FIND AREA OF A ELLIPSE

Find the area of each figure :

Problem 1 :

Solution :

a = 6 m and b = 3 m

Area of an ellipse = πab

= 22/7 × 6 × 3

Area = 56.52 m²

Problem 2 :

Solution :

a = 5 m and b = 7 m

Area of an ellipse = πab

= 22/7 × 5 × 7

Area = 110 m²

Problem 3 :

Solution :

2a = 6 m ==> a = 3 m

2b = 4 m ==> b = 2 m

Area of an ellipse = πab

= 22/7 × 3 × 2

Area = 18.84 m²

Problem 4 :

Solution :

2a = 10 cm ==> a = 5 cm

2b = 7 cm ==> b = 3.5 cm

Area of an ellipse = πab

= 22/7 × 5 × 3.5

Area = 54.95 cm²

Problem 5 :

Solution :

Given, a = 5 cm

b = 4 cm

Area of half ellipse = 1/2 × πab

= 1/2 × 22/7 × 5 × 4

Area = 31.4 cm²

Problem 6 :

Solution :

Given, a = 6.2 m

b = 4.1 m

Area of quarter ellipse = 1/4 × πab

= 1/2 × 22/7 × 6.2 × 4.1

Area = 20.0 m²

Problem 7 :

The circle and the ellipse alongside have the same area. Find the radius of the circle.

Solution :

2a = 18 m ==> a = 9 m

2b = 8 m ==> b = 4 m

Area of an ellipse = πab

= 22/7 × 9 × 4

Area = 113 m²

Area of circle = πr²

113 = 22/7 × r²

r² = (113×7)/22

r² = 35.95

r = 6 m

So, radius of circle is 6 m.

Problem 8 :

A semicircular pond has a radius of 5 m, what is the area of the pond to the nearest m² ?

Solution :

Radius = 5 m

Area of semicircle = (1/2) π r²

= (1/2) x π x 5²

= (25/2)π

= 12.5 π m²

Problem 9 :

A semi circular mirror has a radius of 24 cm. What is the area of the mirror ?

Solution :

Radius = 24 cm

Area of semicircle = (1/2) π r²

= (1/2) x π x 24²

= (144/2)π

= 72π m²

Problem 10 :

You throw a dart at the board shown. Your dart is equally likely to hit any point inside the square board. Are you more likely to get 10 points or 0 points?

Solution :

Side length of board = 6(3)

= 18 inches

Area of board = 18²

= 324

Area of smallest circle which gain 10 points = π r²

Radius = 3 inches

= π (3)²

= 9π

The probability of getting 10 points is = area of smallest circle / area of board

= 9π/324

= π/36

= 0.08722

Area of largest circle :

Radius of largest circle = 3(3)

= 9 inches

Area of circle = π r²

= π (9)²

= 81π

The probability of getting 0 points is

= (area of board - area of largest circle) / area of board

= (324 - 81π)/324

= 69.66/324

= 0.215

You are more likely to get 0 points.

Problem 11 :

Find the areas of the sectors formed by ∠UTV.

Area of sector creating the angle measure 70 degree = (θ/360)π r²

θ = 70

= (70/360) x 3.14 x 8²

= 39.07

Approximately 39 square inches.

Area of sector USV = π r² - area of small sector

= 3.14 (8)² - 39.07

= 3.14(64) - 39.07

= 200.96 - 39.07

= 161.89

Approximately 162 square inches.

Problem 12 :

A rectangular wall has an entrance cut into it. You want to paint the wall. To the nearest square foot, what is the area of the region you need to paint ?

Solution :

Area of wall = area of rectangle - (area of semicircle + area of square)

Area of rectangle = length x width

area of square = side x side

Area of semicircle = (1/2) π r²

= 36 x 26 - [(1/2) x 3.14 x 8² + 16 x 16]

= 936 - [100.48 + 256]

= 936 - 356.48

= 579.52

Approximately 580 square feet.

So, area of the wall to be painted is 580 square feet.

Problem 13 :

Find the indicated measure when the diameter of a circle with an area of 676π square centimeters

Solution :

Area of circle = 676π square centimeters

π r² = 676 π

r² = 676

r = √676

= √(26 x 26)

= 26 cm

Diameter = 2(26)

= 52 cm

So, the required diameter is 52 cm.

Problem 13 :

Find the area of the figure given below,

Solution :

Area of the figure = area of triangle + area of semicircle

= (1/2) x base x height + (1/2) π r²

base = 7 m and height = 7 m

radius = 7/2 ==> 3.5 m

= (1/2) x 7 x 7 + (1/2) π (3.5)²

= 49/2 + 0.5 x 3.14 x 12.25

= 24.5 + 19.23

= 43.73

Approximately 44 square meter.