WORD PROBLEMS ON CONVERSION AS ONE SHAPE TO ANOTHER

Problem 1 :

A steel wire when bent in the form of a square encloses an area of 121 sq. cm. If the same wire is bent into the form of a circle, find the area of the circle.

Solution :

The steel wire is being bent into the shape of square and its area is 121 sq.cm

a² = 121

a = √121

a = 11 cm

The square which is creating the area of 121 sq.cm is bent into the shape of circle.

Perimeter of the square = circumference of the circle

4(11) = 2πr

44 = 2πr

r = (44/2) x (7/22)

r = 7 cm

Area of the circle = πr²

= (22/7) x 7²

= 154 cm²

So, area of the required circle is 154 cm².

Problem 2 :

A wire when bent in the form of an equilateral triangle encloses an area of 121 √3 cm². If the same wire is bent in the form of a circle, find the area of the circle

Solution :

Area of equilateral triangle = 121 √3

(√3/4) a² = 121√3

a² = 121√3 (4/√3)

a² = 121 x 4

a = 11 x 2

a = 22

Perimeter of the equilateral triangle = 3a

= 3(22)

= 66 cm

Perimeter of the equilateral triangle = Circumference of the circle

66 = 2πr

r = 33 x (7/22)

r = 10.5

Area of circle = πr²

= π(10.5)²

= 3.14(110.25)

= 346.185 cm²

So, area of the required circle is 346.185 cm².

Problem 3 :

A wire is bent to form a circle of radius 35 cm. if it is bent in the form of a square, then what will be its area ?

Solution :

Circumference of the circle = Perimeter of the square

2πr = 35

4a = 35

a = 35/4

Area of square = a²

= (35/4)²

= 1225/16

= 76.56 cm²

Problem 4 :

The length of a wire in the form of an equilateral triangle is 44 cm. If it is rebent into the form of a circle, then area of the circle is :

(a) 484 cm²     (b) 176 cm²     (c) 154 cm²     (d) 144 cm²

Solution :

Perimeter of the equilateral triangle = 44

Circumference of circle = 2πr

2πr = 44

r = 44x (1/2) x (7/22)

= 7

Area of the circle = π r²

= 3.14 (7)²

= 153.86

Approximately 154 cm²

Problem 5 :

If a wire is bent into the shape of a square, then the area enclosed by the square is 81 cm². When the same wire is bent into a semi-circular shape, then the area enclosed by the semi circle will be :

Solution :

Area of the square = 81 cm²

a² = 81

a = 9

Area of semicircle = (1/2) π r²

To find the area of semicircle, we need radius of the semicircle.

Perimeter of the square = perimeter of the semicircle

4a = 2π r

4(9) = 2π r

36 / 2 = π r

r = 18 x (7/22)

r = 5.27

Area of semicircle = (1/2) π r²

= (1/2) x (22/7) x 5.27²

= 43.64 cm²

= 44 cm²

Problem 6 :

A wire is in the shape of a circle of radius 21 cm. It is bent to form a square. The side of the square

Solution :

Radius of the circle = 21 cm

Circumference of the circle = Perimeter of the square

2π r = 4a

2 x (22/7) x 21 = 4a

a = (44 x 3)/3

a = 44

So, the required side length of the square is 44 cm.

Problem 7 :

A piece of wire that has been bent in the form of a semicircle including the bounding diameter is straightened and then bent in the form the of a square. The diameter of the semicircle is 14 cm. Which has a larger area, the semi-circle or the square? Also, find the difference between them.

Solution :

Perimeter of the semicircle = perimeter of square

r(π + 2) = 4a

diameter = 14 cm

radius = 7 cm

7 (22/7 + 2) = 4a

4a = 7 x (36/7)

a = 36/4

a = 9

Area of semicircle = (1/2) π r²

= (1/2) x (22/7) x 7²

= 77 cm²

Area of square = a²

= 81 cm²

Difference in area = 81 - 77

= 4 cm²

Square has larger area and its difference is 4 cm².