Q1.
If two circles touch externally, then the distance between their centres is equal to the sum of their radii.
(a) True (b) False (c) Can't say
Solution.
(a) True
Solution
Q2.
Find the area of two quadrants whose circumference is 88 cm.
(a) 390 cm2 (b) 300 cm2
(c) 308 cm2 (d) 380 cm2
Solution.
2πr = 88
(2 x 22 x r)/r = 88
r = 14 cm
Area of two quadrants = 1/2 π r2
1/2 x 22/7 x 14 x 14
= 22 x 14
= 308 cm2
Solution
Q3.
A wheel has diameter 84 cm. find how many complete revolutions must it take to cover 396 meters.
(a) 250 (b) 100
(c) 50 (d) 150
Solution.
circumference of the wheel = 2πr = 2.64 m
total numbers of wheel = 396/2.64 = 150
Solution
Q4.
The side of a square is 20 cm. Find the area of circum scribed circle .
(a) 400 cm2 (b) 498 cm2
(c) 500 cm2 (d) 628 cm2
Solution.
Diameter of the circumscribed circle = Diagonal of the square
AC = 2r = a√2(a = side of square)
2r = 20√2
r = 10√2
= π r
2 = 3.14 x 100 x 2
= 3.14 x 200
= 6.28 x 100
= 628 cm
2
Solution
Q5.
In a given circle, are lenth of a sector is 20 cm and its radius is 6 cm. find the area of this sector ,
(a) 53cm2
(b) 60cm2 (c) 6cm2 (d) 100cm2
Solution.
A = 1/2 lr
A = 1/2 x 20 x 6 = 60cm2
Solution
Q6.
Angle described by minute hand in 15 minutes
(a) 30o (b) 90o (c) 60o (d) 180o
Solution.
Angle described by minute hand in one minute = (360o/60o) = 6o
So, angle described
in 15 minutes = 15 x 6o = 90o
Solution
Q7.
Find the area of shaded portion
(a) 1/2 r2(tanθ - θ π/180)
(b) 1/2 r2(sinθ - θ π/180) (c) r2 (d) 2r
Solution.
Area of segment ABC = Area of triangle OAB - Area of sector OAC
Area of triangle OAB = 1/2 OA x AB
= 1/2 x r x rtanθ
= 1/2 r2 tanθ
Area of sector = θ/360 π r2
= 1/2 r2 . tanθ - θ/360 π r2
1/2 r2(tanθ - θ π/180)
Solution
Q8.
If the area of a sector of a circle is 7/12 of the area of circle, then the angle subtended to sector on center is
(a) 90o (b) 210o (c) 180o (d) 20o
Solution.
(θ/360 π r2)/π r2 = 7/12
θ/360o = 7/12
θ/30o = 7
θ = 210o
Solution
Q9.
Three are two figures a semi- sphere and a both has same circumference and same surface area. find the ratio of its radius and height.
(a) √3 : 4 (b) 1 : √3 (c) 2 : √3 (d) √3 : 3
Solution.
Both solids have same radius (r)
As per question πrl = 2π r2
πr √r2 +
h2 = 2π r2
π2 r2 (r2 + h2 ) = 4π2 r4
r2 + h2 = 4r2
h2 = 3r2
r/h = 1/√3
r : h = 1 : √3
Solution
Q10.
Two cones have same height and radius are in ratio 9 : 8 , find the ratio of volume of both cones.
(a) 16:15 (b) 9:8
(c) 6:19 (d) 81:64
Solution.
v1/v2 = (1/3 π r12h)/(1/3 π r22h)
r12/ r22 = 81/64
v1:v2 = 81:64
Solution
