 cm

Question 16.
In the figure, AP is a tangent to the circle with centre O such that OP = 4 cm and ∠OPA = 30°. Then, AP =
(a) 2√2 cm
(b) 2 cm
(c) 2√3 cm
(d) 3√2 cm

Solution:
(c) In the figure, AP is the tangent to the circle with centre O such that
OP = 4 cm, ∠OPA = 30°
Join OA, let AP = x

Question 17.
AP and AQ are tangents drawn from a point A to a circle with centre O and radius 9 cm. If OA = 15 cm, then AP + AQ =
(a) 12 cm
(b) 18 cm
(c) 24 cm
(d) 36 cm
Solution:
(c) OP is radius, PA is the tangent
OP ⊥ AP

Now in right ∆OAP,
OA² = OP² + AP²
(15)² = (9)² + AP²
225 = 81 + AP²
=> AP² = 225 – 81 = 144 = (12)²
AP = 12 cm
But AP = AQ = 12 cm (tangents from A to the circle)
AP + AQ = 12+ 12 = 24 cm

Question 18.
At one end of a diameter PQ of a circle of radius 5 cm, tangent XPY is drawn to the circle. The length of chord AB parallel to XY and at a distance of 8 cm from P is
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) 8 cm
Solution:
(d) In the figure, PQ is diameter XPY is tangent to the circle with centre O and radius 5 cm
From P, at a distance of 8 cm AB is a chord drawn parallel to XY
To find the length of AB
Join OA

XY is tangent and OP is the radius
OP ⊥ XY or PQ ⊥ XY
AB || XY
OQ is ⊥ AB which meets AB at R
Now in right ∆OAR,
OA² = OR² + AR²
(5)² = (3)² + AR²
25 = 9 + AR²
=> AR² = 25 – 9 = 16 = (4)²
AR = 4 cm
But R is mid-point of AB
AB = 2 AR = 2 x 4 = 8 cm

Question 19.
If PT is tahgent drawn froth a point P to a circle touching it at T and O is the centre of the circle, then ∠OPT + ∠POT =
(a) 30°
(b) 60°
(c) 90°
(d) 180°
Solution:
(c) In the figure, PT is the tangent to the circle with centre O.
OP and OT are joined

PT is tangent and OT is the radius
OT ⊥ PT
Now in right ∆OPT
∠OTP = 90°
∠OPT + ∠POT = 180° – 90° = 90°

Question 20.
In the adjacent figure, if AB = 12 cm, BC = 8 cm and AC = 10 cm, then AD =
(a) 5 cm
(b) 4 cm
(c) 6 cm
(d) 7 cm

Solution:
(d) In the figure, ∆ABC is the circumscribed a circle
AB = 12 cm, BC = 8 cm and AC = 10 cm
Let AD = a, DB = b and EC = c, then
AF = a, BE = b and FC = c

But AB + BC + AC = 12 + 8 + 10 = 30
a + b + b + c + c + a = 30
=> 2 (a + b + c) = 30
a + b + c = 15
Subtracting BC or b + c from this a = 15 – 8 = 7
AD = 7 cm

Question 21.
In the figure, if AP = PB, then
(a) AC = AB
(b) AC = BC
(c) AQ = QC
(d) AB = BC

Solution:
(b) In the figure, AP = PB
But AP and AQ are the tangent from A to the circle

AP = AQ
Similarly PB = BR
But AP = PB (given)
AQ = BR ….(i)
But CQ and CR the tangents drawn from C to the circle
CQ = CR
Adding in (i)
AQ + CQ = BR + CR
AC = BC

Question 22.
In the figure, if AP = 10 cm, then BP =
(a) √91 cm
(b) √127 cm
(c) √119 cm
(d) √109 cm

Solution:
(b) In the figure,
OA = 6 cm, OB = 3 cm and AP = 10 cm

OA is radius and AP is the tangent
OA ⊥ AP
Now in right ∆OAP
OP² = AP² + OA² = (10)² + (6)² = 100 + 36 = 136
Similarly BP is tangent and OB is radius
OP² = OB² + BP²
136 = (3)² + BP2
=> 136 = 9 + BP²
=> BP² = 136 – 9 = 127
BP = √127 cm

Question 23.
In the figure, if PR is tangent to the circle at P and Q is the centre of the circle, then ∠POQ =
(a) 110°
(b) 100°
(c) 120°
(d) 90°

Solution:
(c) In the figure, PR is the tangent to the circle at P.
O is the centre of the circle ∠QPR = 60°

OP is the radius and PR is the tangent OPR = 90°
=> ∠OPQ + ∠QPR = 90°
=> ∠OPQ + 60° = 90°
=> ∠OPQ = 90° – 60° = 30°
OP = OQ (radii of the circle)
∠OQP = 30°
In ∆OPQ,
∠OPQ + ∠OQP + ∠POQ = 180°
=> 30° + 30° + ∠PQR = 180°
=> 60° + ∠POQ = 180°
∠POQ = 180° – 60° = 120°

Question 24.
In the figure, if quadrilateral PQRS circumscribes a circle, then PD + QB =
(a) PQ
(b) QR
(c) PR
(d) PS

Solution:
(a) In the figure, quadrilateral PQRS is circumscribed a circle

PD = PA (tangents from P to the circle)
Similarly QA = QB
PD + QB = PA + QA = PQ

Question 25.
In the figure, two equal circles touch each other at T, if QP = 4.5 cm, then QR =
(a) 9 cm
(b) 18 cm
(c) 15 cm
(d) 13.5 cm

Solution:
(a) In the figure, two equal circles touch, each other externally at T
QR is the common tangent
QP = 4.5 cm

PQ = PT (tangents from P to the circle)
Similarly PT = PR
PQ = PT = PR
Now QR = PQ + PR = 4.5 + 4.5 = 9 cm

Question 26.
In the figure, APB is a tangent to a circle with centre O at point P. If ∠QPB = 50°, then the measure of ∠POQ is
(a) 100°
(b) 120°
(c) 140°
(d) 150°

Solution:
(a) In the figure, APB is a tangent to the circle with centre O

∠QPB = 50°
OP is radius and APB is a tangent
OP ⊥ AB
=> ∠OPB = 90°
=> ∠OPQ + ∠QPB = 90°
∠OPQ + 50° = 90°
=> ∠OPQ = 90° – 50° = 40°
But OP = OQ
∠OPQ = OQP = 40°
∠POQ = 180°- (40° + 40°) = 180° – 80° = 100°

Question 27.
In the figure, if tangents PA and PB are drawn to a circle such that ∠APB = 30° and chord AC is drawn parallel to the tangent PB, then ∠ABC =
(a) 60°
(b) 90°
(c) 30°
(d) None of these

Solution:
(c) In the figure, PA and PB are the tangents to the circle with centre O

∠APB = 30°
Chord AC || BP,
AB is joined
PA = PB
∠PAB = ∠PBA
But ∠PAB + ∠PBA = 180° – 30° = 150°
=> ∠BPA + ∠PBA = 150°
=> 2 ∠PBA = 150°
=> ∠PBA = 75°
AC || BC
∠BAC = ∠PBA = 75°
But ∠PBA = ∠ACB = 75° (Angles in the alternate segment)
∠ABC = 180° – (75° + 75°) = 180° – 150° = 30°

Question 28.
In the figure, PR =
(a) 20 cm
(b) 26 cm
(c) 24 cm
(d) 28 cm

Solution:
(b) In the figure, two circles with centre O and O’ touch each other externally
PQ and RS are the tangents drawn to the circles

OQ and O’S are the radii of these circles and
OQ = 3 cm, PQ = 4 cm O’S = 5 cm and SR = 12 cm
Now in right ∆OQP
OP² = (OQ)² + PQ² = (3)² + (4)² = 9 + 16 = 25 = (5)²
OP = 5 cm
Similarly in right ∆RSO’
(O’R)² = (RS)² + (O’S)² = (12)² + (5)² = 144 + 25 = 169 = (13)²
O’R = 13 cm
Now PR = OP + OO’ + O’R = 5 + (3 + 5) + 13 = 26 cm

Question 29.
Two circles of same radii r and centres O and O’ touch each other at P as shown in figure. If OO’ is produced to meet the circle C (O’, r) at A and AT is a tangent to the circle C (O, r) such that O’Q ⊥ AT. Then AO : AO’ =
(a) \(\frac { 3 }{ 2 }\)
(b) 2
(c) 3
(d) \(\frac { 1 }{ 4 }\)

Solution:
(c) Two circles of equal radii touch each other externally at P. OO’ produced meets at A

From A, AT is the tangent to the circle (O, r)
O’Q ⊥ AT
Now AO : AO’ = 3r : r
= 3 : 1 = \(\frac { 3 }{ 1 }\)

Question 30.
Two concentric circles of radii 3 cm and 5 cm are given. Then length of chord BC which touches the inner circle at P is equal to
(a) 4 cm
(b) 6 cm
(c) 8 cm
(d) 10 cm

Solution:
(c) In the figure, two concentric circles of radii 3 cm and 5 cm with centre O
Chord BC touches the inner circle at P
Draw a tangent AB to the inner circle
Join OQ and OA

OQ is radius and AQB is the tangent
OQ ⊥ AB and OQ bisects AB
AQ = QB
Similarly, BP = PC or P is mid-point of BC
But BQ and BP are tangents from B
QB = BP = AQ
In right ∆OAQ,
OA² = AQ² + OQ²
(5)² = AQ² + (3)²
=> AQ² = (5)² – (3)²
=> AQ² = 25 – 9 = 16 = (4)²
AQ = 4 cm
BC = 2 BP = 2 BQ = 2 AQ = 2 x 4 = 8 cm

Question 31.
In the figure, there are two concentric, circles with centre O. PR and PQS are tangents to the inner circle from point plying on the outer circle. If PR = 7.5 cm, then PS is equal to
(a) 10 cm
(b) 12 cm
(c) 15 cm
(d) 18 cm

Solution:
(c) In the figure, two concentric circles with centre O
From a point P on the outer circle,
PRT and PQS are the tangents are drawn to the inner circle at R and Q respectively
PR = 7.5 cm
Join OR and OQ

PT is chord and OR is radius
R is mid-point of PT
Similarly Q is mid-point of PS
But PR = PQ (tangents from P)
PT = 2 PR and PS = 2 PQ
PS = 2 PQ = 2 PR = 2 x 7.5 = 15 cm

Question 32.
In the figure, if AB = 8 cm and PE = 3 cm, then AE =
(a) 11 cm
(b) 7 cm
(c) 5 cm
(d) 3 cm

Solution:
(c) In the figure, AB and AC are the tangents to the circle from A
DE is another tangent drawn from P
AB = 8 cm, PE = 3 cm

AB = AC (tangents drawn from A to the circle)
Similarly PE = EC and DP = DB
Now AE = AC – CE = AB – PE = 8 – 3 = 5 cm

Question 33.
In the figure, PQ and PR are tangents drawn from P to a circle with centre O. If ∠OPQ = 35°, then
(a) a = 30°, b = 60°
(b) a = 35°, b = 55°
(c) a = 40°, b = 50°
(d) a = 45°, b = 45°

Solution:
(b) In the figure, PQ and PR are the tangents drawn from P to the circle with centre O
∠OPQ = 35°
PO is joined

PQ = PR (tangents from P to the circle)
∠OPQ = ∠OPR
=> 35° = a
=> a = 35°
OQ is radius and PQ is tangent
OQ ⊥ PQ
=> ∠OQP = 90°
In ∆OQP,
∠POQ + ∠QPO = 90°
=> b + 35° = 90°
=> b = 90° – 35° = 55°
a = 35°, b = 55°

Question 34.
In the figure, if TP and TQ are tangents drawn from an external point T to a circle with centre O such that ∠TQP = 60°, then
(a) 25°
(b) 30°
(c) 40°
(d) 60°

Solution:
(b) In the figure, TP and TQ are the tangents drawn from T to the circle with centre O
OP, OQ and PQ are joined
∠TQP = 60°
TP = TQ (Tangents from T to the circle)
∠TQP = ∠TPQ = 60°
∠PTQ = 180° – (60° + 60°) = 180° – 120° = 60°
and ∠POQ = 180° – ∠PTQ = 180° – 60° = 120°
But OP = OQ (radii of the same circle)
∠OPQ = ∠OQP
But ∠OPQ + ∠OQP = 180° – 120° = 60°
But ∠OPQ = 30°

Question 35.
In the figure, the sides AB, BC and CA of triangle ABC, touch a circle at P, Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, then length of BC is [CBSE 2012]

(a) 11 cm
(b) 10 cm
(c) 14 cm
(d) 15 cm
Solution:
(b) In the figure,
PA = 4 cm, BP = 3 cm, AC = 11 cm
AP and AR are the tangents from A to the circle
AP = AR
=> AR = 4 cm
Similarly BP and BQ are tangents
BQ = BP = 3 cm
AC =11 cm
AR + CR = 11 cm
4 + CR =11 cm
CR = 11 – 4 = 7 cm
CQ and CR are tangents to the circle
CQ = CR = 7 cm
Now, BC = BQ + CQ = 3 + 7 = 10 cm

Question 36.
In the figure, a circle touches the side DF of AEDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm, then the perimeter of ∆EDF is [CBSE 2012]
(a) 18 cm
(b) 13.5 cm
(c) 12 cm
(d) 9 cm

Solution:
(a) In ∆DEF
DF touches the circle at H
and circle touches ED and EF Produced at K and M respectively
EK = 9 cm
EK and EM are the tangents to the circle
EM = EK = 9 cm
Similarly DH and DK are the tangent
DH = DK and FH and FM are tangents
FH = FM
Now, perimeter of ∆DEF
= ED + DF + EF
= ED + DH + FH + EF
= ED + DK + EM + EF
= EK + EM
= 9 + 9
= 18 cm

Question 37.
In the figure DE and DF are tangents from an external point D to a circle with centre A. If DE = 5 cm and DE ⊥ DF, then the radius of the circle is [CBSE 2013]

(a) 3 cm
(b) 5 cm
(c) 4 cm
(d) 6 cm
Solution:
(b) If figure, DE and DF are tangents to the circle drawn from D.
A is the centre of the circle.
DE = 5 cm and DE ⊥ DF
Join AE, AF

DE is the tangent and AE is radius
AE ⊥ DE
Similarly, AF ⊥ DF
But ∠D = 90° (given)
AFDE is a square
AE = DE (side of square)
But DE = 5 cm
AE = 5 cm
Radius of circle is 5 cm

Question 38.
In the figure, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches sides BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90° and DS = 5 cm, then the radius of the circle (in cm) is [CBSE 2013]

(a) 11
(b) 18
(c) 6
(d) 15
Solution:
(a) In the figure, a circle touches the sides of a quadrilateral ABCD
∠B = 90°, OP = OQ = r
AB = 29 cm, AD = 23 cm, DS = 5 cm
∠B = 90°
BA is tangent and OQ is radius
∠OQB = 90°
Similarly OP is radius and BC is tangents
∠OPB = 90°
But ∠B = 90° (given)
PBQO is a square
DS = 5 cm
But DS and DR are tangents to the circles
DR = 5 cm
But AD = 23 cm
AR = 23 – 5= 18 cm
AR = AQ (tangents to the circle from A)
AQ = 18 cm
But AB = 29 cm
BQ = 29 – 18 = 11 cm
OPBQ is a square
OQ = BQ = 11 cm
Radius of the circle = 11 cm

Question 39.
In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle (in cm) is
(a) 4
(b) 3
(c) 2
(d) 1
Solution:
(c)

Question 40.
Two circles touch each other externally at P. AB is a common tangent to the circle touching them at A and B. The value of ∠APB is
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
(d) We have, AT = TP and TB = TP (Lengths of the tangents from ext. point T to the circles)
∠TAP = ∠TPA = x (say)
and ∠TBP = ∠TPB = y (say)
Also, in triangle APB,
x + x + x + y + y = 180°
=> 2x + 2y = 180°
=> x + y = 90°
=> ∠APB = 90°

Question 41.
In the figure, PQ and PR are two tangents to a circle with centre O. If ∠QPR= 46, then ∠QOR equals
(a) 67°
(b) 134°
(c) 44°
(d) 46°

Solution:
(b) ∠OQP = 90°
[Tangent is ⊥ to the radius through the point of contact]
∠ORP = 90°
∠OQP + ∠QPR + ∠ORP + ∠QOR = 360° [Angle sum property of a quad.]
90° + 46° + 90° + ∠QOR = 360°
∠QOR = 360° – 90° – 46° – 90° = 134°

Question 42.
In the figure, QR is a common tangent to the given circles touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm, then the length of QR (in cm) is [CBSE2014]
(a) 3.8
(b) 7.6
(c) 5.7
(d) 1.9

Solution:
(b) In the figure, QR is common tangent to the two circles touching each other externally at T
Tangent at T meets QR at P
PT = 3.8 cm
PT and PQ are tangents from P
PT = PQ = 3.8 cm
Similarly PT and PR are tangents
PT = PR = 3.8 cm
QR = 3.8 + 3.8 = 7.6 cm

Question 43.
In the figure, a quadrilateral ABCD is drawn to circumscribe a circle such that its sides AB, BC, CD and AD touch the circle at P, Q, R and S respectively. If AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm, then x =
(a) 10
(b) 9
(c) 8
(d) 7 (CBSE 2014)

Solution:
(b) In the given figure,
ABCD is a quadrilateral circumscribe a circle and its sides AB, BC, CD and DA touch the circle at P, Q, R and S respectively
AB = x cm, BC = 7 cm, CR = 3 cm, AS = 5 cm
CR and CQ are tangents to the circle from C
CR = CQ = 3 cm
BQ = BC – CQ = 7 – 3 = 4 cm
BQ = and BP are tangents from B
BP = BQ = 4 cm
AS and AP are tangents from A
AP = AS = 5 cm
AB = AP + BP = 5 + 4 = 9 cm
x = 9 cm

Question 44.
If angle between two radii of a circle is 130°, the angle between the tangent at the ends of radii is (NCERT Exemplar)
(a) 90°
(b) 50°
(c) 70°
(d) 40°
Solution:
(b) O is the centre of the circle.
Given, ∠POQ = 130°
PT and QT are tangents drawn from external point T to the circle.

∠OPT = ∠OQT = 90° [Radius is perpendicular to the tangent at point of contact]
In quadrilateral OPTQ,
∠PTQ + ∠OPT + ∠OQT + ∠POQ = 360°
=> ∠PTQ + 90° + 90° + 130° = 360°
=> ∠PTQ = 360° – 310° = 50°
Thus, the angle between the tangents is 50°.

Question 45.
If two tangents inclined at a angle of 60° are drawn to a circle of radius 3 cm, then length of each tangent is equal to [NCERT Exemplar]
(a) \(\frac { 3\surd 3 }{ 2 }\) cm
(b) 6 cm
(c) 3 cm
(d) 3√3 cm
Solution:
(d) Let P be an external point and a pair of tangents is drawn from point P and angle between these two tangents is 60°.
Join OA and OP.

Also, OP is a bisector of line ∠APC
∠APO = ∠CPO = 30°
Also, OA ⊥ AP
Tangent at any point of a circle is perpendicular to the radius through the point of contact.

Hence, the length of each tangent is 3√3 cm

Question 46.
If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other circle is [NCERT Exemplar]
(a) 3 cm
(b) 6 cm
(c) 9 cm
(d) 1 cm
Solution:
(b) Let O be the centre of two concentric circles C₁ and C₂, whose radii are r₁ = 4 cm and r₂ = 5 cm.
Now, we draw a chord AC of circle C₂, which touches the circle C₁ at B.
Also, join OB, which is perpendicular to AC. [Tangent at any point of circle is perpendicular to radius throughly the point of contact]

Now, in right angled ∆OBC, by using Pythagoras theorem,
OC² = BC² + BO² [(hypotenuse)² = (base)² + (perpendicular)²]
=> 5² = BC² + 4²
=> BC² = 25 – 16 = 9
=> BC = 3 cm
Length of chord AC = 2 BC = 2 x 3 = 6 cm

Question 47.
At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is [NCERT Exemplar]
(a) 4 cm
(b) 5 cm
(b) 6 cm
(d) 8 cm
Solution:
(d) First, draw a circle of radius 5 cm having centre O.
A tangent XY is drawn at point A.

A chord CD is drawn which is parallel to XY and at a distance of 8 cm from A.
Now, ∠OAY = 90°
[Tangent and any point of a circle is perpendicular to the radius through the point of contact]
∠OAY + ∠OED = 180°
[sum of cointerior is 180°]
=> ∆OED = 180°
Also, AE = 8 cm, Join OC
Now, in right angled ∆OBC
OC² = OE² + EC²
=> EC² = OC² – OE² [by Pythagoras theorem]
EC² = 5² – 3² [OC = radius = 5 cm, OE = AE – AO = 8 – 5 = 3 cm]
EC² = 25 – 9 = 16
=> EC = 4 cm
Hence, length of chord CD = 2 CE = 2 x 4 = 8 cm
[Since, perpendicular from centre to the chord bisects the chord]

Question 48.
From a point P which is at a distance 13 cm from the centre O of a circle of radius 5 cm, the pair of tangent PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is [NCERT Exemplar]
(a) 60 cm²
(b) 65 cm²
(c) 30 cm²
(d) 32.5 cm²
Solution:
(a) Firstly, draw a circle of radius 5 cm having centre O.
P is a point at a distance of 13 cm from O.
A pair of tangents PQ and PR are drawn.
Thus, quadrilateral PQOR is formed.
OQ ⊥ QP [since, AP is a tangent line]
In right angled ∆PQO,
OP² = OQ² + QP²
=> 13² = 5² + QP²
=> QP² = 169 – 25 = 144 = 12²
=> QP = 12 cm

Now, area of ∆OQP = \(\frac { 1 }{ 2 }\) x QP x QO = \(\frac { 1 }{ 2 }\) x 12 x 5 = 30 cm²
Area of quadrilateral QORP = 2 ∆OQP = 2 x 30 = 60 cm²

Question 49.
If PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to
(a) 25°
(b) 30°
(c) 40°
(d) 50°
Solution:
(a) Given, PA and PB are tangent lines.
PA = PB [Since, the length of tangents drawn from an ∠PBA = ∠PAB = θ [say]

In ∆PAB,
∠P + ∠A + ∠B = 180°
[since, sum of angles of a triangle = 180°
50°+ θ + θ = 180°
2θ = 180° – 50° = 130°
θ = 65°
Also, OA ⊥ PA
[Since, tangent at any point of a circle is perpendicular to the radius through the point of contact]
∠PAO = 90°
=> ∠PAB + ∠BAO = 90°
=> 65° + ∠BAO = 90°
=> ∠BAO = 90° – 65° = 25°

Question 50.
The pair of tangents AP and AQ drawn from an external point to a circle with centre O are perpendicular to each other and length of each tangent is 5 cm. The radius of the circle is [NCERT Exemplar]
(a) 10 cm
(b) 7.5 cm
(c) 5 cm
(d) 2.5 cm
Solution:
(c)

Question 51.
In the figure, if ∠AOB = 125°, then ∠COD is equal to [NCERT Exemplar]

(a) 45°
(b) 35°
(c) 55°
(d) 62\(\frac { 1 }{ 2 }\)°
Solution:
(c) We know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
∠AOB + ∠COD = 180°
=> ∠COD = 180° – ∠AOB = 180° – 125° = 55°

Question 52.
In the figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to [NCERT Exemplar]

(a) 20°
(b) 40°
(c) 35°
(d) 45°
Solution:
(b) Given, AB || PR
∠ABQ = ∠BQR = 70° [alternate angles]
Also QD is perpendicular to AB and QD bisects AB.
In ∆QDA and ∆QDB
∠QDA = ∠QDB [each 90°]
AD = BD
QD = QD [common side]
∆ADQ = ∆BDQ [by SAS similarity criterion]
Then, ∠QAD = ∠QBD …(i) [c.p,c.t.]
Also, ∠ABQ = ∠BQR [alternate interior angle]
∠ABQ = 70° [∠BQR = 70°]
Hence, ∠QAB = 70° [from Eq. (i)]
Now, in ∆ABQ,
∠A + ∠B + ∠Q = 180°
=> ∠Q = 180° – (70° + 70°) = 40°

Hope given RD Sharma Class 10 Solutions Chapter 8 Circles MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.2
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

Mark the correct alternative in each of the following :

Question 1.
If 6 θ is an acute angle such that cos θ =

Solution:

Question 2.

Solution:

Question 3.


Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.
If 6 is an acute angle such that tan2 6 = 8 \(\frac { 8 }{ 7 }\), then the value of

Solution:

Question 9.
If 3 cos θ = 5 sin 6, then the value of

Solution:

Question 10.
If tan2 45° – cos2 30° = x sin 45° cos 45°, then x =

Solution:

Question 11.
The value of cos217° – sin2 73° is

Solution:
cos2 17° – sin2 73° = cos2 (90° – 73°) – sin2 73°
= sin2 73° – sin2 73° = 0 (c)

Question 12.

Solution:

Question 13.

Solution:

Question 14.
If A and B are complementary angles then
(a) A – sin B
(b) cos A = cos B
(c) A = tan B
(d) sec A = cosec B
Solution:
∵ A and B are complementary angles
∴ A + B = 90°
⇒ A – 90° – B
sec A = sec (90° – B) = cosec B (d)

Question 15.
If x sin (90° – θ) cot (90° – 6) – cos (90° – θ), then x =
(a) 0
(b) 1
(c) -1
(d) 2,
Solution:

Question 16.
If x tan 45° cos 60° = sin 60° cot 60°, then x is equal to

Solution:

Question 17.
If angles A, B, C of a AABC form an increasing AP, then sin B =

Solution:

Question 18.
If 6 is an acute angle such that sec2 θ =

Solution:

Question 19.
The value of tan 1° tan 2° tan 3s tan 89° is
(a) 1
(b) -1
(c) 0
(d) None of these
Solution:
tan 1° tan 2° tan 3° tan 44° tan 45° tan 46° tan 89°
= tan 1° tan 2° tan 3° tan 44° tan 45° tan
(90° – 44°) tan (90° – 43°) tan (90° – 1°)’
= tan 1° tan 2° tan 3° tan 44° tan 45° cot
44° cot 43°….. cot 1°

Question 20.
The value of cos 1″ cos 2° cos 3° cos 180° is
(a) 1
(b) 0
(c) -1
(d) None of these
Solution:
None of these, because we deal here only an angle O<θ ≤ 90° (d)

Question 21.
The value of tan 10° tan 15° tan 75° tan 80° is
(a) -1
(b) 0
(c) 1
(d) None of these
Solution:
tan 10° tan 15° tan 75° tan 80°
= tan 10° tan 15° tan (90° – 15°) tan (90° – 10°) .
= tan 10° tan 15° cot 15° cot 10°
= tan 10° cot 10° tan 15° cot 15°
{∵ tan θ cot θ = 1}
= 1×1 = 1 (C)

Question 22.
The value of

Solution:

Question 23.
If 6 and 20 – 45° arc acute angles such that sin 0 = cos (20 – 45°), then tan 0 is equal to

Solution:

Question 24.
If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2sin 3θ-√3 tan 3θ is equal to
(a) 1
(b) 0
(c) -1
(d) 1 + √3
Solution:

Question 25.
If A + B = 90°, then
Solution:

Question 26.

Solution:

Question 27.

Solution:

Question 28.
sin 2A = 2 sin A is true when A =
(a) 0°
(b) 30°
(c) 45°
(d) 60°
Solution:

Question 29.

Solution:

Question 30.
If A, B and C are interior angles of a

Solution:

Question 31.
If cos θ = \(\frac { 2 }{ 3 }\), then 2 sec2 θ + 2 tan2 θ-7 is equal to
(a) 1
(b) 0
(c) 3
(d) 4
Solution:

Question 32.
tan 5° x tan 30° x 4 tan 85° is equal to

Solution:

Question 33.

(a)-2
(b) 2
(c) 1
(d) 0
Solution:

Question 34.
In the figure, the value of cos Φ is

Solution:

Question 35.
In the figure, AD = 4 cm, BD = 3 cm and CB = 12 cm, find cot θ.

Solution:

Hope given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3

Other Exercises

• RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.1
• RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2
• RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3

Question 1.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution:
Steps of construction :
(i) Draw a circle with O centre and 6 cm radius.
(ii) Take a point P, 10 cm away from the centre O.
(iii) Join PO and bisect it at M.
(iv) With centre M and diameter PO, draw a circle intersecting the given circle at T and S.
(v) Join PT and PS.
Then PT and PS are the required tangents.

Question 2.
Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 3 cm.
(ii) Draw a diameter and produce it to both sides.
(iii) Take two points P and Q on this diameter with a distance of 7 cm each from the centre O.
(iv) Bisect PO at M and QO at N
(v) With centres M and N, draw circle on PO and QO as diameter which intersect the given circle at S, T and S’, T’ respectively.
(vi) Join PS, PT, QS’ and QT’.
Then PS, PT, QS’ and QT’ are the required tangents to the given circle.

Question 3.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle. [CBSE 2013]
Solution:
Steps of construction :
(i) Draw a line segment AB = 8 cm.
(ii) With centre A and radius 4 cm and with centre B and radius 3 cm, circles are drawn.
(iii) Bisect AB at M.
(iv) With centre M and diameter AB, draw a circle which intersects the two circles at S’, T’ and S, T respectively.
(v) Join AS, AT, BS’and BT’.
Then AS, AT, BS’ and BT’ are the required tangent.

Question 4.
Draw two tangents to a circle of raidus 3.5 cm from a point P at a distance of 6.2 cm from its centre.
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 3.5 cm
(ii) Take a point P which is 6.2 cm from O.
(iii) Bisect PO at M and draw a circle with centre M and diameter OP which intersects the given circle at T and S respectively.
(iv) Join PT and PS.
PT and PS are the required tangents to circle.

Question 5.
Draw a pair of tangents to a circle of radius 4.5 cm, which are inclined to each other at an angle of 45°.            [CBSE 2013]
Solution:
Steps of construction :
Angle at the centre 180° – 45° = 135°
(i) Draw a circle with centre O and radius 4.5 cm.

(ii) At O, draw an angle ∠TOS = 135°
(iii) At T and S draw perpendicular which meet each other at P.
PT and PS are the tangents which inclined each other 45°.

Question 6.
Draw a right triangle ABC in which AB = 6 cm, BC = 8 cm and ∠B = 90°. Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle.
Solution:
Steps of Construction :
Draw a line segment BC = 8 cm
From B draw an angle of 90°
Draw an arc \(\breve { BA }\)  = 6cm cutting the angle at A.
Join AC.
ΔABC is the required A.
Draw ⊥ bisector of BC cutting BC at M.
Take M as centre and BM as radius, draw a circle.
Take A as centre and AB as radius draw an arc cutting the circle at E. Join AE.
AB and AE are the required tangents.
Justification :
∠ABC = 90°                                            (Given)
Since, OB is a radius of the circle.
∴ AB is a tangent to the circle.
Also AE is a tangent to the circle.

Question 7.
Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to the smaller circle from a point on the larger circle. Also, measure its length.                      [CBSE 2016]
Solution:
Given, two concentric circles of radii 3 cm and 5 cm with centre O. We have to draw pair of tangents from point P on outer circle to the other.

Steps of construction :       
(i) Draw two concentric circles with centre O and radii 3 cm and 5 cm.
(ii) Taking any point P on outer circle. Join OP.
(iii) Bisect OP, let M’ be the mid-point of OP.
Taking M’ as centre and OM’ as radius draw a circle dotted which cuts the inner circle as M and P’.
(iv) Join PM and PP’. Thus, PM and PP’ are the required tangents.
(v) On measuring PM and PP’, we find that PM = PP’ = 4 cm.
Actual calculation:
In right angle ΔOMP, ∠PMO = 90°
∴ PM² = OP² – OM²
[by Pythagoras theorem i.e. (hypotenuse)² = (base)² + (perpendicular)²]
⇒ PM² = (5)² – (3)² = 25 – 9 = 16
⇒ PM = 4 cm
Hence, the length of both tangents is 4 cm.

Hope given RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2

Other Exercises

• RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.1
• RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2
• RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3

Question 1.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are (2/3) of the corresponding sides of it.
Solution:
Steps of construction :
(i) Draw a line segment BC = 5 cm.
(ii) With centre B and radius 4 cm and with centre C and radius 6 cm, draw arcs intersecting each other at A.
(iii) Join AB and AC. Then ABC is the triangle.
(iv) Draw a ray BX making an acute angle with BC and cut off 3 equal parts making BB₁ = B₁B₂= B₂B₃.
(v) Join B₃C.
(vi) Draw B’C’ parallel to B₃C and C’A’ parallel to CA then ΔA’BC’ is the required triangle.

Question 2.
Construct a triangle similar to a given ΔABC such that each of its sides is (5/7)^th of the corresponding sides of ΔABC. It is given that AB = 5 cm, BC = 7 cm and ∠ABC = 50°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm.
(ii) Draw a ray BX making an angle of 50° and cut off BA = 5 cm.
(iii) Join AC. Then ABC is the triangle.
(iv) Draw a ray BY making an acute angle with BC and cut off 7 equal parts making BB, =B₁B₂=B₂B₃=B₃B₄=B₄B_s=B₅B₆=B₆B₇
(v) Join B₇ and C
(vi) Draw B₅C’ parallel to B₇C and C’A’ parallel to CA.
Then ΔA’BC’ is the required triangle.

Question 3.
Construct a triangle similar to a given ∠ABC such that each of its sides is \(\frac { 2 }{ 3 }\)rd of the corresponding sides of ΔABC. It is given that BC = 6 cm, ∠B = 50° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm.
(ii) Draw a ray BX making an angle of 50° and CY making 60° with BC which intersect each other at A. Then ABC is the triangle.
(iii) From B, draw another ray BZ making an acute angle below BC and intersect 3 equal parts making BB₁ =B₁B₂ = B₂B₂
(iv) Join B₃C.
(v) From B₂, draw B₂C’ parallel to B₃C and C’A’ parallel to CA.
Then ΔA’BC’ is the required triangle.

Question 4.
Draw a ΔABC in which BC = 6 cm, AB = 4 cm and AC = 5 cm. Draw a triangle similar to ΔABC with its sides equal to \(\frac { 3 }{ 4 }\)th of the corresponding sides of ΔABC.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm.
(ii) With centre B and radius 4 cm and with centre C and radius 5 cm, draw arcs’intersecting eachother at A.
(iii) Join AB and AC. Then ABC is the triangle,
(iv) Draw a ray BX making an acute angle with BC and cut off 4 equal parts making BB₁=  B₁B₂ = B₂B₃ = B₃B₄.
(v) Join B₄ and C.
(vi) From B₃C draw C₃C’ parallel to B₄C and from C’, draw C’A’ parallel to CA.
Then ΔA’BC’ is the required triangle.

Question 5.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac { 7 }{ 5 }\) of the corresponding sides of the first triangle.
Solution:
Steps of construction :
(i) Draw a line segment BC = 5 cm.
(ii) With centre B and radius 6 cm and with centre C and radius 7 cm, draw arcs intersecting eachother at A.
(iii) Join AB and AC. Then ABC is the triangle.
(iv) Draw a ray BX making an acute angle with BC and cut off 7 equal parts making BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇.
(v) Join B₅ and C.
(vi) From B₇, draw B₇C’ parallel to B₅C and C’A’ parallel CA. Then ΔA’BC’ is the required triangle.

Question 6.
Draw a right triangle ABC in which AC = AB = 4.5 cm and ∠A = 90°. Draw a triangle similar to ΔABC with its sides equal to (\(\frac { 5 }{ 4 }\))th ot the corresponding sides of ΔABC.
Solution:
Steps of construction :
(i) Draw a line segment AB = 4.5 cm.
(ii) At A, draw a ray AX perpendicular to AB and cut off AC = AB = 4.5 cm.
(iii) Join BC. Then ABC is the triangle.
(iv) Draw a ray AY making an acute angle with AB and cut off 5 equal parts making AA₁ = A₁A₂ = A₂A₃ =A₃A₄ = A₄A₅
(v) Join A₄ and B.
(vi) From 45, draw 45B’ parallel to  A₄B and  B’C’ parallel to BC.
Then ΔAB’C’ is the required triangle.

Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm. Then construct another triangle whose sides are \(\frac { 5 }{ 3 }\) times the corresponding sides of the given triangle.  (C.B.S.E. 2008)
Solution:
Steps of construction :
(i) Draw a line segment BC = 5 cm.
(ii) At B, draw perpendicular BX and cut off BA = 4 cm.
(iii )join Ac , then ABC is the triangle
(iv) Draw a ray BY making an acute angle with BC, and cut off 5 equal parts making BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅
(v) Join B₃ and C.
(vi) From B₅, draw B₅C’ parallel to B₃C and C’A’ parallel to CA.
Then ΔA’BC’ is the required triangle.

Question 8.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are \(\frac { 3 }{ 2 }\) times the corresponding sides of the isosceles triangle.
Solution:
Steps of construction :
(i) Draw a line segment BC = 8 cm and draw its perpendicular bisector DX and cut off DA = 4 cm.
(ii) Join AB and AC. Then ABC is the triangle.
(iii) Draw a ray DY making an acute angle with OA and cut off 3 equal parts making DD₁ = D₁D₂ =D₂D₃ = D₃D₄
(iv) Join D₂
(v) Draw D₃A’ parallel to D₂A and A’B’ parallel to AB meeting BC at C’ and B’ respectively.
Then ΔB’A’C’ is the required triangle.

Question 9.
Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a trianglewhose sides are \((\frac { 3 }{ 4 } )\)th of the corresponding sides of the ΔABC.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm.
(ii) At B, draw a ray BX making an angle of 60° with BC and cut off BA = 5 cm.
(iii) Join AC. Then ABC is the triangle.
(iv) Draw a ray BY making an acute angle with BC and cut off 4 equal parts making BB₁= B₁B₂  B₂B₃=B₃B₄.
(v) Join B₄ and C.
(vi) From B₃, draw B₃C’ parallel to B₄C and C’A’ parallel to CA.
Then ΔA’BC’ is the required triangle.

Question 10.
Construct a triangle similar to ΔABC in which AB = 4.6 cm, BC = 5.1 cm,∠A = 60° with scale factor 4 : 5.
Solution:
Steps of construction :
(i) Draw a line segment AB = 4.6 cm.
(ii) At A, draw a ray AX making an angle of 60°.
(iii) With centre B and radius 5.1 cm draw an arc intersecting AX at C.
(iv) Join BC. Then ABC is the triangle.
(v) From A, draw a ray AX making an acute angle with AB and cut off 5 equal parts making AA₁ = A₁A₂ = A₂A₃ = A₃A₄=A₄A₅.
(vi) Join A₄ and B.
(vii) From A₅, drawA₅B’ parallel to A₄B and B’C’ parallel to BC.
Then ΔC’AB’ is the required triangle.

Question 11.
Construct a triangle similar to a given ΔXYZ with its sides equal to \((\frac { 3 }{ 2 })\) th of the corresponding sides of ΔXYZ. Write the steps of construction.                      [CBSE 1995C]
Solution:
Steps of construction :
(i) Draw a triangle XYZ with some suitable data.
(ii) Draw a ray YL making an acute angle with XZ and cut off 5 equal parts making YY₁= Y₁Y₂ = Y₂Y₃ = Y₃Y₄.
(iii) Join Y₄ and Z.
(iv) From Y₃, draw Y₃Z’ parallel to Y₄Z and Z’X’ parallel to ZX.
Then ΔX’YZ’ is the required triangle.

Question 12.
Draw a right triangle in which sides (other than the hypotenuse) are of lengths 8 cm and 6 cm. Then construct another triangle whose sides are \(\frac { 3 }{ 4 }\) times the corresponding sides of the first triangle.
Solution:
(i) Draw right ΔABC right angle at B and BC = 8 cm and BA = 6 cm.
(ii) Draw a line BY making an a cut angle with BC and cut off 4 equal parts.
(iii) Join 4C and draw 3C’ || 4C and C’A’ parallel to CA.
The BC’A’ is the required triangle.

Question 13.
Construct a triangle with sides 5 cm, 5.5 cm and 6.5 cm. Now construct another triangle, whose sides are 3/5 times the corresponding sides of the given triangle. [CBSE 2014]
Solution:
Steps of construction:
(i) Draw a line segment BC = 5.5 cm.
(ii) With centre B and radius 5 cm and with centre C and radius 6.5 cm, draw arcs which intersect each other at A
(iii) Join BA and CA.
ΔABC is the given triangle.
(iv) At B, draw a ray BX making an acute angle and cut off 5 equal parts from BX.
(v) Join C5 and draw 3D || 5C which meets BC at D.
From D, draw DE || CA which meets AB at E.
∴ ΔEBD is the required triangle.

Question 14.
Construct a triangle PQR with side QR = 7 cm, PQ = 6 cm and ∠PQR = 60°. Then construct another triangle whose sides are 3/5 of the corresponding sides of ΔPQR. [CBSE 2014]
Solution:
Steps of construction:
(i) Draw a line segment QR = 7 cm.
(ii) At Q draw a ray QX making an angle of 60° and cut of PQ = 6 cm. Join PR.
(iii) Draw a ray QY making an acute angle and cut off 5 equal parts.
(iv) Join 5, R and through 3, draw 3, S parallel to 5, R which meet QR at S.
(v) Through S, draw ST || RP meeting PQ at T.
∴ ΔQST is the required triangle.

Question 15.
Draw a ΔABC in which base BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct another triangle whose sides are \(\frac { 3 }{ 4 }\) of the corresponding sides of ΔABC.    [CBSE 2017]
Solution:
Steps of construction:

1. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
2.  Draw a ray BX, which makes an acute angle ∠CBX below the line BC.
3. Locate four points B₁, B₂, B₃and B₄ on BX such that BB₁ = B₁B₂=B₂B₃ = B₃B₄.
4. Join B₄C and draw a line through B₃ parallel to B₄C intersecting BC to C’.
5. Draw a line through C’ parallel to the line CA to intersect BA at A’.

Question 16.
Draw a right triangle in which the sides (other than the hypotenuse) arc of lengths 4 cm and 3 cm. Now, construct another triangle whose sides are \(\frac { 5 }{ 3 }\) times the corresponding sides of the given triangle. [CBSE 2017]
Solution:
Steps of construction:

1. Draw a right triangle ABC in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. ∠B = 90°.
2. Draw a line BX, which makes an acute angle ∠CBX below the line BC.
3. Locate 5 points B₁, B₂, B₃, B₄ and B₅ on BX such that BB₁ = B₁B₂=B₂B₃=B₃B₄=B₄B₅.
4. Join B₃ to C and draw a line through B₅ parallel to B₃C, intersecting the extended line segment BC at C’.
5. Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’.

Question 17.
Construct a ΔABC in which AB = 5 cm, ∠B = 60°, altitude CD = 3 cm. Construct a ΔAQR similar to ΔABC such that side of ΔAQR is 1.5 times that of the corresponding sides of ΔACB.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5 cm.
(ii) At A, draw a perpendicular and cut off AE = 3 cm.
(iii) From E, draw EF || AB.
(iv) From B, draw a ray making an angle of 60 meeting EF at C.
(v) Join CA. Then ABC is the triangle.
(vi) From A, draw a ray AX making an acute angle with AB and cut off 3 equal parts making A A₁= A₁A₂ = A₂A₃.
(vii) Join A₂ and B.
(viii) From A , draw A^B’ parallel to A₂B and B’C’ parallel toBC.
Then ΔC’AB’ is the required triangle.

Hope given RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.1
• RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.2
• RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS
• RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

Answer each of the following questions either in one word or one sentence or as per requirement of the questions :
Question 1.
In the figure, PA and PB are tangents to the circle drawn from an external point P. CD is a third tangent touching the circle at Q. If PB = 10 cm and CQ = 2 cm, what is the length PC ?

Solution:
In the figure, PA and PB are the tangents to the circle drawn from P
CD is the third tangent to the circle drawn at Q
PB = 10 cm, CQ = 2 cm

PA and PB are tangents to the circle
PA = PB = 10 cm
Similarly CQ and CA are tangents to the circle
CQ = CA = 2 cm
PC = PA – CA = 10 – 2 = 8 cm

Question 2.
What is the distance between two parallel tangents of a circle of radius 4 cm ?
Solution:
TT’ and SS’ are two tangents of a circle with centre O and radius 4 cm and TT’ || SS’
OP and OQ are joined

Now OP is the radius and TPT’ is the tangent
OP ⊥ TPT’
Similar OQ ⊥ SS’
But TT’ || SS’
POQ is the diameter
Which is 4 x 2 = 8 cm
Distance between the two parallel tangents is 8 cm

Question 3.
The length of tangent from a point A at a distance of 5 cm from the centre of the circle is 4 cm. What is the radius of the circle ?
Solution:
PA is a tangent to the circle from P at a distance of 5 cm from the centre O
PA = 4 cm
OA is joined and let OA = r

Now in right ∆OAP,
OP² = OA² + PA²
=> (5)² = r² + (4)²
=> 25 = r + 16
=> r² = 25 – 16 = 9 = (3)²
r = 3
Radius of the circle = 3 cm

Question 4.
Two tangents TP and TQ are drawn from an external point T to a circle with centre O as shown in the following figure. If they are inclined to each other at an angle of 100°, then what is the value of ∠POQ ?

Solution:
TP and TQ are the tangents from T to the circle with centre O and ∠PTQ = 100°
OT, OP and OQ are joined
OP and OQ are radius
OP ⊥ PT and OQ ⊥ QT
Now in quadrilateral OPTQ,
∠POQ + ∠OPT + ∠PTQ + ∠OQT = 360° (Sum of angles of a quadrilateral)
=> ∠POQ + 90° + 100° + 90° = 360°
=> ∠POQ + 280° = 360°
=> ∠POQ = 360° – 280° = 80°
Hence ∠POQ = 80°

Question 5.
What is the distance between two parallel tangents to a circle of radius 5 cm?
Solution:
In a circle, the radius is 5 cm and centre is O

TT’ and SS’ are two tangents at P and Q to the circle
Such that TT’ || SS’
Join OP and OQ
OP is radius and TPT’ is the tangent
OP ⊥ TT’
Similarly OQ ⊥ SS’
POQ is the diameter of the circle
Now length of PQ = OP + OQ = 5 + 5 = 10 cm
Hence distance between the two parallel tangents = 10 cm

Question 6.
In Q. No. 1, if PB = 10 cm, what is the perimeter of ∆PCD ?
Solution:
In the figure, PB = 10 cm, CQ = 2 cm

PA and PB are tangents to the give from P
PA = PB = 10 cm
Similarly, CA and CQ are the tangents
CA = CQ = 2 cm
and DB and DQ are the tangents
DB = DQ
Now, perimeter of ∆PCD
PC + PD + CQ + DQ
= PC + CQ + PD + DQ
= PC + CA + PD + DB {CQ = CA and DQ = DB}
= PA + PB = 10 + 10 = 20 cm

Question 7.
In the figure, CP and CQ are tangents to a circle with centre O. ARB is another tangent touching the circle at R. If CP = 11 cm and BC = 7 cm, then find the length of BR. (C.B.S.E. 2009)

Solution:
Given : In the figure, CP and CQ are tangents to a circle with centre O
ARB is a third tangent to the circle at R
CP = 11 cm, BC = 7 cm

To find : The length of BR
BQ and BR are tangents to the circle drawn from B
BQ = BR ….(i)
Similarly CQ = CP
=> BC + BQ = CP = 11 (CP = 11 cm and BC = 7 cm)
=> 7 + BQ = 11
=> BQ = 11 – 7
BQ = 4 cm
But BQ = BR
BR = 4 cm

Question 8.
In the figure, ∆ABC is circumscribing a circle. Find the length of BC. (C.B.S.E. 2009)

Solution:
∆ABC is circumscribing a circle which touches it at P, Q and R
AC = 11 cm, AR = 4 cm, BR = 3 cm
Now we have to find BC
AR and AQ are tangents to the circle from A
AQ = AR = 4 cm
Then CQ = AC – AQ = 11 – 4 = 7 cm
Similarly,
CP and CQ are tangents from C
CP = CQ = 7 cm
and BP and BR are tangents from B
BP = BR = 3 cm
Now BC = BP + CP = 3 + 7 = 10 cm

Question 9.
In the figure, CP and CQ are tangents from an external point C to a circle with centre O. AB is another tangent which touches the circle at R. If CP = 11 cm and BR = 4 cm, find the length of BC. [CBSE 2010]

Solution:
CP and CQ are the tangents to the circle from C.
AB is another tangent to the same circle which touches at R and meets the first two tangents at A and B. O is the centre of the circle.
OC is joined
CP = 11 cm, BR = 4 cm
CP and CQ are tangents to the circle
CP = CQ = 11 cm
Similarly from B, CR and BQ are the tangents
BQ = BR = 4 cm
Now BC = CQ – BQ = 11 – 4 = 7 cm

Question 10.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution:
Two concentric circles with centre O, have radii 5 cm and 3 cm
AB is a chord which touches the smaller circle at P
OP is joined which is radius of smaller circle

P is mid-point of AB
OP = 3 cm and OA = 5 cm
Now in right ∆OAP
OA² = OP² + AP²
(5)² = (3)² + AP²
=> 25 = 9 + AP²
=> AP² = 25 – 9 = 16 = (4)²
AP = 4 cm
AB = 2 AP = 2 x 4 cm = 8 cm

Question 11.
In the given figure, PA and PB are tangents to the circle with centre O such that ∠APB = 50°. Write the measure of ∠OAB. [CBSE 2015]

Solution:
In the given figure,
PA and PB are tangents to the circle from P
PA = PB
∠APB = 50°, OA is joined
To find ∠OAB
In ∆PAB
PA = PB

Question 12.
In the figure, PQ is a chord of a circle and PT is the tangent at P such that ∠QPT = 60°. Then, find ∠PRQ. [NCERT Exemplar]

Solution:
∠OPQ = ∠OQP = 30°, i.e., ∠POQ = 120°
Also, ∠PRQ = \(\frac { 1 }{ 2 }\) reflex ∠POQ

Question 13.
In the figure, PQL and PRM are tangents to the circle with centre O at the points Q and R respectively and S is a point on the circle such that ∠SQL = 50° and ∠SRM = 60°. Then, find ∠QSR. [NCERT Exemplar]

Solution:
Here ∠OSQ = ∠OQS = 90° – 50° = 40°
and ∠RSO = ∠SRO = 90° – 60° = 30°.
Therefore, ∠QSR = 40° + 30° = 70°

Question 14.
In the figure, BOA is a diameter of a circle and the tangent at a point P meets BA produced at T. If ∠PBO = 30°, then find ∠PTA. [NCERT Exemplar]

Solution:
As ∠BPA = 90°,
∠PAB = ∠OPA = 60°
Also OP ⊥ PT.
Therefore, ∠APT = 30°
and ∠PTA = 60° – 30° = 30°

Hope given RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.2
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

Answer each of the following questions either in one word or one sentence or as per requirement of the questions :

Question 1.
Write the maximum and minimum values of sin θ.
Solution:
Maximum value of sin θ = 1
and minimum value of sin θ = 0

Question 2.
Write the maximum and minimum values of cos 0.
Solution:
Maximum value cos θ=1 and minimum value of cos θ = θ

Question 3.
What is the maximum value of \(\frac { 1 }{ sec\theta }\) ?
Solution:
Maximum value of \(\frac { 1 }{ sec\theta }\) or cos θ = 1

Question 4.
What is the maximum value of \(\frac { 1 }{ cosec\theta }\)
Solution:
Maximum value of \(\frac { 1 }{ cosec\theta }\) or sin θ = 1

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.
If tan A = \(\frac { 3 }{ 4 }\) and A + B = 90°, then what is the value of cot B ?
Solution:

Question 11.
If A + B = 90°, cos B = \(\frac { 3 }{ 5 }\), what is the value of sin A.
Solution:

Question 12.
Write the acute angle θ satisfying √3 sin θ = cos θ.
Solution:

Question 13.
Write the Value of cos 1° cos 2° cos 3° ……. cos 179° cos 180°.
Solution:

Question 14.
Write the Value of tan 10′ tan 15° tan 75° tan 80°.
Solution:

Question 15.
If A + B = 90° and tan A = \(\frac { 3 }{ 4 }\) what is cot B?
Solution:

Question 16.
If tan A = \(\frac { 5 }{ 12 }\), find the value of (sin A + cos A) sec A. (C.B.S.E. 2008)
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS are helpful to complete your math homework.

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These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.1

Other Exercises

• RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.1
• RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2
• RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3

Question 1.
Determine a point which divides a line segment of length 12 cm internally in the ratio 2 : 3. Also justify your construction.
Solution:
Steps of construction :
(i) Draw a line segment AB = 12 cm.
(ii) Draw a ray AX at A making an acute angle with AB.
(iii) From B, draw another ray BY parallel to AX.
(iv) Cut off 2 equal parts from AX and 3 equal parts from BY.
(v) Join 2 and 3 which intersects AB at P.
P is the required point which divides AB in the ratio of 2 : 3 internally.

Question 2.
Divide a line segment of length 9 cm internally in the ratio 4 : 3. Also, give justification of the construction.
Solution:
Steps of construction :
(i) Draw a line segment AB = 9 cm.
(ii) Draw a ray AX making an acute angle with AB.
(iii) From B, draw another ray BY parallel to AX.
(iv) Cut off 4 equal parts from AX and 3 parts from BY.
(v) Join 4 and 3 which intersects AB at P.
P is the required point which divides AB in the ratio of 4 : 3 internally.

Question 3.
Divide a line segment of length 14 cm internally in the ratio 2 : 5. Also justify your construction.
Solution:
Steps of construction :
(i) Draw a line segment AB = 14 cm.
(ii) Draw a ray AX making an acute angle with AB.
(iii) From B, draw another ray BY parallel to AX.
(iv) From AX, cut off 2 equal parts and from B, cut off 5 equal parts.
(v) Join 2 and 5 which intersects AB at P.
P is the required point which divides AB in the ratio of 2 : 5 internally.

Question 4.
Draw a line segment of length 8 cm and divide it internally in the ratio 4 : 5.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8 cm.
(ii) Draw a ray AX making an acute angle with ∠BAX = 60° withAB.

(iii) Draw a ray BY parallel to AX by making an acute angle ∠ABY = ∠BAX.
(iv) Mark four points A₁, A₂, A₃, A₄ on AX and five points B₁, B₂, B₃, B₄, B_s on BY in such a way that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ .
(v) Join A₄B₅.
(vi) Let this line intersect AB at a point P.
Thus, P is the point dividing the line segment AB internally in the ratio of 4 : 5.

Hope given RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.1
• RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2
• RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities VSAQS
• RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS

Mark the correct alternative in each of the folloiwng :
Question 1.
If sec θ + tan θ = x, then sec θ =

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.
sec⁴ A – sec² A is equal to
(a) tan² A – tan⁴ A
(b) tan⁴ A – tan² A
(c) tan⁴ A + tan² A          
(d) tan² A + tan⁴ A
Solution:

Question 6.
cos⁴ A – sin⁴ A is equal to
(a) 2 cos² A + 1        
(b) 2 cos² A – 1
(c) 2 sin² A – 1           
 (d) 2 sin² A + 1
Solution:
cos⁴ A – sin⁴ A = (cos² A + sin² A) (cos² A – sin² A)
= 1 (cos² A – sin² A) = cos² A – (1 – cos² A)
= cos² A – 1 + cos² A
= 2 cos² A – 1            (b)

Question 7.

Solution:

Question 8.

Solution:

Question 9.
The value of (1 + cot θ – coscc θ) (1 + tan θ + sec θ) is
(a) 1                          
(b) 2
(c) 4                          
(d) 0
Solution:

Question 10.

Solution:

Question 11.
(cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) is equal 
(a) 0                          
(b) 1
(c) -1
(d) None of these
Solution:

Question 12.
If x = a cos θ and y = b sin θ, then b²x² + a²y² =
(a) a²b²                             
(b) ab
(c) a⁴b⁴                      
(d) a² + b²
Solution:
x = a cos θ, y = b sin θ                       …(i)
bx = ab cos θ, ay = ab sin θ          ….(ii)
Adding (i) and (ii) we get,
b²x²+ a²y² = a²b² cos²  θ + a²b² sin²  θ
= a²b² (cos²  θ + sin²  θ)
= a²b² x 1
= a²b²                         (a)

Question 13.
If x = a sec θ and y-b tan θ, then b²x² – a²y²
(a) ab
(b) a² – b²
(c) a² + b²
(d) a²b²
Solution:
x = a sec θ and y = b tan θ
b²x² – a²y² = b² (a sec θ)² – a² (b tan θ)²
= a²b² sec² θ – a²b² tan²  θ
= a²b² (sec²  θ – tan²  θ)
= a²b² x 1
= a²b²                       (d)

Question 14.

Solution:

Question 15.
2 (sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) is equal to
(a) 0                
(b) 1
(c) -1               
(d) None of these
Solution:
.

Question 16.
If a cos θ + b sin θ = 4 and a sin θ – b cos θ = 3, then a² + b² =
(a) 7                 
(b) 12
(c) 25                         
(d) None of these
Solution:
a cos θ + b sin θ = 4
a sin θ – b cos θ = 3
Squaring and adding
a² cos²  θ + b²  sin² θ  + 2ab sin θ cos θ=16
a² sin² θ + b²  cos² θ  – 2ab sin θ  cos θ = 9
a² (cos² θ + sin² θ) + b² (sin² θ + cos² θ) = 25 (∵ sin² θ + cos² θ=1)
⇒ a² x 1 + b² x 1 = 25
⇒  a² + b² = 25                                           (c)

Question 17.
If a cot θ + b cosec θ = p and b cot θ + a cosec θ = q, then p² – q² =
(a)   a² – b²                  
(b) b² – a²
(c)  a² + b²                  
(d)  b – a
Solution:
a cot θ + b cosec θ = p
b cot θ + a cosec θ = q
Squaring and subtracting,
p² – q²  = (a cot θ + b cosec θ)² – (b cot θ + a cosec θ)²
= a² cot² θ + b² cosec² θ + 2ab cot θ cosec θ – (b² cot² θ + a² cosec² θ + 2ab cot θ cosec θ)
= a² cot² θ +  b²  cosec² θ + lab cot θ cosec θ – b² cot²  θ – a² cosec² θ – lab cot θ cosec θ
= a² (cot² θ – cosec² θ) + b² (cosec² θ – cot^{2 θ})
= -a² (cosec² θ – cot² θ) + b² (cosec² θ – cot² θ)
= -a² x 1 + b² x 1 = b² – a²                                    (b)

Question 18.
The value of sin² 29° + sin² 61° is
(a) 1
(b) 0
(c) 2sin² 29“                 
(d)  2cos² 61°
Solution:
sin² 29° + sin² 61° = sin² 29° + sin² (99° – 29°)
= sin² 29 + cos² 29°              (a)
(sin² θ + cos² θ=1)

Question 19.
If x = r sin θ cos φ, y = r sin θ sin φ and z – r cos θ, then
(a) x² +y² + z² = r²      
(b)   x² +y² – z² = r²
(c) x²– y²+ z² = r²      
(d)   z² + y² – x² = r²
Solution:

Question 20.
If sin θ + sin²  θ=1, then cos²  θ + cos⁴  θ
(a) -1                         
(b) 1
(c) 0                           
(d) None of these
Solution:
sin θ + sin² θ=1
⇒ sin θ = 1- sin² θ
⇒ sin θ = cos² θ
cos² θ + cos⁴ θ = sin θ + sin² θ     {∵ cos² θ = sin θ}
⇒  cos² θ + cos⁴ θ=1                                 (b)
{∵ sin θ + sin² θ = 1 (given)}

Question 21.
If a cos θ + b sin θ = m and a sin θ – b cos θ = it, then a² + b² =
(a) m² – it²        
(b) m²n²
(c) n² – m²        
(d) m² + n²
Solution:
a cos θ + b sin θ = m
a sin θ – b cos θ = n
Squaring and adding
a² cos² θ + b² sin² θ   + lab sin  θ cos θ = m²
a² sin² θ + b² cos² θ   – 2ab sin  θ cos θ = n²
a² (cos² θ + sin² θ) + b² (sin² θ + cos² θ) = m² + n²     {sin² θ + cos² θ=1}
⇒   a² + 1 + b² x 1 = m² – n²
⇒  a² + b² = m² + n²
Hence a² + b² = m² + n²          (d)

Question 22.
If cos A + cos² A = 1, then sin² A + sin⁴ A = 
(a) -1                         
(b) 0                            
(c) 1                           
(d) None of these
Solution:
cos A + cos² A = 1
⇒ cos A = 1 – cos² A
⇒  cos A = sin² A
Now, sin² A + sin⁴ A = cos A + cos² A = 1 (∵ cos A + cos² A = 1) (given)
∴ sin² A + sin⁴ A = 1                              (c)

Question 23.

Solution:

Question 24.

Solution:

Question 25.
9sec² A – 9tan² A is equal to
(a) 1
(b) 9
(c) 8                           
(d) 0
Solution:
9sec² A – 9tan² A = 9 (sec² A – tan² A)
= 9 x 1       (∵ sec² A – tan² A = 1)
= 9                        (b)

Question 26.
(1 + tan θ + sec θ) (1 + cot θ – cosec θ) =
(a) 0
(b) 1
(c) 1                           
(d) -1
Solution:

Question 27.
(sec A + tan A) (1 – sin A) =
(a) sec A
(b) sin A
(c) cosec A                        
(d) cos A
Solution:

Question 28.

Question 29.
If sin θ cos θ = 0, then the value of sin⁴ θ + cos⁴ θ is

Solution:

Question 30.
The value of sin (45° + θ) – cos (45° – θ) is equal to
(a) 2 cos θ                  
(b) 0
(c) 2 sin θ
(d) 1
Solution:
sin (45° + θ) – cos (45° – θ)
= sin (45° + θ) – sin (90° – 45° + θ)
= sin (45° + θ) – sift (45° + θ)
= 0                                                     (b)

Question 31.
If ΔABC is right angled at C, then the value of cos (A + B) is

Solution:

Question 32.

Solution:
cos (9θ) = sin θ
⇒ sin (90° – 9θ) = sin θ
⇒ 90° – 90 = θ
⇒ 9θ = 90°
⇒  θ= 10
tan 6θ = tan 6
= tan 60° = \(\sqrt { 3 } \)        (b)

Question 33.
If cos (α + β) =0 , then sin (α – β) can be reduced to 
(a) cos β
(b) cos 2β
(c) sin α
(d) sin 2α
Solution:
cos (α + β) = 0
⇒ α + β = 90°                       [∵ cos 90° = 0]
⇒  θ = 90° – β                                          …(i)
sin (α – β) = sin (90° – β – β) [using (i)]
= sin (90° – 2β)
= cos 2β [∵ sin (90° – θ) = cos θ]         (b)

Hope given RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.2

Other Exercises

• RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.1
• RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.2
• RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS
• RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

Question 1.
If PT is a tangent at T to a circle whose centre is O and OP = 17 cm, OT = 8 cm. Find the length of the tangent segment PT.
Solution:
PT is the tangent to the circle with centre O, at T
Radius OT = 8 cm, OP = 17 cm

PT is the tangent segment
Now in right ∆OPT,
OP² = OT² + PT² (Pythagoras Theorem)
=> (17)² = (8)² + PT²
=> 289 = 64 + PT²
=> PT² = 289 – 64 = 225 = (15)²
PT = 15 cm

Question 2.
Find the length of a tangent drawn to a circle with radius 5 cm, from a point 13 cm from the centre of the circle.
Solution:
From a point P outside the circle with centre O, PT is the tangent to the circle and radius
OT = 5 cm, OP = 15 cm

OT ⊥ PT
Now in right ∆OPT,
OP² = OT² + PT² (Pythagoras Theorem)
(13)² = (5)² + PT²
=> 169 = 25 + PT²
=> PT² = 169 – 25 = 144 = (12)²
PT = 12 cm

Question 3.
A point P is 26 cm away from the centre O of a circle and the length PT of the tangent drawn from P to the circle is 10 cm. Find the radius of the circle.
Solution:
From a point P outside the circle of centre 0 and radius OT, PT is the tangent to the circle
OP = 26 cm, PT = 10 cm

Now in right ∆OPT
Let r be the radius
OP² = OT² + PT² (Pythagoras Theorem)
=> (26)² = r² + (10)²
=> 676 = r² + 100
=> 676 – 100 = r²
=> r² = 576 = (24)²
r = 24
Hence radius of the circle = 24 cm

Question 4.
If from any point on the common chord of two intersecting circles, tangents be drawn to the circles, prove that they are equal.
Solution:
Given : QR is the common chord of two circles intersecting each other at Q and R
P is a point on RQ when produced From PT and RS are the tangents drawn to tire circles with centres O and C respectively

To prove : PT = PS
Proof: PT is the tangent and PQR is the secant to the circle with centre O
PT² = PQ x PR ….(i)
Similarly PS is the tangent and PQR is the secant to the circle with centre C
PS² = PQ x PR ….(ii)
From (i) and (ii)
PT² = PS²
PT = PS
Hence proved.

Question 5.
If the sides of a quadrilateral touch a circle, prove that the sum of a pair of opposite sides is equal to the sum of the other pair.
Solution:
Given : The sides of a quadrilateral ABCD touch the circle at P, Q, R and S respectively

To prove : AB + CD = AP + BC
Proof : AP and AS are the tangents to the circle from A
AP = AS ….(i)
Similarly BP = BQ ……(ii)
CR = CQ ….(iii)
and DR = DS ….(iv)
Adding, we get
AP + BP + CR + DR = AS + BQ + CQ + DS
=> (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
=> AB + CD = AD + BC
Hence proved.

Question 6.
Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle. [NCERT Exemplar]
Solution:
Let C₁ and C₂ be the two circles having same centre O. AC is a chord which touches the C₁ at point D

Join OD.
Also, OD ⊥ AC
AD = DC = 4 cm
[perpendicular line OD bisects the chord]
In right angled ∆AOD,
OA² = AD² + DO²
[by Pythagoras theorem, i.e.,
(hypotenuse)² = (base)² + (perpendicular)²]
=> DO² = 5² – 4² = 25 – 16 = 9
=> DO = 3 cm
Radius of the inner circle OD = 3 cm

Question 7.
A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ. [NCERT Exemplar]
Solution:
Given : Chord PQ is parallel tangent at R.
To prove : R bisects the arc PRQ.

Proof: ∠1 = ∠2 [alternate interior angles]
∠1 = ∠3
[angle between tangent and chord is equal to angle made by chord in alternate segment]
∠2 = ∠3
=> PR = QR
[sides opposite to equal angles are equal]
=> PR = QR
So, R bisects PQ.

Question 8.
Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A. [NCERT Exemplar]
Solution:
Given, AB is a diameter of the circle.
A tangent is drawn from point A.
Draw a chord CD parallel to the tangent MAN.

So, CD is a chord of the circle and OA is a radius of the circle.
∠MAO = 90°
[Tangent at any point of a circle is perpendicular to the radius through the point of contact]
∠CEO = ∠MAO [corresponding angles]
∠CEO = 90°
Thus, OE bisects CD
[perpendicular from centre of circle to chord bisects the chord]
Similarly, the diameter AB bisects all. Chord which are parallel to the tangent at the point A.

Question 9.
If AB, AC, PQ are the tangents in the figure, and AB = 5 cm, find the perimeter of ∆APQ.

Solution:
Given : AB, AC and PQ are the tangents to the circle as shown in the figure above and AB = 5 cm
To find : The perimeter of ∆APQ
Proof: PB and PX are the tangents to the circle
PB = PX
Similarly QC and QX are the tangents from
QC = QX
and AB and AC are the tangents from A
AB = AC
Now perimeter of ∆APQ
= AP + PQ + AQ
= AP + PX + QX + AQ
= AP + PB + QC + AQ { PB = PX and QC = QX}
= AB + AC
= AB + AB (AB=AC)
= 2 AB = 2 x 5 = 10 cm

Question 10.
Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.
Solution:
Given : PQ and RS are parallel tangents of a circle
RMP is the intercept of the tangent between PQ and RS
RO and PQ are joined

Question 11.
In the figure, PQ is tangent at a point R of the circle with centre O. If ∠TRQ = 30°, find m ∠PRS
Solution:
In the figure,
PRQ is tangent to the circle with centre O at R

RT and RS are joined such that ∠TRQ = 30°
Let ∠PRS = x°
Now ∠SRX = 90° (angle in a semicircle)
But ∠TRQ + ∠SRT + ∠PRS = 180° (Angles of a line)
=> 30° + 90° + x° = 180°
=> 120° + x° = 180°
=> x° = 180° – 120° = 60°
∠PRS = 60°

Question 12.
If PA and PB are tangents from an outside point P, such that PA = 10 cm and ∠APB = 60°. Find the length of chord AB.
Solution:
PA and PB are the tangents from a point PQ outside the circle with centre O
PA = 10 cm and ∠APB = 60°

Tangents drawn from a point outside the circle are equal
PA = PB = 10 cm ∠PAB = ∠PBA
(Angles opposite to equal sides)
But in ∆APB,
∠APB + ∠PAB + ∠PBA = 180° (Angles of a triangle)
=> 60° + ∠PAB + ∠PAB = 180°
=> 2 ∠PAB = 180° – 60° = 120°
∠PAB = 60°
∠PBA = ∠PAB = 60°
PA = PB = AB = 10 cm
Hence length of chord AB = 10 cm

Question 13.
In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC. [NCERT Exemplar]
Solution:
Let O be the centre of the given circle. Suppose, the tangent at P meets BC at Q.
Join BP.

To prove : BQ = QC
[angles in alternate segment]
Proof : ∠ABC = 90°
[tangent at any point of circle is perpendicular to radius through the point of contact]
In ∆ABC, ∠1 + ∠5 = 90°
[angle sum property, ∠ABC = 90°]
∠3 = ∠1
[angle between tangent and the chord equals angle made by the chord in alternate segment]
∠3 + ∠5 = 90° ……..(i)
Also, ∠APB = 90° [angle in semi-circle]
∠3 + ∠4 = 90° …….(ii)
[∠APB + ∠BPC = 180°, linear pair]
From Eqs. (i) and (ii), we get
∠3 + ∠5 = ∠3 + ∠4
∠5 = ∠4
=> PQ = QC
[sides opposite to equal angles are equal]
Also, QP = QB
[tangents drawn from an internal point to a circle are equal]
=> QB = QC
Hence proved.

Question 14.
From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of ∆PCD.
Solution:
PA and PB are the tangents drawn from a point P out side the circle with centre O
CD is another tangents to the circle at point E which intersects PA and PB at C and D respectively

PA = 14 cm
PA and PB are the tangents to the circle from P
PA = PB = 14 cm
Now CA and CE are the tangents from C
CA = CE ….(i)
Similarly DB and DE are the tangents from D
DB = DE ….(ii)
Now perimeter of ∆PCD
= PC + PD + CD
= PC + PD + CE + DE
= PC + CE + PD + DE
= PC + CA + PD = DB {From (i) and (ii)}
= PA + PB
= 14 + 14
= 28 cm

Question 15.
In the figure, ABC is a right triangle right-angled at B such that BC = 6 cm and AB = 8 cm. Find the radius of its incircle. [CBSE 2002]

Solution:
In right ∆ABC, ∠B = 90°
BC = 6 cm, AB = 8 cm
Let r be the radius of incircle whose centre is O and touches the sides A B, BC and CA at P, Q and R respectively
AP and AR are the tangents to the circle AP = AR
Similarly CR = CQ and BQ = BP
OP and OQ are radii of the circle
OP ⊥ AB and OQ ⊥ BC and ∠B = 90° (given)
BPOQ is a square
BP = BQ = r
AR = AP = AB – BD = 8 – r
and CR = CQ = BC – BQ = 6 – r
But AC² = AB² + BC² (Pythagoras Theorem)
= (8)² + (6)² = 64 + 36 = 100 = (10)²
AC = 10 cm
=> AR + CR = 10
=> 8 – r + 6 – r = 10
=> 14 – 2r = 10
=> 2r = 14 – 10 = 4
=> r = 2
Radius of the incircle = 2 cm

Question 16.
Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc. [NCERT Exemplar]
Solution:
Let mid-point of an arc AMB be M and TMT’ be the tangent to the circle.
Join AB, AM and MB.

Since, arc AM = arc MB
=> Chord AM = Chord MB
In ∆AMB, AM = MB
=> ∠MAB = ∠MBA ……(i)
[equal sides corresponding to the equal angle]
Since, TMT’ is a tangent line.
∠AMT = ∠MBA
[angle in alternate segment are equal]
∠AMT = ∠MAB [from Eq. (i)]
But ∠AMT and ∠MAB are alternate angles, which is possible only when AB || TMT’
Hence, the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.
Hence proved

Question 17.
From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP = diameter of the circle show that ∆APB is equilateral.
Solution:
Given : From a point P outside the circle with centre O, PA and PB are the tangerts to the circle such that OP is diameter.
AB is joined.

To prove: APB is an equilateral triangle
Const : Join OP, AQ, OA
Proof : OP = 2r
=> OQ + QP = 2r
=> OQ = QP = r (OQ = r)
Now in right ∆OAP,
OP is its hypotenuse and Q is its mid point
OA = AQ = OQ
(mid-point of hypotenuse of a right triangle is equidistances from its vertices)
∆OAQ is equilateral triangle ∠AOQ = 60°
Now in right ∆OAP,
∠APO = 90° – 60° = 30°
=> ∠APB = 2 ∠APO = 2 x 30° = 60°
But PA = PB (Tangents from P to the circle)
=> ∠PAB = ∠PBA = 60°
Hence ∆APB is an equilateral triangle.

Question 18.
Two tangents segments PA and PB are drawn to a circle with centre O such that ∠APB = 120°. Prove that OP = 2 AP. [CBSE 2014]
Solution:
Given : From a point P. Out side the circle with centre O, PA and PB are tangerts drawn and ∠APB = 120°
OP is joined To prove : OP = 2 AP
Const: Take mid point M of OP and join AM, join also OA and OB.

Proof : In right ∆OAP,
∠OPA = \(\frac { 1 }{ 2 }\) ∠APB = \(\frac { 1 }{ 2 }\) x 120° = 60°
∠AOP = 90° – 60° = 30°
M is mid point of hypotenuse OP of ∆OAP
MO = MA = MP
∠OAM = ∠AOM = 30° and ∠PAM = 90° – 30° = 60°
∆AMP is an equilateral triangle
MA = MP = AP
But M is mid point of OP
OP = 2 MP = 2 AP
Hence proved.

Question 19.
If ∆ABC is isosceles with AB = AC and C (0, r) is the incircle of the ∆ABC touching BC at L. Prove that L bisects BC.
Solution:
Given: In ∆ABC, AB = AC and a circle with centre O and radius r touches the side BC of ∆ABC at L.

To prove : L is mid point of BC.
Proof : AM and AN are the tangents to the circle from A
AM = AN
But AB = AC (given)
AB – AN = AC – AM
BN = CM
Now BL and BN are the tangents from B
BL = BN
Similarly CL and CM are tangents
CL = CM
But BM = CM (proved)
BL = CL
L is mid point of BC.

Question 20.
AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects AB at a point D. Prove that BC = BD. [NCERT Exemplar]
Solution:
To prove, BC = BD
Join BC and OC.

Given, ∠BAC = 30°
=> ∠BCD = 30°
[angle between tangent and chord is equal to angle made by chord in the alternate segment]
∠ACD = ∠ACO + ∠OCD
∠ACD = 30° + 90° = 120°
[OC ⊥ CD and OA = OC = radius => ∠OAC = ∠OCA = 30°]
In ∆ACD,
∠CAD + ∠ACD + ∠ADC = 180°
[since, sum of all interior angles of a triangle is 180°]
=> 30° + 120° + ∠ADC = 180°
=> ∠ADC = 180° – (30° + 120°) = 30°
Now, in ∆BCD,
∠BCD = ∠BDC = 30°
=> BC = BD
[since, sides opposite to equal angles are equal]

Question 21.
In the figure, a circle touches all the four sides of a quadrilateral ABCD with AB = 6 cm, BC = 7 cm, and CD = 4 cm. Find AD. [CBSE 2002]
Solution:
A circle touches the sides AB, BC, CD and DA of a quadrilateral ABCD at P, Q, R and S respectively.
AB = 6 cm, BC = 7 cm, CD = 4cm

Let AD = x
AP and AS are the tangents to the circle
AP = AS
Similarly,
BP = BQ
CQ = CR
and OR = DS
AB + CD = AD + BC
=> 6 + 4 = 7 + x
=> 10 = 7 + x
=>x = 10 – 7 = 3
AD = 3 cm

Question 22.
Prove that the perpendicular at the point contact to the tangent to a circle passes through the centre of the circle.
Solution:
Given : TS is a tangent to the circle with centre O at P, OP is joined

To prove : OP is perpendicular to TS which passes through the centre of the circle
Construction : Draw a line OR which intersect the circle at Q and meets the tangent TS at R
Proof: OP = OQ
(radii of the same circle) and OQ < OR => OP < OR
Similarly we can prove that OP is less than all lines which can be drawn from O to TS
OP is the shortest
OP is perpendicular to TS
Perpendicular through P, will pass through the centre of the circle
Hence proved.

Question 23.
Two circles touch externally at a point P. From a point T on the tangent at P, tangents TQ and TR are drawn to the circles with points of contact Q and R respectively. Prove that TQ = TR.

Solution:
Given : Two circles with centres O and C touch each other externally at P. PT is its common tangent
From a point T on PT, TR and TQ are the tangents drawn to the circles

To prove : TQ = TR
Proof : From T, TR and TP are two tangents to the circle with centre O
TR = TP ….(i)
Similarly, from T,
TQ and TP are two tangents to the circle with centre C
TQ = TP ….(ii)
From (i) and (ii)
TQ = TR
Hence proved.

Question 24.
A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the ∆ABC. [NCERT Exemplar]
Solution:
Given : Two tangents are drawn from an external point A to the circle with centre O, Tangent BC is drawn at a point R, radius of circle equals to 5 cm.

To find : Perimeter of ∆ABC.
Proof : ∠OPA = 90°
[Tangent at any point of a circle is perpendicular to the radius through the point of contact]
OA² = OP² + PA² [by Pythagoras Theorem]
(13)² = 5² + PA²
=> PA² = 144 = 12²
=> PA = 12 cm
Now, perimeter of ∆ABC = AB + BC + CA = (AB + BR) + (RC + CA)
= AB + BP + CQ + CA [BR = BP, RC = CQ tangents from internal point to a circle are equal]
= AP + AQ = 2AP = 2 x (12) = 24 cm
[AP = AQ tangent from internal point to a circle are equal]
Hence, the perimeter of ∆ABC = 24 cm.

Question 25.
In the figure, a circle is inscribed in a quadrilateral ABCD in which ∠B = 90°. If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius r of the circle.

Solution:
In the figure, O is the centre of the circle inscribed in a quadrilateral ABCD and ∠B = 90°
AD = 23 cm, AB = 29 cm, DS = 5 cm

OP = OQ (radii of the same circle)
AB and BC are tangents to the circle and OP and OQ are radii
OP ⊥ BC and OQ ⊥ AB
∠OPB = ∠OQB = 90°
PBQO is a square
DS and DR are tangents to the circle
DR = DS = 5 cm
AR = AD – DR = 23 – 5 = 18 cm
AR and AQ are the tangents to the circle
AQ = AR = 18 cm But AB = 29 cm
BQ = AB – AQ = 29 – 18 = 11 cm
Side of square PBQO is 11 cm
OP = 11 cm
Hence radius of the circle = 11 cm

Question 26.
In the figure, there are two concentric circles with centre O of radii 5 cm and 3 cm. From an external point P, tangents PA and PB are drawn to these circles. If AP =12 cm, find the length of BP. [CBSE 2010]

Solution:
Two concentric circles with centre O with radii 5 cm and 3 cm respectively from a
point P, PA and BP are tangents drawn to there circles

AP = 12 cm
To find BP
In right ∆OAP,
OP² = OA² + AP² (Pythagoras Theorem)
= (5)² + (12)² = 25 + 144
= 169 = (13)²
OP = 13 cm
Now in right ∆OBP,
OP² = OB² + BP²
=> (13)² = (3)² + BP²
=> 169 = 9 + BP²
=> BP² = 169 – 9 = 160 = 16 x 10
BP = √(16 x 10) = 4√10 cm

Question 27.
In the figure, AB is a chord of length 16 cm of a circle of radius 10 cm. The tangents at A and B intersect at a point P. Find the length of PA. [CBSE 2010]

Solution:
In the figure, AB is the chord of the circle with centre O and radius 10 cm.
Two tangents from P are drawn to the circle touching it at A and B respectively
AB is joined with intersects OP at L

Question 28.
In the figure, PA and PB are tangents from an external point P to a circle with centre O. LN touches the circle at M. Prove that PL + LM = PN + MN. [CBSE 2010]

Solution:
Given : In the figure, PA and PB are the tangents to the circle with centre O from a point P outside it LN touches it at M

To prove : PL + LM = PN + MN
Prove : PA and PB are tangents to the circle from P
PA = PB
Similarly from L, LA and LM are tangents
LA = LM
Similarly NB = NM
Now PA = PB => PL + LA = PN + NB
PL + LM = PN + NM
Hence proved.

Question 29.
In the figure, BDC is a tangent to the given circle at point D such that BD = 30 to the circle and meet when produced at A making BAC a right angle triangle. Calculate (i) AF (ii) radius of the circle.

Solution:
In the figure, BDC is a tangent to the given circle with centre O and D is a point such that
BD = 30 cm and CD = 7 cm

BE and CF are other two tangents drawn from B and C respectively which meet at A on producing this and ∆BAC is a right angle so formed
To find : (i) AF and (ii) radius of the circle
Join OE and OF
OE = OF radii of the circle
OE ⊥ AB and OF ⊥ AC
OEAF is a square
BD and BE are the tangents from B
BE = BD = 30 cm and similarly
CF = CD = 7 cm
Let r be the radius of the circle
OF = AF = AE = r
AB = 30 + r and AC = 7 + r and BC = 30 + 7 = 37 cm
Now in right ∆ABC

Question 30.
If d₁, d₂ (d₂ > d₁) be the diameters of two concentric circles and c be the length of a chord of a circle which is tangent to the other circle, prove that \({ d }_{ 2 }^{ 2 }={ c }^{ 2 }+{ d }_{ 1 }^{ 2 }\). [NCERT Exemplar]
Solution:
Let AB be a chord of a circle which touches the other circle at C. Then ∆OCB is right triangle.

Question 31.
In the given figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find ∠RQS. [CBSE 2015, NCERT Exemplar]

Solution:
In the given figure,
PQ and PR are tangents to the circle with centre O drawn from P
∠RPQ = 30°
Chord RS || PQ is drawn
To find ∠RQS
PQ = PR (tangents to the circle)
∠PRQ = ∠PQR But ∠RPQ = 30°

Question 32.
From an external point P, tangents PA = PB are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB. [CBSE 2016]
Solution:

PA = PB [tangents drawn from external point are equal]
∠PBA = ∠PAB = 50° [angles equal to opposite sides]
∠APB = 180° – 50° – 50° = 80° [angle-sum property of a A]
In cyclic quad. OAPB
∠AOB + ∠APB = 180° [sum of opposite angles of a cyclic quadrilateral is 180°]
∠AOB + 80° = 180°
∠AOB = 180°- 80° = 100°

Question 33.
In the figure, two tangents AB and AC are drawn to a circle with centre O such that ∠BAC = 120°. Prove that OA = 2AB.

Solution:
Given : In the figure, O is the centre of the circle.
AB and AC are the tangents to the circle from A such that
∠BAC = 120° .
To prove : OA = 2AB
Proof : In ∆OAB and ∆OAC
∠OBA = ∠OCA – 90° (OB and OC are radii)
OA = OA (common)
OB = OC (radii of the circle)
∆OAB ~ ∆OAC
∠OAB = ∠OAC = 60°
Now in right ∆OAB,

Question 34.
The lengths of three concecutive sides of a quadrilateral circumscribing a circle are 4 cm, 5 cm, and 7 cm respectively. Determine the length of the fourth side.
Solution:
In quadrilateral ABCD which is circumerscribing it
BC = 4 cm, CD = 5 cm and DA = 7 cm

We know that if a quad, is circumscribed in a circle, then
AB + CD = AD + BC
=> AB + 5 = 4 + 7
=> AB + 5 = 11
AB = 11 – 5 = 6
AB = 6 cm

Question 35.
The common tangents AB and CD to two circles with centres O and O’ intersect at E between their centres. Prove that the points O, E and O’ are collinear. [NCERT Exemplar]
Solution:
Joint AO, OC and O’D, O’B
Now, in ∆EO’D and ∆EO’B
O’D = O’B [radius]
O’E = O’E [common side]
ED = EB

Question 36.
In the figure, common tangents PQ and RS to two circles intersect at A. Prove that PQ = RS.

Solution:
Given : Two common tangents PQ and RS intersect each other at A.
To prove : PQ = RS
Proof: From A, AQ and AR are two tangents are drawn to the circle with centre O.

AP = AR ….(i)
Similarly AQ and AS are the tangents to the circle with centre C
AQ = AS ….(ii)
Adding (i) and (ii)
AP + AQ = AR + AS
=> PQ = RS
Hence proved.

Question 37.
Two concentric circles are of diameters 30 cm and 18 cm. Find the length of the chord of the larger circle which touches the smaller circle. [CBSE 2014]
Solution:
Let R be the radius of outer circle and r be the radius if small circle of two concentric circle
AB is the chord of the outer circle and touches the smaller circle at P
Join OP, OA

Question 38.
AB and CD are common tangents to two circles of equal radii. Prove that AB = CD. [NCERT Exemplar]
Solution:
Given : AB and CD are tangents to two circles of equal radii.
To prove :

Construction : Join OA, OC, O’B and O’D
Proof: Now, ∠OAB = 90°
[tangent at any point of a circle is perpendicular to radius through the point of contact]
Thus, AC is a straight line.
Also, ∠OAB + ∠OCD = 180°
AB || CD
Similarly, BD is a straight line and ∠O’BA = ∠O’DC = 90°
Also, AC = BD
[radii of two circles are equal] In quadrilateral ABCD,
∠A = ∠B = ∠C = ∠D = 90°
andAC = BD
ABCD is a rectangle
Hence, AB = CD
[opposite sides of rectangle are equal]

Question 39.
A triangle PQR is drawn to circumscribe a circle of radius 8 cm such that the segments QT and TR, into which QR is divided by the point of contact T, are of lengths 14 cm and 16 cm respectively. If area of ∆PQR is 336 cm², find the sides PQ and PR. [CBSE 2014]
Solution:
∆PQR is circumscribed by a circle with centre O and radius 8 cm
T is point of contact which divides the line segment OT into two parts such that
QT = 14 cm and TR = 16 cm

Area of ∆PQR = 336 cm²
Let PS = x cm
QT and QS are tangents to the circle from Q
QS = QT = 14 cm
Similarly RU and RT are tangents to the circle
RT = RU = 16 cm
Similarly PS and PU are tangents from P
PS = PU = x
Now PQ = x + 14 and PR = x + 16 and QR = 14 + 16 = 30 cm
Now area of ∆PQR = Area of ∆POQ + area of ∆QOR + area of ∆POR

Question 40.
In the figure, the tangent at a point C of a circle and a diameter AB when extended itersect at P. If ∠PCA = 110°, find ∠CBA. [NCERT Exemplar]

Solution:
Here, AB is a diameter of the circle from point C and a tangent is drawn which meets at a point P.
Join OC. Here, OC is radius.

Since, tangent at any point of a circle is perpendicular to the radius through point of contact circle.
OC ⊥ PC
Now, ∠PCA = 110° [given]
=> ∠PCO + ∠OCA = 110°
=> 90° + ∠OCA = 110°
=> ∠OCA = 20°
OC = OA = Radius of circle
∠OCA = ∠OAC = 20°
[since, two sides are equal, then their opposite angles are equal]
Since, PC is a tangent, so
∠BCP = ∠CAB = 20°
[angles in a alternate segment are equal]
In ∆PBC, ∠P + ∠C + ∠A= 180°
∠P = 180° – (∠C + ∠A)
∠P = 180° – (110° + 20°)
∠P = 180° – 130° = 50°
In ∆PBC,
∠BPC + ∠PCB + ∠PBC = 180°
[sum of all interior angles of any triangle is 180°]
=> 50° + 20° + ∠PBC = 180°
=> ∠PBC = 180° – 70°
∠PBC = 110°
Since, ∆PB is a straight line.
∠PBC + ∠CBA = 180°
∠CBA = 180° – 110° = 70°

Question 41.
AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in the figure. Prove that ∠BAT = ∠ACB. [NCERT Exemplar]

Solution:
Since, AC is a diameter line, so angle in semicircle makes an angle 90°.
∠ABC = 90° [by property]
In ∆ABC,
∠CAB + ∠ABC + ∠ACB = 180°
[ sum of all interior angles of any triangle is 180°]
=> ∠CAB + ∠ACB = 180° – 90° = 90° ……….(i)
Since, diameter of a circle is perpendicular to the tangent.
i.e. CA ⊥ AT
∠CAT = 90°
=> ∠CAB + ∠BAT = 90° …….(ii)
From Eqs. (i) and (ii),
∠CAB + ∠ACB = ∠CAB + ∠BAT
=> ∠ACB = ∠BAT
Hence proved.

Question 42.
In the given figure, a ∆ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 8 cm and 6 cm respectively. Find the lengths of sides AB and AC, when area of ∆ABC is 84 cm². [CBSE 2015]

Solution:
In the given figure,
In ∆ABC is circle is inscribed touching it at D, E and F respectively.
Radius of the circle (r) = 4cm
OD ⊥ BC, then
OD = 4 cm, BD = 8 cm, DC = 6 cm
Join OE and OF

Question 43.
In the given figure, AB is a diameter of a circle with centre O and AT is a tangent. If ∠AOQ = 58°, find ∠ATQ. [CBSE 2015]

Solution:
In the given figure,
AB is the diameter, AT is the tangent
and ∠AOQ = 58°
To find ∠ATQ
Arc AQ subtends ∠AOQ at the centre and ∠ABQ at the remaining part of the circle
∠ABQ = \(\frac { 1 }{ 2 }\) ∠AOQ = \(\frac { 1 }{ 2 }\) x 58° = 29°
Now in ∆ABT,
∠BAT = 90° ( OA ⊥ AT)
∠ABT + ∠ATB = 90°
=> ∠ABT + ∠ATQ = 90°
=> 29° + ∠ATQ = 90°
=> ∠ATQ = 90°- 29° = 61°

Question 44.
In the figure, OQ : PQ = 3:4 and perimeter of ∆POQ = 60 cm. Determine PQ, QR and OP.

Solution:
In the figure, OQ : PQ = 3:4

Perimeter of ∆POQ = 60 cm
To find PQ, QR and OP
OQ : PQ = 3 : 4
Let OQ = 3x and PQ = 4x
Now in right ∆OPQ,
OP² = OQ² + PQ² = (3x)² + (4x)² = 9x² + 16x² = 25x² = (5x)²
OP = 5x
But OQ + QP + OP = 60 cm
3x + 4x + 5x = 60
=> 12x = 60
x = 5
PQ = 4x = 4 x 5 = 20 cm
QR = 2 OQ = 2 x 3x = 6 x 5 = 30 cm
OP = 5x = 5 x 5 = 25 cm

Question 45.
Equal circles with centre O and O’ touch each other at X. OO’ produced to meet a circle with centre O’, at A. AC is a tangent to the circle whose centre is O. O’ D is perpendicular to AC. Find the value of \(\frac { DO’ }{ CO }\)

Solution:
Two equal circles with centre O and O’ touch each other externally at X
OO’ produced to meet at A

AC is the tangent of circle with centre O,
O’D ⊥ AC is drawn OC is joined
AC is tangent and OC is the radius
OC ⊥ AC
O’D ⊥ AC
OC || O’D
Now O’A = \(\frac { 1 }{ 2 }\) A x or \(\frac { 1 }{ 2 }\) AO
Now in O’AD and AOAC
∠A = ∠A (common)
∠AO’D = ∠AOC (corresponding angles)

Question 46.
In the figure, BC is a tangent to the circle with centre O. OE bisects AP. Prove that ∆AEO ~ ∆ABC.

Solution:
Given : In the figure, BC is a tangent to the circle with centre O at B.
AB is diameter AC is joined which intersects the circle at P
OE bisects AP

To prove : ∆AEO ~ ∆ABC
Proof: In ∆OAE and ∆OPE
OE = OE (common)
OA = OP ‘ (radii of the same circle)
EA = EP (given)
∆OAE = ∆OPE (SSS axiom)
∠OEA = ∠OEP
But ∠OEA + ∠OEP = 180°
∠OEA = 90°
Now in ∆AEO and ∆ABC
∠OEA = ∠ABC (each 90°)
∠A = ∠A (common)
∆AEO ~ ∆ABC (AA axiom)
Hence proved.

Question 47.
In the figure, PO ⊥ QO. The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OT are right bisectors of each other.

Solution:
Given : In the figure, O is the centre of the circle
PO ⊥ QO
They tangents at P and Q intersect each other at T

To prove : PQ and OT are right bisector of each other
Proof : PT and QT are tangents to the circle
PT = QT
OP and OQ are radii of the circle and ∠POQ = 90° ( PO ⊥ QO)
OQTP is a square Where PQ and OT are diagonals
Diagonals of a square bisect each other at right angles
PQ and OT bisect each other at right angles
Hence PQ and QT are right bisectors of each other.

Question 48.
In the figure, O is the centre of the circle and BCD is tangent to it at C. Prove that ∠BAC + ∠ACD = 90°.

Solution:
Given : In the figure, O is the centre of the circle BCD is a tangent, CP is a chord

prove : ∠BAC + ∠ACD = 90°
Proof: ∠ACD = ∠CPA (Angles in the alternate segment)
But in ∆ACP,
∠ACP = 90° (Angle in a semicircle)
∠PAC + ∠CPA = 90°
=> ∠BAC + ∠ACD = 90°
(∠ACD = ∠CPA proved)
Hence proved.

Question 49.
Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines. [NCERT Exemplar]
Solution:
Given : Two tangents PQ and PR are drawn from an external point P to a circle with centre O.
To prove : Centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

Construction : Join OR, and OQ.
In ∆POR and ∆POQ
∠PRO = ∠PQO = 90°
[tangent at any point of a circle is perpendicular to the radius through the point of contact]
OR = OQ [radii of same circle]
Since, OP is common.
∆PRO = ∆PQO [RHS]
Hence, ∠RPO = ∠QPO [by CPCT]
Thus, O lies on angle bisector of PR and PQ.
Hence proved.

Question 50.
In the figure, there are two concentric circles with centre O. PRT and PQS are tangents to the inner circle from a point P lying on the outer circle. If PR = 5 cm, find the lengths of PS. [CBSE 2017]

Solution:
Construction : Join OS and OP.
Consider ∆POS
We have,
PO = OS

∆POS is an isosceles triangle.
We know that in an isosceles triangle, if a line drawn perpendicular to the base of the triangle from the common vertex of the equal sides, then that line will bisect the base (unequal side).
And PQ = PR = 5 cm
[PRT and PQS are tangents to the inner circle to the inner circle from a point P lying on the outer circle]
We have, PQ = QS
It is given that, PQ = 5 cm
QS = 5 cm
From the figure, we have
PS = PQ + QS
=> PS = 5 + 5
=> PS = 10 cm

Question 51.
In the figure, PQ is a tangent from an external point P to a circle with centre O and OP cuts the circle at T and QOP is a diameter. If ∠POR = 130° and S is a point on the circle, find ∠1 + ∠2. [CBSE 2017]

Solution:
Construction : Join RT.

Given, ∠POR = 130°
∠POQ = 180°- (∠POR) = 180° – 130° = 50°
Since, PQ is a tangent
∠PQO = 90°
Now, In ∆POQ,
∠POQ + ∠PQO + ∠QPO = 180°
=> 50° + 90° + ∠1 = 180°
=> ∠1 = 180° – 140°
=> ∠1 = 40°
Now, In ∆RST
∠RST = \(\frac { 1 }{ 2 }\) ∠ROT
[Angle which is subtended by on arc at the centre of a circle is double the size of the angle subtended at any point on the circumference]
=> ∠2 = \(\frac { 1 }{ 2 }\) x 130° = 65°
Now ∠1 + ∠2 = 40° + 65° = 105°

Question 52.
In the figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC = PD. [CBSE 2017]

Solution:
PA = PB = 12 cm …(i)
QC = AC = 3cm …(ii)
QD = BD = 3 cm …(iii)
[Tangents drawn from an external point are equal]
To find : PC + PD
= (PA – AC) + (PB – BD)
= (12 – 3) + (12 – 3) [From (i), (ii), and (iii)]
= 9 + 9 = 18 cm

Hope given RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.2 are helpful to complete your math homework.

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