Maths RD Sharma Solutions Class 10

RD Sharma Class 10 Solutions Chapter 2 Polynomials MCQS

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)

Polynomial Root calculator. Welcome to our step-by-step math solver! Solve · Simplify · Factor

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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RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

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.

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RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

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.

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RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3

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.

RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2

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.