PROBLEMS ON FINDING AREA OF CIRCLE

Find the area of :

Problem 1 :

Solution :

Area of a circle = πr²

= 3.14 × (7)²

= 3.14 × 49

= 153.86

So, the area is 153.86 cm².

Problem 2 :

Solution :

r = d/2

= 18/2

= 9

Area of a circle = πr²

= 3.14 × (9)²

= 3.14 × 81

= 254.34

So, the area is 254.34 cm².

Problem 3 :

Solution :

r = d/2

= 10.2/2

= 5.1

Area of a semicircle = πr²/2

= (3.14 × (5.1)²)/2

= 40.85

So, the area is 40.85 cm²

Problem 4 :

A circular rug is laid on a tiled floor. Find :

a) the area of the rug

b) the visible area of the tiled floor.

Solution :

a) Area of the rug = πr²

r = d/2

= 2.4/2

r = 1.2

= 3.14 × (1.2)²

= 3.14 × 1.44

= 4.5216

So, area of the rug is 4.5216 m²

b) The visible area = l × w

= 5.2 × 3.5

= 18.2

So, the visible area is 18.2 m²

Problem 5 :

A circular table top has a diameter of 1.6 m. A rectangular tablecloth 2 m by 2 m is placed over the table top. What area of the tablecloth overlaps the table?

Solution :

Diameter of a circular table top = 1.6 m

A rectangular tablecloth = 2 m by 2 m

r = d/2

= 1.6/2

r = 0.8

area of a circular table = πr²

= 22/7 × (0.8)²

= 3.14 × 0.64

= 2

Area of a circular table is 2 m².

A rectangular tablecloth = 2 × 2

= 4

Area of rectangular cloth is 4 m².

So, area of the tablecloth overlaps the table = 4 – 2

= 2 m²

Problem 6 :

The diagram shows plans for a garden which is 40 m by 40 m. it consists of 4 quarter circles of lawn with a flower bed in the middle as shown.

Find :

a) The perimeter of the garden

b) The total area of lawn

c) The total area of the garden

d) The area of the flower bed

e) The length of edging around the flower bed. 

Solution :

a) The perimeter of the garden = 4 × 40

= 160 m

b) Area of lawn = 4 area of quarter circle

Area of lawn = 4(πr²/4) ==> πr²

r = D/2

= 40/2

r = 20

= π × (20)²

= 400π m

= 400(3.14)

Approximately 1257 m²

c) Area of the garden a = a²

= 40 × 40

= 1600 m

d) Area of the flower bed = area of garden - area of lawn

= 1600 - 1257

= 343 m²

So, area of the flower bed is 343 m²

e) Length of edge

= 4(Perimeter of quarter (circle curve only))

= 4 (2πr/4)

= 2πr

= 2 (3.14) (20)

= 125.6

Approximately 126 m.

Problem 7 :

You are placing non-slip tape along the perimeter of the bottom of a semicircle-shaped doormat. How much tape will you save applying the tape to the perimeter of the inside semicircle of the doormat? Justify your answer.

Solution :

Diameter of semicircle = 30 inches

Radius = 30/2

R = 15 inches

Radius of semicircular not related tape = 15 - 3

r = 12 inches

Area of the tape applied = (1/2)πR² - (1/2)πr²

= (1/2)π (R² - r²)

= (1/2)π (15² - 12²)

= (1/2)π (225 - 144)

= (1/2)π (81)

= 40.5π square inches

Problem 8 :

You need to carry a circular cake through a 32-inch wide doorway without tilting it. The circumference of the cake is 100 inches. Will the cake fit through the doorway? Explain.

Solution :

Circumference of cake = 100 inches

2πr = 100

2 x 3.14 x r = 100

r = 100/(2 x 3.14)

= 15.92

Approximately 16 inches

Diameter of the cake = 32 inches

So, the cake will fit through the doorway.

Problem 9 :

Estimate the radius of the Big Ben clock face in London.

Solution :

2πr = 44

2 x 3.14 x r = 44

r = 44/(2 x 3.14)

= 7

So, the required radius is 7 m.

Problem 10 :

A desktop is shaped like a semicircle with a diameter of 28 inches. What is the area of the desktop?

Solution :

Diameter = 28 inches

Radius = 28/2 ==> 14 inches

Area of desktop = πr²

= 3.14 x 14²

= 3.14 x 196

= 615.44 square inches

Problem 11 :

An ecologist is studying an algal bloom that has formed on the entire surface of a circular pond. What is the area of the surface of the pond covered by the algal bloom?

Solution :

Diameter = 28 inches

Radius = 28/2 ==> 14 inches

Area of pond = πr²

= 3.14 x 14²

= 3.14 x 196

= 615.44 square meter

Approximately 615 square meter 

So, area of the pond is 615 square meter.