Problem 1 :
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) 8800cm2
Solution :
Radius = 56 cm
Circumference of circle = 2πr
= 2 × 22/7 × 56
Circumference of circle = 352 cm
Perimeter of a square = 4a
352 = 4a
a = 352/4
a = 88 cm
Area of square = a2
= 88 cm × 88 cm
Area of square = 7744 cm2
So, option c) is correct.
Problem 2 :
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 semi-circular 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 square = 81 cm2
Area of square = a2
81 = a2
a = 9 cm
Perimeter of square = 4a
= 4(9)
= 36 cm
Perimeter of semi-circular = πr + 2r
36 = r(π + 2)
36 = r(22/7 + 2)
36 = r(36/7)
r = 7 cm
area of the semi circular = 1/2(πr2)
= 1/2(22/7 × 72)
= 1/2(22/7 × 49)
= 77 cm2
So, option c) is correct.
Problem 3 :
The circumference of a circle is 100cm. The side of a square inscribed inthe circle is
(a) 50√2 cm (b) 100/π cm (c) 50√2/π cm (d) 100√2/π cm
Solution :
circumference of a circle = 100cm.
2πr = 100
r = 100/2π
r = 50/π cm
Diagonal of square = Diameter of circle
√2 a = 2r
√2 a = 2(50/π)
√2 a = 100/π
a = 100/π × 1/√2
a = 100/π × (1/√2 × √2/√2)
a = 100/π × √2/2
a = 50√2/π
So, option c) is correct.
Problem 4 :
If an arc of a circle forms 90º at the centre of the circle, then the ratio of its length to the circumference of the circle is
(a) 1 : 4 (b) 3 : 4 (c) 1 : 3 (d) 2 : 3
Solution :
θ = 90º
Length of the arc = θ/360º × 2πr
= 90º/360º × 2πr
= 1/4 × 2πr
Circumference of a circle = 2πr
Ratio of its length to the circumference of the circle
= Length of the arc/circumference of a circle
= 1/4 × 2πr/2πr
= 1/4
So, ratio of its length to the circumference of the circle is 1 : 4.
So, option a) is correct.
Problem 5 :
The area of a circle is 220cm2. The area of square inscribed in it is
(a) 49 cm2 (b) 70 cm2 (c) 140 cm2 (d) 150 cm2
Solution :
area of a circle = 220cm2
area of a circle = πr2
220 = πr2
220 = 22/7 × r2
220 × 7/22 = r2
70 = r2
r = √70
Diagonal of square = Diameter of circle
√2 a = 2r
√2 a = 2(√70)
a = 2(√70/√2)
a = 2(√35)
area of square = a2
= 22(√35)2
= 140
So, option c) is correct.
Problem 6 :
The area of the circle that can be inscribed in a square of side 6 cm is
(a) 36 π cm2 (b) 18 π cm2 (c) 12 π cm2 (d) 9 π cm2
Solution :
Sides of a square = 6 cm
Radius of the circle = 6/2
r = 3
Area of the circle = πr2
= π(3)2
= 9π
So, option d) is correct.
Problem 7 :
The area of the square that can be inscribed in a circle of radius 8 cm is
(a) 256 cm2 (b) 128 cm2 (c) 64√2 cm2 (d) 64 cm2
Solution :
radius of circle = 8 cm
diameter of the circle = AC
= 2 × OC
= 2 × 8
= 16 cm
Let 'a' be the side of square.
Using Pythagorean theorem.
AC2 = AB2 + BC2
(16)2 = a2 + a2
256 = 2a2
a2 = 256/2
a2 = 128
So, area of square is 128 cm2.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM