RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1

RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1

Other Exercises

Question 1.
Fill in the blanks using the correct word given in brackets :
(i) All circles are …………… (congruent, similar).
(ii) All squares are ………… (similar, congruent).
(iii) All …………….. triangles are similar (isosceles, equilaterals) :
(iv) Two triangles are similar, if their corres-ponding angles are ……… (proportional, equal)
(v) Two triangles are similar, if their corres-ponding sides are ……………. (proportional, equal)
(vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are …………. and (b) their corres-ponding sides are …………….. (equal, proportional).
Solution:
(i) All circles are similar.
(ii) All squares are similar.
(iii) All equilaterals triangles are similar.
(iv) Two triangles are similar, if their corresponding angles are equal.
(v) Two triangles are similar, if their corresponding sides are proportional.
(vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.

Question 2.
Write the truth value (T/F) of each of the following statements :
(i) Any two similar figures are congruent.
(ii) Any two congruent figures are similar.
(iii) Two polygons are similar, if their corresponding sides are proportional.
(iv) Two polygons are similar if their corresponding angles are proportional.
(v) Two triangles are similar if their corresponding sides are proportional.
(vi) Two triangles are similar if their corresponding angles are proportional.
Solution:
(i) False : In some cases, the similar polygons can be congruent.
(ii) True.
(iii) False : Its corresponding angles must be equal also.
(iv) False : Angle are equal not proportional.
(v) True.
(vi) False : Sides should be proportional and corresponding angles should be equal.

 

Hope given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1 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.

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS

Other Exercises

Question 1.
If the circumference and the area of a circle are numerically equal, then diameter of the circle is
(a) π/2
(b) 2π
(c) 2
(d) 4
Solution:
Let r be the radius of the circle, then Circumference = 2πr
and area = πr²
But 2πr= πr²
∴ 2r = r²
⇒ r = 2
Diameter = 2r = 2 x 2 = 4 (d)

Question 2.
If the difference between the circum-ference and radius of a circle is 37 cm., then using π =  \(\frac { 22 }{ 7 }\)  the circumference (in cm) of the circle is
(a) 154
(b) 44
(c) 14
(d) 7 [CBSE 2013]
Solution:
Difference between circumference and radius of a circle = 37 cm
∴  2πr-r = 37
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 1

Question 3.
A wire can be bent in the form of a circle of radius 56 cm. If it is bent in the form of a square, then its area will be
(a) 3520 cm²
(b) 6400 cm²
(c) 7744 cm²
(d) 8800 cm²
Solution:
Radius of a circular wire (r) = 56 cm
Circumference = 2πr = 2 x \(\frac { 22 }{ 7 }\) x 56 cm = 352 cm
Now perimeter of square = 352 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 2

Question 4.
 If a wire is bent into the shape of a square, then the area of the square is 81 cm². When wire is bent into a semicircular shape, then the area of the semi-circle will be
(a) 22 cm²
(b) 44 cm²
(c) 77 cm²
(d) 154 cm²
Solution:
Area of a square wire = 81 cm²
∴ Side of square = \(\sqrt { Area } \)  = \(\sqrt { 81 } \)  cm = 9 cm ard per in eret of square =4a = 4 x 9 = 36cm
Perimeter of semicircular wire whose bent = 36 cm
Let r be the radius, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 3

Question 5.
A circular park has a path of uniform width around it. The difference between the outer and inner circumferences of the circular path is 132 m. Its width is
(a) 20 m
(b) 21 m
(c) 22 m
(d) 24 m
Solution:
Let R and r be the radii of the outer and inner circles of the park, then 2πR – 2πr = 132
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 4

Question 6.
The radius of a wheel is 0.25 m. The number of revolutions it will make to travel a distance of 11 km will be
(a) 2800
(b) 4000
(c) 5500
(d) 7000
Solution:
Radius of the wheel (r) = 0.25 m = 25 cm
Circumference of the wheel
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 5

Question 7.
The ratio of the outer and inner perimeters of a circular path is 23 : 22. If the path is 5 metres wide, the diameter of the inner circle is
(a) 55 m
(b) 110 m
(c) 220 m
(d) 230 m
Solution:
Ratio in the outer and inner perimeter of a circular path = 23 : 22
Width of path = 5 m
Let R and r be the radii of outer and inner path then R- r = 5 m ….(i)
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 6
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 7

Question 8.
The circumference of a circle is 100 cm. The side of a square inscribed in the circle is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 8
Solution:
Circumference of a circle (c) = 100 cm
Diagonal of square which is inscribed in the circle
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 9

Question 9.
The area of the incircle of an equilateral triangle of side 42 cm is :
(a) 22 73 cm²
(b) 231 cm²
(c) 462 cm²
(d) 924 cm²
Solution:
Side of an equilateral triangle (a) = 42 cm
Radius of inscribed circle = \(\frac { 1 }{ 3 }\) x altitude
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 10

Question 10.
The area of incircle of an equilateral triangle is 154 cm2. The perimeter of the triangle is
(a) 71.5 cm
(b) 71.7 cm
(c) 72.3 cm
(d) 72.7 cm
Solution:
Area of incircle of an equilateral triangle = 154 cm²
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 11

Question 11.
The area of the largest triangle that can be inscribed in a semi-circle of radius r. is
(a) r²
(b) 2 r²
(c) r³
(d) 2r³
Solution:
The largest triangle inscribed in a semi-circle of radius r, can be ΔABC as shown in the figure, whose base = AB = 2r
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 12

Question 12.
The perimeter of a triangle is 30 cm and the circumference of its incircle is 88 cm. The area of the triangle is
(a) 70 cm²
(b) 140 cm²
(c) 210 cm²
(d) 420 cm²
Solution:
The perimeter of a triangle = 30 cm
and circumference of its incircle = 88 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 13

Question 13.
The area of a circle is 220 cm², the area of a square inscribed in it is
(a) 49 cm²
(b) 70 cm²
(c) 140 cm²
(d) 150 cm²
Solution:
Area of a circle = 220 cm²
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 14

Question 14.
If the circumference of a circle increases from 4π to 8π, then its area is
(a) halved
(b) doubled
(c) tripled
(d) quadrupled
Solution:
In first case circumference of a circle = 4π
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 15

Question 15.
If the radius of a circle is diminished by 10%, then its area is diminished by
(a) 10% 
(b) 19%
(c) 20%
(d) 36%
Solution:
Let in first case radius of a circle = r
Then area = πr²
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 16
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 17

Question 16.
If the area of a square is same as the area of a circle, then the ratio of their perimeter, in terms of 7t, is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 18
Solution:
Let side of square = a
Perimeter = 4 a
Then area = a²
∴ Area of circle = a²
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 19

Question 17.
The area of the largest triangle that can be inscribed in a semi-circle of radius r is
(a) 2r
(b) r²
(c) r
(d) \(\sqrt { r } \)
Solution:
Radius of semicircule = r
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 20

Question 18.
The ratio of the areas of a circle and an equilateral triangle whose diameter and a side are respectively equal, is
(a) π : \(\sqrt { 2 } \)
(b) π : \(\sqrt { 3 } \)
(c) \(\sqrt { 3 } \) :π
(d) \(\sqrt { 2 } \) : π
Solution:
Let side of equilateral triangle = a
Then area = \(\frac { \sqrt { 3 } }{ 4 }\) a²
Diameter of circle = a
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 21

Question 19.
If the sum of the areas of two circles with radii r1 and r2 is equal to the area of a circle of radius r, then r + r
(a) >r²
(b) = r²
(c) < r²
(d) None of these
Solution:
Sum of area of two circles with radii r1  and r2
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 22

Question 20.
If the perimeter of a semi-circular protractor is 36 cm, then its diameter is
(a) 10 cm
(b) 12 cm
(c) 14 cm
(d) 16 cm
Solution:
Perimeter of a semicircle = 36 cm
Let d be its diameter, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 23

Question 21.
The perimeter of the sector OAB shown in the fiugre, is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 24
Solution:
Radius of sector of 60° = 7 cm
∴ Perimeter = arc AB + 2 r
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 25

Question 22.
 If the perimeter of a sector of a circle of radius 6.5 cm is 29 cm, then its area is
(a) 58 cm²
(b) 52 cm²
(c) 25 cm²
(d) 56 cm²
Solution:
Radius of a sector = 6.5 cm
and perimeter = 29 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 26

Question 23.
If the area of a sector of a circle bounded by an arc of length 5K cm is equal to 20K cm², then its radius is
(a) 12 cm
(b) 16 cm
(c) 8 cm
(d) 10 cm
Solution:
Let r be the radius, then
Length of the arc of sector of θ angle = 5π
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 27

Question 24.
The area of the circle that can be inscribed in a square of side 10 cm is
(a) 40π cm²
(b) 30π cm²
(c) 100π cm²
(d) 25π cm²
Solution:
Side of square = 10 cm
∴ Diameter of the inscribed circle = 10 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 28

Question 25.
If the difference between the circumference
(a) 154 cm²
(b) 160 cm²
(c) 200 cm²
(d) 150 cm²
Solution:
Let r be the radius of a circle then circum-ference = 2πr
∴  2πr-r = 37
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 29

Question 26.
The area of a circular path of uniform width h surrounding a circular region of radius r is
(a) π (2r + h) r
(b) π (2r + h) h
(c) π (h + r)r
(d) π (h + r) A
Solution:
Let r be the radius of inner circle h is the width of circular path
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 30

Question 27.
 If AB is a chord of length 5\(\sqrt { 3 } \)  cm of a circle with centre O and radius 5 cm, then area of sector OAB is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 31
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 32
Solution:
Radius of the circle (r) = 5 cm
AB is a chord = 5\(\sqrt { 3 } \)
Draw OM ⊥ AB which bisects the chord AB at M
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 33

Question 28.
The area of a circle whose area and circumference are numerically equal, is
(a) 2π sq. units
(b) 4π sq. units
(c) 6π sq. units
(d) 8π sq. units
Solution:
Let radius of the circle = r
∴ Area = πr²
and circumference = 2πr
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 34

Question 29.
If diameter of a circle is increased by 40%, then its area increases by
(a) 96%
(b) 40%
(c) 80%
(d) 48%
Solution:
Let the diameter of a circle in first case = 2r
Then radius = r
Area = πr²
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 35

Question 30.
In the figure, the shaded area is
(a) 50 (π – 2) cm²
(b) 25 (π – 2) cm²
(c) 25 (π + 2) cm²
(d) 5 (π – 2) cm²
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 36
Solution:
In the figure, ∠AOB = 90°
and radius of the circle = 10 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 37

Question 31.
In the figure, the area of the segment PAQ is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 38
Solution:
a is the radius of the circle arc PAQ subtends angle 90° at the centre
∴ Area of segment PAQ
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 39

Question 32.
In the figure, the area of segment ACB is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 40
Solution:
r is the radius of the circle and arc ACB subtends angle of 120° at the centre
Area of segment ACB = r
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 41

Question 33.
If the area of a sector of a circle bounded by an arc of length 5π cm is equal to 20rc cm², then the radius of the circle is
(a) 12 cm
(b) 16 cm
(c) 8 cm
(d) 10 cm
Solution:
Length of arc = 5π cm
area of sector = 20π cm²
Let the angle at the centre be θ
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 42

Question 34.
In the figure, the ratio of the areas of two sectors S1 and S2 is
(a) 5 : 2
(b) 3 : 5
(c) 5 : 3
(d) 4 : 5
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 43
Solution:
Let r be the radius of the circle
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 44
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 45

Question 35.
If the area of a sector of a circle is \(\frac { 5 }{ 18 }\) of the area of the circle, then the sector angle is equal to
(a) 60°
(b) 90°
(c) 100°
(d) 120°
Solution:
Area of sector of a circle = \(\frac { 5 }{ 18 }\) x area of circle
Let θ be its angle at the centre and r be radius
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 46

Question 36.
If the area of a sector of a circle is \(\frac { 7 }{ 20 }\) of the area of the circle, then the sector angel is equal to
(a) 110°
(b) 130°
(c) 100°
(d) 126°
Solution:
Area of sector of a circle = \(\frac { 7 }{ 20 }\) of the area of the circle
Let r be the radius and θ be its angle at the centre
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 47

Question 37.
In the figure, if ABC is an equilateral triangle, then shaded area is equal to?
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 48
Solution:
ΔABC is an equilateral triangle inscribed in a circle with centre O and radius r
BO and CO are joined
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 49

Question 38.
 In the figure, the ara of the shaded region is
(a) 3π cm²
(b) 6π cm²
(c) 9π cm²
(d) 7π cm²
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 50
Solution:
In the figure, ∠B = ∠C = 90°, ∠D = 60?
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 51
∴ ∠A= 360° – (90° + 90° + 60°) = 360° – 240° = 120°
Radius of the sector = 3 cm
∴Area of shaded portion
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 52

Question 39.
If the perimeter of a circle is equal to that of a square, then the ratio of their areas is
(a) 13 : 22
(b) 14 : 11
(c) 22 : 13
(d) 11 :14
Solution:
Let side of square = a units
∴ Area = a² sq. units
and perimeter = 4a units
Now perimeter of circle = 4a units
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 53
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 54

Question 40.
The radius of a circle is 20 cm. It is divided into four parts of equal area by drawing three concentric circles inside it. Then, the radius of the largest of three concentric circles drawn is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 55
Solution:
Radius of circle (R) = 20 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 56
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 57

Question 41.
The area of a sector whose perimeter is four times its radius r units, is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 58
Solution:
Radius of sector = r
Perimeter = 4r
and length of arc = 4r – 2r = 2r
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 59

Question 42.
If a chord of a circle of radius 28 cm makes an angle of 90° at the centre, then the area of the major segment is
(a) 392 cm²
(b) 1456 cm²
(c) 1848 cm²
(d) 2240 cm²
Solution:
A chord AB makes an angle of 90° at the centre
Radius of the circle = 28 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 60
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 61

Question 43.
If the area of a circle inscribed in an equilateral triangle is 48π square units, then perimeter of the trianlge is
(a) 17\(\sqrt { 3 } \) units
(b) 36 units
(c) 72 units
(d) 48\(\sqrt { 3 } \)  units
Solution:
Area of a circle inscribed in an equilateral triangle = 48π sq. units
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 62

Question 44.
The hour hand of a clock is 6 cm long. The area swept by it between 11.20 am and 11.55 am is
(a) 2.75 cm²
(b) 5.5 cm²
(c) 11 cm²
(d) 10 cm²
Solution:
Length of hour hand of a clock (r) = 6 cm
Time 11.20 am to 11.55 am = 35 minutes
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 63
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 64

Question 45.
ABCD is a square of side 4 cm. If ? is a point in the interior of the square such that ΔCED is equilateral, then area of ΔACEis
(a) 2 (\(\sqrt { 3 } \) – 1) cm²
(b) 4 (\(\sqrt { 3 } \) -1) cm²
(c) 6(\(\sqrt { 3 } \)-1)cm²
(d) 8(\(\sqrt { 3 } \)-1)cm²
Solution:
Side of square ABCD = 4 cm
and side of equilateral ΔCED = 4 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 65
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 66

Question 46.
If the area of a circle is equal to the sum of the areas of two circles of diameters 10 cm and 24 cm, then diameter of the larger circle (in cm) is
(a) 34
(b) 26

(c) 17
(d) 14

Solution:
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 67

Question 47.
If π is taken as 22/7, the-distance (in metres) covered by a wheel of diameter 35 cm, in one revolution, is (a) 2.2
(b) 1.1
(b) 9.625
(d) 96.25   [CBSE 2013]
Solution:
Diameter of a wheel = 35 cm = \(\frac { 35 }{ 100 }\) m
Circumference of the wheel = πd
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 68

Question 48.
ABCD is a rectangle whose three vertices are B (4, 0), C (4, 3) and D (0, 3). The length of one of its diagonals is
(a) 5
(b) 4

(c) 3
(d) 25 [CBSE 2014]
Solution:
Three vertices of a rectangle ABCD are B (4,0), C (4, 3) and D (0, 3) length of one of its diagonals
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 69

Question 49.
Area of the largest triangle that can be inscribed in a semi-circle of radius r units is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 70
Solution:
Take a point C on the circumference of the semi-circle and join it by the end points of diameter A and B.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 71

Question 50.
If the sum of the areas of two circles with radii r1 and r2 is equal to the area of a circle of radius r, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 72
Solution:
According to the given condition,
Area of circle = Area of first circle + Area of second circle.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 73

Question 51.
If the sum of the circumference of two circles with radii r, and r2 is equal to the circumference of a circle of radius r, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 74
Solution:
According to the given condition, Circumference of circle = Circumference of first circle + Circumference of second circle
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 75

Question 52.
 If the circumference of a circle and the perimeter of a square are equal, then
(a) Area of the circle = Area of the square
(b) Area of the circle < Area of the square
(c) Area of the circle > Area of the square

(d) Nothing definite can be said
Solution:
According to the given condition, Circumference of a circle = Perimeter of square 2 πr = 4 a
[where, r and a are radius of circle and side of square respectively]
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 76

Question 53.
 If the perimeter of a circle is equal to that of a square, then the ratio of their areas is
(a) 22 : 7
(b) 14 : 11
(c) 7 : 22
(d) 11 : 14
Solution:
Let radius of circle be r and side of a square be a
According to the given condition,
Perimeter of a circle = Perimeter of a square
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 77

Hope given RD Sharma Class 10 Solutions Chapter 13 Areas Related to 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.

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

Other Exercises

Mark the correct alternative in each of the following :
Question 1.
The distance between the points (cosθ, sinθ) and (sinθ, -cosθ) is
(a) √3
(b) √2
(c) 2
(d) 1
Solution:
(b) Distance between (cosθ, sinθ) and (sinθ, -cosθ)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 1
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 2

Question 2.
The distance between the points (a cos 25°, 0) and (0, a cos 65°) is
(a) a
(b) 2a
(c) 3a
(d) None of these
Solution:
(a) Distance between (a cos 25°, 0) and (0, a cos 65°)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 3

Question 3.
If x is a positive integer such that the distance between points P (x, 2) and Q (3, -6) is 10 units, then x =
(a) 3
(b) -3
(c) 9
(d) -9
Solution:
(c) Distance between P (x, 2) and Q (3, -6) = 10 units
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 4
=> x (x – 9) + 3 (x – 9) = 0
(x – 9) (x + 3) = 0
Either x – 9 = 0, then x = 9 or x + 3 = 0, then x = -3
x is positive integer
Hence x = 9

Question 4.
The distance between the points (a cosθ + b sinθ, 0) and (0, a sinθ – b cosθ) is
(a) a² + b²
(b) a + b
(c) a² – b²
(d) √(a²+b²)
Solution:
(d) Distance between (a cosθ + b sinθ, 0) and (0, a sinθ – b cosθ)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 5

Question 5.
If the distance between the points (4, p) and (1, 0) is 5, then p =
(a) ±4
(b) 4
(c) -4
(d) 0
Solution:
(a) Distance between (4, p) and (1, 0) = 5
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 6
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 7

Question 6.
A line segment is of length 10 units. If the coordinates of its one end are (2, -3) and the abscissa of the other end is 10, then its ordinate is
(a) 9, 6
(b) 3, -9
(c) -3, 9
(d) 9, -6
Solution:
(b) Let the ordinate of other end = y
then distance between (2, -3) and (10, y) = 10 units
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 8

Question 7.
The perimeter of the triangle formed by the points (0, 0), (1, 0) and (0, 1) is
(a) 1 ± √2
(b) √2 + 1
(c) 3
(d) 2 + √2
Solution:
(d) Let the vertices of ∆ABC be A (0, 0), B(1, 0) and C (0, 1)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 9
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 10

Question 8.
If A (2, 2), B (-4, -4) and C (5, -8) are the vertices of a triangle, then the length of the median through vertices C is
(a) √65
(b) √117
(c) √85
(d) √113
Solution:
(c) Let mid point of A (2, 2), B (-4, -4) be D whose coordinates will be
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 11

Question 9.
If three points (0, 0), (3, √3) and (3, λ) form an equilateral triangle, then λ =
(a) 2
(b) -3
(c) -4
(d) None of these
Solution:
(d) Let the points (0, 0), (3, √3) and (3, λ) from an equilateral triangle
AB = BC = CA
=> AB² = BC² = CA²
Now, AB² = (x2 – x1)² + (y2 – y1
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 12

Question 10.
If the points (k, 2k), (3k, 3k) and (3, 1) are collinear, then k
(a) \(\frac { 1 }{ 3 }\)
(b) \(\frac { -1 }{ 3 }\)
(c) \(\frac { 2 }{ 3 }\)
(d) \(\frac { -2 }{ 3 }\)
Solution:
(b) Let the points A (k, 2k), B (3k, 3k) and C (3, 1) be the vertices of a ∆ABC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 13
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 14

Question 11.
The coordinates of the point on x-axis which are equidistant from the points (-3, 4) and (2, 5) are
(a) (20, 0)
(b) (-23, 0)
(c) (\(\frac { 4 }{ 5 }\) , 0)
(d) None of these
Solution:
(d)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 15

Question 12.
If (-1, 2), (2, -1) and (3, 1) are any three vertices of a parallelogram, then
(a) a = 2, b = 0
(b) a = -2, b = 0
(c) a = -2, b = 6
(d) a = 0, b = 4
Solution:
(d) In ||gm ABCD, diagonals AC and AD bisect each other at O
O is mid-point of AC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 16

Question 13.
If A (5, 3), B (11, -5) and P (12, y) are the vertices of a right triangle right angled at P, then y =
(a) -2, 4
(b) -2, 4
(c) 2, -4
(d) 2, 4
Solution:
(c) A (5, 3), B (11, -5) and P (12, y) are the vertices of a right triangle, right angle at P
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 17
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 18

Question 14.
The area of the triangle formed by (a, b + c), (b, c + a) and (c, a + b) is
(a) a + b + c
(b) abc
(c) (a + b + c)²
(d) 0
Solution:
(d) Vertices of a triangle are (a, b + c), (b, c + a) and (c, a + b)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 19

Question 15.
If (x, 2), (-3, -4) and (7, -5) are coliinear, then x =
(a) 60
(b) 63
(c) -63
(d) -60
Solution:
(c) Area of triangle whose vertices are (x, 2), (-3, -4) and (7, -5)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 20
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 21

Question 16.
If points (t, 2t), (-2, 6) and (3, 1) are collinear, then t =
(a) \(\frac { 3 }{ 4 }\)
(b) \(\frac { 4 }{ 3 }\)
(c) \(\frac { 5 }{ 3 }\)
(d) \(\frac { 3 }{ 5 }\)
Solution:
(b) The area of triangle whose vertices are (t, 2t), (-2, 6) and (3, 1)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 22

Question 17.
If the area of the triangle formed by the points (x, 2x), (-2, 6) and (3, 1) is 5 square units, then x =
(a) \(\frac { 2 }{ 3 }\)
(b) \(\frac { 3 }{ 5 }\)
(c) 2
(d) 5
Solution:
(c) Area of triangle whose vertices are (x, 2x), (-2, 6) and (3, 1)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 23

Question 18.
If points (a, 0), (0, b) and (1, 1) are collinear, then \(\frac { 1 }{ a }\) + \(\frac { 1 }{ b }\) =
(a) 1
(b) 2
(c) 0
(d) -1
Solution:
(a) The area of triangle whose vertices are (a, 0), (0, b) and (1, 1)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 24
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 25

Question 19.
If the centroid of a triangle is (1, 4) and two of its vertices are (4, -3) and (-9, 7), then the area of the triangle is
(a) 183 sq. units
(b) \(\frac { 183 }{ 2 }\) sq. units
(c) 366 sq. units
(d) \(\frac { 183 }{ 4 }\) sq. units
Solution:
(b) Centroid of a triangle = (1, 4)
and two vertices of the triangle are (4, -3) and (-9, 7)
Let the third vertex be (x, y), then
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 26
= \(\frac { 183 }{ 2 }\) sq. units

Question 20.
The line segment joining points (-3, -4) and (1, -2) is divided by y-axis in the ratio
(a) 1 : 3
(b) 2 : 3
(c) 3 : 1
(d) 2 : 3
Solution:
(c) The point lies on y-axis
Its abscissa will be zero
Let the point divides the line segment joining the points (-3, -4) and (1, -2) in the ratio m : n
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 27

Question 21.
The ratio in which (4, 5) divides the join of (2, 3) and (7, 8) is
(a) -2 : 3
(b) -3 : 2
(c) 3 : 2
(d) 2 : 3
Solution:
(d) Let the point (4, 5) divides the line segment joining the points (2, 3) and (7, 8) in the ratio m : n
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 28

Question 22.
The ratio in which the X-axis divides the segment joining (3, 6) and (12, -3) is
(a) 2 : 1
(b) 1 : 2
(c) -2 : 1
(d) 1 : -2
Solution:
(a) The point lies on x-axis
Its ordinate is zero
Let this point divides the line segment joining the points (3, 6) and (12, -3) in the ratio m : n
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 29

Question 23.
If the centroid of the triangle formed by the points (a, b), (b, c) and (c, a) is at the origin, then a3 + b3 + c3 =
(a) abc
(b) 0
(c) a + b + c
(d) 3 abc
Solution:
(d) Centroid of the triangle formed by the points (a, b), (b, c) and (c, a) is origin (0, 0)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 30

Question 24.
If points (1, 2), (-5, 6) and (a, -2) are collinear, then a =
(a) -3
(b) 7
(c) 2
(d) -2
Solution:
(b) The area of a triangle whose vertices are (1, 2), (-5, 6) and (a, -2)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 31
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 32

Question 25.
If the centroid of the triangle formed by (7, x), (y, -6) and (9, 10) is at (6, 3), then (x, y) =
(a) (4, 5)
(b) (5, 4)
(c) (-5, -2)
(d) (5, 2)
Solution:
(d) Centroid of (7, x), (y, -6) and (9, 10) is (6, 3)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 33

Question 26.
The distance of the point (4, 7) from the x-axis is
(a) 4
(b) 7
(c) 11
(d) √65
Solution:
(b) The distance of the point (4, 7) from x-axis = 7

Question 27.
The distance of the point (4, 7) from the y-axis is
(a) 4
(b) 7
(c) 11
(d) √65
Solution:
(a) The distance of the point (4, 7) from y-axis = 4

Question 28.
If P is a point on x-axis such that its distance from the origin is 3 units, then the coordinates of a point Q on OY such that OP = OQ, are
(a) (0, 3)
(b) (3, 0)
(c) (0, 0)
(d) (0, -3)
Solution:
(a) P is a point on x-axis and its distance from 0 is 3
Co-ordinates of P will be (3, 0)
Q is a point on OY such that OP = OQ
Co-ordinates of Q will be (0, 3)

Question 29.
If the point (x, 4) lies on a circle whose centre is at the origin and radius is 5, then x =
(a) ±5
(b) ±3
(c) 0
(d) ±4
Solution:
(b) Point A (x, 4) is on a circle with centre O (0, 0) and radius = 5
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 34

Question 30.
If the point P (x, y) is equidistant from A (5, 1) and B (-1, 5), then
(a) 5x = y
(b) x = 5y
(c) 3x = 2y
(d) 2x = 3y
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 35

Question 31.
If points A (5, p), B (1, 5), C (2, 1) and D (6, 2) form a square ABCD, then p =
(a) 7
(b) 3
(c) 6
(d) 8
Solution:
(c) Vertices of a square are A (5, p), B (1, 5), C (2, 1) and D (6, 2)
The diagonals bisect each other at O
O is the mid-point of AC and BD
O is mid-point of BD, then
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 36

Question 32.
The coordinates of the circumcentre of the triangle formed by the points O (0, 0), A (a, 0) and B (0, b) are
(a) (a, b)
(b) (\(\frac { a }{ 2 }\) , \(\frac { b }{ 2 }\))
(c) (\(\frac { b }{ 2 }\) , \(\frac { a }{ 2 }\))
(d) (b, a)
Solution:
(b) Let co-ordinates of C be (x, y) which is the centre of the circumcircle of ∆OAB
Radii of a circle are equal
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 37
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 38

Question 33.
The coordinates of a point on x-axis which lies on the perpendicular bisector of the line segment joining the points (7, 6) and (-3, 4) are
(a) (0, 2)
(b) (3, 0)
(b) (0, 3)
(d) (2, 0)
Solution:
(d) The given point P lies on x-axis
Let the co-ordinates of P be (x, 0)
The point P lies on the perpendicular bisector of of the line segment joining the points A (7, 6), B (-3, 4)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 39

Question 34.
If the centroid of the triangle formed by the points (3, -5), (-7, 4), (10, -k) is at the point (k, -1), then k =
(a) 3
(b) 1
(c) 2
(d) 4
Solution:
(c) O (k, -1) is the centroid of triangle whose vertices are
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 40

Question 35.
If (-2, 1) is the centroid of the triangle having its vertices at (x, 0), (5, -2), (-8, y), then x, y satisfy the relation
(a) 3x + 8y = 0
(b) 3x – 8y = 0
(c) 8x + 3y = 0
(d) 8x = 3y
Solution:
(-2, 1) is the centroid of triangle whose vertices are (x, 0), (5, -2), (-8, y)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 41

Question 36.
The coordinates of the fourth vertex of the rectangle formed by the points (0, 0), (2, 0), (0, 3) are
(a) (3, 0)
(b) (0, 2)
(c) (-2, 3)
(d) (3, 2)
Solution:
(c) Three vertices of a rectangle are A (0, 0), B (2, 0), C (0, 3)
Let fourth vertex be D (x, y)
The diagonals of a rectangle bisect eachother at O
O is the mid-point of AC, then
Coordinates of O will be (\(\frac { 0+0 }{ 2 }\) , \(\frac { 0+3 }{ 2 }\))
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 42

Question 37.
The length of a line segment joining A (2, -3) and B is 10 units. If the abscissa of B is 10 units, then its ordinates can be
(a) 3 or -9
(b) -3 or 9
(c) 6 or 27
(d) -6 or-27
Solution:
(a) Abscissa of B is 10 and co-ordinates of A are (2, -3)
Let ordinates of B be y, then
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 43

Question 38.
The ratio in which the line segment joining P(x1, y1) and Q (x2, y2) is divided by x-axis is
(a) y1 : y2
(b) -y1 : y2
(c) x1 : x2
(d) -x1 : x2
Solution:
(b) Let a point A on x-axis divides the line segment joining the points P (x1, y1), Q (x2, y2) in the ratio m1 : m2 and
let co-ordinates of A be (x, 0)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 44

Question 39.
The ratio in which the line segment joining points A (a1, b1) and B (a2, b2) is divided by y-axis is
(a) -a1 : a2
(b) a1 : a2
(c) b1 : b2
(d) -b1 : b2
Solution:
(a) Let the point P on y-axis, divides the line segment joining the point A (a1, b1) and B (a2, b2) is the ratio m1 : m2 and
let the co-ordinates of P be (0, y), then
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 45

Question 40.
If the line segment joining the points (3, -4) and (1, 2) is trisected at points P
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 46
Solution:
(b)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 47
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 48

Question 41.
If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (-2, 5), then the coordinates of the other end of the diameter are [CBSE 2012]
(a) (-6, 7)
(b) (6, -7)
(c) (6, 7)
(d) (-6, -7)
Solution:
(a) Let AB be the diameter of a circle with centre O
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 49

Question 42.
The coordinates of the point P dividing the line segment joining the points A (1, 3) and B (4, 6) in the ratio 2 : 1 are
(a) (2, 4)
(b) (3, 5)
(c) (4, 2)
(d) (5, 3) [CBSE 2012]
Solution:
(b) Point P divides the line segment joining the points A (1, 3) and B (4, 6) in the ratio 2 : 1
Let coordinates of P be (x, y), then
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 50

Question 43.
In the figure, the area of ∆ABC (in square units) is [CBSE 2013]
(a) 15
(b) 10
(c) 7.5
(d) 2.5
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 51
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 52

Question 44.
The point on the x-axis which is equidistant from points (-1, 0) and (5, 0) is
(a) (0, 2)
(b) (2, 0)
(c) (3, 0)
(d) (0, 3) [CBSE 2013]
Solution:
(c) Let the point P (x, 0) is equidistant from the points A (-1, 0), B (5, 0)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 53

Question 45.
If A (4, 9), B (2, 3) and C (6, 5) are the vertices of ∆ABC, then the length of median through C is
(a) 5 untis
(b) √10 units
(c) 25 units
(d) 10 units [CBSE 2014]
Solution:
(b) A (4, 9), B (2, 3) and C (6, 5) are the vertices of ∆ABC
Let median CD has been drawn C (6, 5)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 54

Question 46.
If P (2, 4), Q (0, 3), R (3, 6) and S (5, y) are the vertices of a paralelogram PQRS, then the value of y is
(a) 7
(b) 5
(c) -7
(d) -8 [CBSE 2014]
Solution:
(a) P (2, 4), Q (0, 3), R (3, 6) and S (5, y) are the vertices of a ||gm PQRS
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 55
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 56

Question 47.
If A (x, 2), B (-3, -4) and C (7, -5) are collinear, then the value of x is
(a) -63
(b) 63
(c) 60
(d) -60 [CBSE 2014]
Solution:
(a) A (x, 2), B (-3, -4) and C (7, -5) are collinear, then area ∆ABC = 0
Now area of ∆ABC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 57

Question 48.
The perimeter of a triangle with vertices (0, 4) and (0, 0) and (3, 0) is
(a) 7 + √5
(b) 5
(c) 10
(d) 12 [CBSE 2014]
Solution:
(d) A (0, 4) and B (0, 0) and C (3, 0) are the vertices of ∆ABC
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 58

Question 49.
If the point P (2, 1) lies on the line segment joining points A (4, 2) and B (8, 4), then
(a) AP = \(\frac { 1 }{ 3 }\) AB
(b) AP = BP
(c) BP = \(\frac { 1 }{ 3 }\) AB
(d) AP = \(\frac { 1 }{ 2 }\) AB
Solution:
(d) Given that, the point P (2, 1) lies on the line segment joining the points (4, 2) and B (8, 4), which shows in the figure below:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 59
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 60

Question 50.
A line intersects the y-axis and x-axis at P and Q, respectively. If (2, -5) is the mid-point of PQ, then the coordinates of P and Q are, respectively
(a) (0, -5) and (2, 0)
(b) (0, 10) and (-4, 0)
(c) (0, 4) and (-10, 0)
(d) (0, -10) and (4, 0)
Solution:
(d) Let the coordinates of P and Q (0, y) and (x, 0), respectively.
So, the mid-point of P (0, y) and Q (x, 0) is
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS 61
2 = \(\frac { x + 0 }{ 2 }\) and -5 = \(\frac { y + 0 }{ 2 }\)
=> 4 = x and -10 = y
=> x = 4 and y = -10
So, the coordinates of P and Q are (0, -10) and (4, 0).

Hope given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry 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.

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS

Other Exercises

Answer each of the following questions either in one word or one sentence or as per requirement of the questions :
Question 1.
Write the distance between the points A (10 cosθ, 0) and B (0, 10 sinθ).
Solution:
Distance between the points A (10 cosθ, 0) and B (0, 10 sinθ)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 1

Question 2.
Write the perimeter of the triangle formed by the points O (0, 0), A (a, 0), and B (0, b).
Solution:
The vertices of a ∆OAB, O (0, 0), A (a, 0), and B (0, b)
Now length of OA
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 2

Question 3.
Write the ratio in which the line segment joining points (2, 3) and (3, -2) is divided by x-axis.
Solution:
The required point is on x-axis
Its ordinate will be 0
Let the point be (x, 0) and let this point divides the join of the points (2, 3) and (3, -2) in the ratio m : n
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 3

Question 4.
What is the distance between the points (5 sin 60°, 0) and (0, 5 sin 30°) ?
Solution:
Distance between the given points
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 4

Question 5.
If A (-1, 3), B (1, -1) and C (5, 1) are the vertices of a triangle ABC, what is the length of the median through vertex A ?
Solution:
The vertices of ∆ABC are A (-1, 3), B (1, -1) and C (5, 1)
Let AD be the median
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 5
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 6

Question 6.
If the distance between points (x, 0) and (0, 3) is 5, what are the value of x ?
Solution:
Distance between (x, 0) and (0, 3) = 5
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 7

Question 7.
What is the area of the triangle formed by the points O (0, 0), A (6, 0) and B (0, 4) ?
Solution:
The vertices of the triangle OAB are O (0, 0), A (6, 0) and B (0, 4)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 8
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 9

Question 8.
Write the coordinates of the point dividing line segment joining points (2, 3) and (3, 4) internally in the ratio 1 : 5.
Solution:
Let the coordinates of the required point be (x, y), then
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 10

Question 9.
If the centroid of the triangle formed by points P (a, b), Q (b, c) and R (c, a) is at the origin, what is the value of a + b + c ?
Solution:
Vertices of ∆PQR are P (a, b), Q (b, c) and R (c, a) and its centroid = O (0, 0)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 11

Question 10.
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 12
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 13
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 14

Question 11.
Write the coordinates of a point on x- axis which is equidistant from the points (-3, 4) and (2, 5).
Solution:
The point is on x-axis
Its ordinates of the point P is (x, 0)
P is equidistant from A (-3, 4) and B (2, 5)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 15

Question 12.
If the mid-point of the segment joining A (x, y + 1) and B (x + 1, y + 2) is C (\(\frac { 3 }{ 2 }\) , \(\frac { 5 }{ 2 }\)) find x, y.
Solution:
C (\(\frac { 3 }{ 2 }\) , \(\frac { 5 }{ 2 }\)) is mid point of the line segment
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 16

Question 13.
Two vertices of a triangle have co-ordinates (-8, 7) and (9, 4). If the centroid of the triangle is at the origin, what are the co-ordinates of the third vertex ?
Solution:
Two vertices of a triangle are (-8, 7) and (9, 4)
Let the third vertex be (x, y)
Centroid of the triangle is (0, 0)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 17

Question 14.
Write the coordinates the reflections of points (3, 5) in x and y-axes.
Solution:
Reflection of P (3, 5) in x-axis is will be (3, -5)
and reflection of P in y-axis will be (-3, 5)

Question 15.
If points Q and R reflections of point P (-3, 4) in X and Y axes respectively, what is QR ?
Solution:
Reflection of point P (-3, 4) in X-axis will be Q with coordinates Q (-3, -4) and reflection in Y-axis will be R (3, 4)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 18

Question 16.
Write the formula for the area of the triangle having its vertices at (x1, y1), (x2, y2) and (x3, y3).
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 19

Question 17.
Write the condition of collinearity of points (x1, y1), (x2, y2) and (x3, y3).
Solution:
Three points (x1, y1), (x2, y2) and (x3, y3). are said to be collinear if the area of the triangle formed by these point = 0 i.e.,
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 20

Question 18.
Find the values of x for which the distance between the point P (2, -3) and Q (x, 5) is 10.
Solution:
Distance between P (2, -3) and Q (x, 5) = 10
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 21
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 22

Question 19.
Write the ratio in which the line segment joining the points A (3, -6) and B (5, 3) is divided by X-axis.
Solution:
The point lies on x-axis
Its ordinate will be = 0
Let the point P (x, 0) divides the line segment joining the points A (3, -6) and B (5, 3) in the ratio m : n.
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 23

Question 20.
Find the distance between the points (\(\frac { -8 }{ 5 }\) , 2) and (\(\frac { 2 }{ 5 }\) , 2). (C.B.S.E. 2009)
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 24
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 25

Question 21.
Find the value of a so that the point (3, a) lies on the line represented by 2x – 3y + 5 = 0. (C.B.S.E. 2009)
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 26

Question 22.
What is the distance between the points A (c, 0) and B (0, – c) ? [CBSE 2010]
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 27

Question 23.
If P (2, 6) is the mid-point of the line segment joining A (6, 5) and B (4, y), find y. [CBSE 2010]
Solution:
P (2, 6) is the mid-point of the line segment A (6, 5) and b (4, y)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 28

Question 24.
If the distance between the points (3, 0) and (0, y) is 5 units and y is positive, then what is the value of y ? [CBSE 2010]
Solution:
Distance between (3, 0) and (0, y) is 5 units
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 29

Question 25.
If P (x, 6) is the mid-point of the line segment joining A (6, 5) and B (4, y), find y. [CBSE 2010]
Solution:
P (x, 6) is the mid-point of the line segment joining the points A (6, 5), B (4, y)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 30

Question 26.
If P (2, p) is the mid-point of the line segment joining the points A (6, -5) and B (-2, 11), find the value of p. [CBSE 2010]
Solution:
P (2, p) is the mid-point of the line segment joining the points A (6, -5) and B (-2, 11)
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 31

Question 27.
If A (1, 2), B (4, 3) and C (6, 6) are the three vertices of a parallelogram ABCD, find the coordinates of fourth vertex D. [CBSE 2010]
Solution:
vertices of a parallelogram Let co-ordinates of D be (x, y)
Diagonals AC and BD bisect each other at O
Co-ordinates of O will be
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 32
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 33

Question 28.
What is the distance between the points A (sinθ – cosθ, 0) and B (0, sinθ + cosθ)? [CBSE 2015]
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 34

Question 29.
What are the coordinates of the point where the perpendicular bisector of the line segment joining the points A (1, 5) and B (4, 6) cuts the y-axis?
Solution:
Firstly, we plot the points of the line segment on the paper and join them.
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 35
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 36
Now, we draw a straight line on paper passes through the mid-point P.
We see that the perpendicular bisector cuts the y-axis at the point (0, 13).
Hence, the required point is (0, 13).

Question 30.
Find the area of the triangle with vertices (a, b + c), (b, c + a) and (c, a + b).
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 37

Question 31.
If the points A (1, 2), O (0, 0) and C (a, b) are collinear, then find a : b.
Solution:
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 38
=> 2a = b
Hence, the required relation is 2a = b

Question 32.
Find the coordinates of the point which is equidistant from the three vertices A (2x, 0), O (0, 0) and B (0, 2y) of ∆AOB.
Solution:
Let the coordinate of the point which is equidistant from the three vertices O (0, 0), A (0, 2y) and B (2x, 0) is P (h, k).
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 39
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 40
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 41

Question 33.
If the distance between the points (4, k), and (1, 0) is 5, then what can be the possible value of k? [CBSE 2017]
Solution:
Let the points x (4, k) and y (1, 0)
It is given that the distance xy is 5 units.
By using the distance formula,
RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS 42

Hope given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry 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.

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS

Other Exercises

Answer each of the following questions either in one word or one sentence or as per requirement of the questions :
Question 1.
What is the ratio of the areas of a circle and an equilateral triangle whose diameter and a side are respectively equal?
Solution:
Diameter of a circle and side of an equilateral triangle are same
Let the diameter of the circle = a
Then radius (r) = \(\frac { a }{ 2 }\)
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 1

Question 2.
If the circumference of two circles are in the ratio 2 : 3, what is the ratio of their areas ?
Solution:
Let R and r be the radii of two circles, then the ratio between their circumferences = 2πR : 2πr
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 2
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 3

Question 3.
Write the area of the sector of a circle whose radius is r and length of the arc is l.
Solution:
Let arc l subtends angle 9 at the centre of the circle
Now radius of a circle = r
and length of arc =l
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 4

Question 4.
What is the length (in terms of π) of the arc that subtends an angle of 36° at the centre of a circle of radius 5 cm?
Solution:
Radius of the circle = 5 cm
Angle at the center = 36°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 5

Question 5.
What is the angle subtended at the centre of a circle of radius 6 cm by an arc of length 3π cm ?
Solution:
Let the arc subtends angle θ at the centre of a circle
Radius of circle (r) = 6 cm
Length of arc = 3π cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 6

Question 6.
What is the area of a sector of a circle of radius 5 cm formed by an arc of length 3.5 cm ?
Solution:
Radius of the circle (r) = 5 cm
Length of arc (l) = 3.5 cm
Let angle 9 be subtended by the arc at the centre
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 7

Question 7.
In a circle of radius 10 cm, an arc subtends an angle of 108° at the centre. What is the area of the sector in terms of π ?
Solution:
Radius of the circle = 10 cm
Angle at the centre = 108°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 8

Question 8.
If a square is inscribed in a circle, what is the ratio of the areas of the circle and the square ?
Solution:
A square ABCD is inscribed in a circle with centre O
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 9
Let the radius of the circle = r
Then its area = πr²
Now diagonal of the square = diameter of the circle = 2r
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 10

Question 9.
Write the formula for the area of a sector of angle θ (in degrees) of a circle of radius r.
Solution:
Area of a sector of a circle whose radius = r
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 11

Question 10.
 Write the formula for the area of a segment in a circle of radius r given that the sector angle is 0 (in degrees).
Solution:
Radius of the circle = r
and angle subtended by the sector at the centre = θ
Area of the segment
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 12

Question 11.
If the adjoining figure is a sector of a circle of radius 10.5 cm, what is the perimeter of the sector ? (Take π= 22/7)
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 13
Solution:
Radius of the circle = 10.5 cm
Angle at the centre of the circle = 60°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 14

Question 12.
If the diameter of a semi-circular protractor is 14 cm then find its perimeter. (C.B.S.E. 2009)
Solution:
Diameter of semicircular protractor = 14 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 15
∴ Radius (r) =  \(\frac { 14 }{ 2 }\) = 7 cm
Now perimeter of protractor
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 16

Question 13.
An arc subtends an angle of 90° at the centre of the circle of radius 14 cm. Write the area of minor sector thus formed in terms of π.
Solution:
AB is an arc of the circle with centre O and radius 14 cm and subtends an angle of 90° at the centre O.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 17

Question 14.
Find the area of the largest triangle that can be inscribed in a semi-circle of radius r units. [CBSE 2015]
Solution:
Radius of semicircle = r
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 18
In semicircle ΔABC is the largest triangle whose base is AC = 2 x r = 2r units
and height OB = r units
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 19

Question 15.
Find the area of a sector of circle of radius 21 cm and central angle 120°.
Solution:
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 20

Question 16.
What is the area of a square inscribed in a circle of diameter p cm?
Solution:
Diameter AC of the circle is p.
Also AC is diagonal of square ABCD.
Each angle of square is of 90°
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 21

Question 17.
Is it true to say that area of a segment of a circle is less than the area of its corresponding sector? Why?
Solution:
False.
It is true only in the case of minor segment. But in case of major segment, area is always greater than the area of sector.

Question 18.
If the numerical value of the area of a circle is equal to the numerical value of its circumference, find its radius.
Solution:
∵ Numerical value of area of circle = Numerical value of circumference
∴  πr² = 2πr
or r = 2 units

Question 19.
How many revolutions a circular wheel of radius r metres makes in covering a distance of s metres?
Solution:
Radius of circular of wheel (r) = r m
Circumference of a circular wheel = 2πr
Distance to be covered = Sm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 22

Question 20.
Find the ratio of the area of the circle circumscribing a square to the area of the circle inscribed in the square.
Solution:
Let each side of of square = x
∴ Diameter of inner circle = x
Radius r = \(\frac { x }{ 2 }\)
Diameter of outer circle = AD
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 23
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS 24

Hope given RD Sharma Class 10 Solutions Chapter 13 Areas Related to 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.