PROBLEMS ON FINDING AREA OF 2D SHAPES

Find the areas of following figures :

Problem 1 :

Solution :

Area of square = a²

Side length of square (a) = 4 cm

Area of square = 4²

= 16 cm²

Problem 2 :

Solution :

Area of rectangle = length x width

Length = 12 cm and width = 4 cm

Area of rectangle = 12 x 4

= 48 cm²

Problem 3 :

Solution :

Area of triangle = (1/2) x base x height 

Base = 10 cm and height = 15 cm

Area of triangle = (1/2) x 10 x 15

= 5 x 15

= 75 cm²

Problem 4 :

Solution :

Area of trapezoid = (1/2) x h (a + b)

h = 8 cm, a = 18 cm and b = 12 cm

Area of trapezoid = (1/2) x 8 x (18 + 12)

= 4 x 30

= 120 cm²

Problem 5 :

Solution :

Area of circle = πr²

radius of the circle = 15 cm

= π(15)²

= 225 π cm²

Problem 6 :

Solution :

Area of trapezoid = (1/2) x h (a + b)

h = 7 mm, a = 8 mm and b = 13 mm

Area of trapezoid = (1/2) x 7 x (8 + 13)

= (1/2) x 7 x 21

= 3.5 x 21

= 73.5 mm²

Problem 7 :

Solution :

Area of circle = πr²

Diameter of the circle = 18 cm

radius of the circle = 18/2 ==> 9 cm

= π(9)²

= 81 π cm²

Problem 8 :

Solution :

Area of rhombus = (1/2) x d₁ x d₂

d₁ = 6 m and d₂ = 7 m

Area of rhombus = (1/2) x 6 x 7

= 3 x 7

= 21 cm²

Problem 9 :

Solution :

Area of rhombus = (1/2) x d₁ x d₂

d₁ = 15 cm and d₂ = 10 cm

Area of rhombus = (1/2) x 15 x 10

= 15 x 5

= 75 cm²

Use Heron’s formula to find the area of the following triangles correct to 2 decimal places.

Problem 10 :

Solution :

Area of scalene triangle = √s(s - a)(s - b)(s - c)

s = (a + b + c)/2

a = 5 cm, b = 12 cm and c = 16 cm

s = (5 + 12 + 16)/2

= 33/2

s = 16.5

Area = √16.5(16.5 - 5)(16.5 - 12)(16.5 - 16)

= √16.5 x 11.5 x 4.5 x 0.5

= √426.9375

= 20.66 cm²

Problem 11 :

Solution :

Area of scalene triangle = √s(s - a)(s - b)(s - c)

s = (a + b + c)/2

a = 8 cm, b = 6 cm and c = 3 cm

s = (8 + 6 + 3)/2

= 17/2

s = 8.5

Area = √8.5(8.5 - 8)(8.5 - 6)(8.5 - 3)

= √8.5 x 0.5 x 2.5 x 5.5

= √58.4375

= 7.64 cm²

Problem 12 :

Solution :

Area of ellipse = π a b

a = 4 mm and b = 9 mm

Area of ellipse = π (4) (9)

= 36 π cm²

Problem 13 :

Solution :

Area of ellipse = π a b

a = 5 mm and b = 12 mm

Area of ellipse = π (5) (12)

= 60 π cm²

Problem 14 :

Solution :

Area of sector = (θ/360) π r²

radius of sector (r) = 12 cm and θ = 30

 = (30/360) π (12)²

= 12 π cm²

Problem 15 :

Solution :

Area of sector = (θ/360) π r²

radius of sector (r) = 6 mm and θ = 345

 = (345/360) π (6)²

= 34.5 π mm²

Problem 16 :

What will be the area of semicircle whose perimeter is 36 cm ?

a) 77 cm²   b) 68 cm²    c) 38 cm²

Solution :

Perimeter of the semicircle = π r + 2r

π r + 2r = 36

3.14r + 2r = 36

5.14r = 36

r = 36/5.14

r = 7

Area of semicircle = (1/2)π r²

= (1/2) x 3.14 x 7²

= 76.93

Approximately 77 cm²

Problem 17 :

If a wire is bent into the shape of square then the area of the square is 81 sq.cm. When the wire is bent into a semicircular shape, then the area of the semicircle will be 

a) 22 cm²   b) 44 cm²    c) 77 cm²     d) 154 cm²

Solution :

Area of square = 81 sq.cm

Let x be the side length of square, then x² = 81

x = 9

Side length of square = 9 cm

Perimeter of square = 4(9)

= 36 cm

Circumference of semicircle = π r + 2r

 π r + 2r = 36

3.14r + 2r = 36

5.14 r = 36

r = 36/5.14

r = 7

Area of semicircle = (1/2)π r²

= (1/2) x 3.14 x 7²

= 76.93

Approximately 77 cm²

So, option c is correct.

Problem 18 :

The circumference and area of a circle are numerically equal, then the diameter is equal to 

a) π/2    b) 2π    c)  2     d) 4

Solution :

Circumference of circle = 2πr

Area of circle = πr²

2πr = πr²

r = 2

Diameter of the circle = 2(r)

= 2(2)

= 4

So, option d is correct.