Unit-6:Mensuration
(a)Perimeter and area related to circles
(b)Surface areas and volumes
Q1.
If two circles touch internally, then the distances between their centers is equal to the difference of their radius.
(a) True (b) False (c) Can't say
Solution.
(a) True
Solution
Q2.
Find the area of a quadrant of a circle whose circumference is 44 cm.
(a) 98 cm2 (b) 38.5 cm2
(c) 40 cm2 (d) 42.5 cm2
Solution.
2 π r = 44
2 x 22/7 x r = 44
r = 7
Area of quadrant = 1/4 π r2
= 1/4 x 22/7 x 49
= 154/4
= 77/2
= 38.5 cm2
Solution
Q3.
If the perimeter of a semi-circular figure is 88 cm, find the diameter of the figure.(take π = 22/7).
(a) 15.89 (b) 14.22 (c) 16 (d) 17.11
Solution.
Let the radius of figure is r cm, perimeter of semi-circle = 1/2(2πr) + 2r
1/2(2πr) + 2r = 88
r(22/7 + 2) = 88
r(22 + 14/7) = 88
r(36/7) = 88
9r/7 = 22
r = 17.11
Solution
Q4.
Two circles touch internally. the sum of their area is 116π cm2 and distance between their centres is 6 cm.
Find the radius of larger circle.
(a) 5 (b) 10 (c) 15 (d) 6
Solution.
Sum of areas = 116 π
π r2 + π R2 = 116π
r2 + R2 = 116
R - r = 6
(R + r)2 + (R - r)2 = 2(R2 + r2)
R + r = 14
Solve -
R + r = 14
R - r = 6
2R = 20
R = 10
Solution
Q5.
In the given figure find the perimeter of shaded portion.
(a) 2r (b) r2
(c) rtanθ + θ/360o x 2 π r (d) r[secθ + tanθ + θ π/180o - 1]
Solution.
AO = OC = r
Perimeter = AB + BD + arc(CA)
tanθ = AB/AO
AB = AOtanθ = rtanθ
secθ = OB/OA
OB = rsecθ
BC = BO - r = rsecθ - r
CA =
θ/360o x 2 π r = θ/180 x π r
= r[secθ + tanθ + θ π/180o - 1]
Solution
Q6.
The area of the largest triangle that can b inscribed in a semi-circle of radius r is .
(a) 2r (b) r2
(c) 4r (d) r3
Solution.
Area of triangle = 1/2 oC x AB
= 1/2 x r x 2r
= r
2
Solution
Q7.
An arc of a circle is of length 10 π cm and the sector it bounds has an area of 40 π cm2. Find the radius of the circle.
(a) 5 cm (b) 40 cm (c) 10 cm (d) 8 cm
Solution.
θ/360o x 2 π r / θ/360o x π r2
=
10 π / 40 π
2 / r
=
1 / 4
r = 8 cm
Solution
Q8.
Find the shaded area ABC.
(a) 2π r2 (b) r2 (c) r2/4 (π - 2)
(d) 4π
Solution.
Area of segment ABC = Area of sector OACBO(a quadrant of circle) - Area of triangle OAB
= θ/360 π r2 - 1/2
x base x height
= 90o/360o πr2 - 1/2 x r x r
= πr2/4 - (1/2)r2
= πr2 /4 - 2r2 /4
= r2 /4(π - 2)
Solution
Q9.
The maximum volume of a cone that can be curved out of a solid hemisphere of radius r, is
(a) πr3/3
(b) πr3 (c) r3/3 (d) πr2/2
Solution.
V = (1/3) π r2h
V = ( 1/3 )π r2 x r
= (1/3) π r3
Solution
Q10.
The volume of two spheres are in the ratio 125:27, the ratio of their surface area is.
(a) 20:25 (b) 9:25
(c) 25:3 (d) 25:9
Solution.
4/3 π r31 / 4/3 π r32
=
125 / 27
r31 / r32
=
125 / 27
r1 / r2
=
5 / 3
Surface area of sphere = 4π r
2
4π r21 / 4π r22
=
r21 / r22
=
(5)2 / (3)2
=
25 / 9
25:9
Solution
