AREA OF COMPOUND SHAPES RECTANGLES AND TRIANGLES

Two or more basic shapes put together is known as compound shapes.

Find the area of the following compound shapes (not drawn to scale)

The dashed lines are perpendicular.

Problem 1 :

Solution :

By observing the figure,

Area of a rectangle A = l × w

Length of a rectangle (l) = 4 cm

Width of a rectangle (w) = 9 cm

Area of a rectangle A = 4 × 9

= 36 cm²

Area of a triangle A = 1/2 (b × h)

Base of a triangle (b) = 5 cm

Height of a triangle (h) = 6 cm

= 1/2 (5 × 6)

= 1/2 (30)

= 15 cm²

Area of Compound Shapes = Area of a rectangle + Area of a triangle  

= 36 + 15

= 51 cm²

Problem 2 :

Solution :

By observing the figure,

Area of a rectangle A = l × w

Length of a rectangle (l) = 3 cm

Width of a rectangle (w) = 6 cm

Area of a rectangle A = 3 × 6

= 18 cm²

Area of a triangle A = 1/2 (b × h)

Base of a triangle (b) = 3 cm

Height of a triangle (h) = 6 cm

= 1/2 (3 × 6)

= 1/2 (18)

= 9 cm²

Area of Compound Shapes = Area of a rectangle + Area of a triangle  

= 18 + 9

= 27 cm²

Problem 3 :

Solution :

By observing the figure,

Area of a rectangle A = l × w

Length of a rectangle (l) = 8 cm

Width of a rectangle (w) = 6 cm

Area of a rectangle A = 8 × 6

= 48 cm²

Area of a triangle A = 1/2 (b × h)

Base of a triangle (b) = 1 cm

Height of a triangle (h) = 6 cm

= 1/2 (1 × 6)

= 1/2 (6)

= 3 cm²

Area of a triangle A = 1/2 (b × h)

Base of a triangle (b) = 8 cm

Height of a triangle (h) = 3 cm

= 1/2 (8 × 3)

= 1/2 (24)

= 12 cm²

Area of Compound Shapes = Area of a rectangle + Area of a triangle + Area of a triangle  

= 48+ 3 + 12

= 63 cm²

Problem 4 :

Solution :

By observing the figure,

Area of a rectangle A = l × w

Length of a rectangle (l) = 4 cm

Width of a rectangle (w) = 4 cm

Area of a rectangle A = 4 × 4

= 16 cm²

Area of a triangle A = 1/2 (b × h)

Base of a triangle (b) = 4 cm

Height of a triangle (h) = 4 cm

= 1/2 (4 × 4)

= 1/2 (16)

= 8 cm²

Area of Compound Shapes = Area of a rectangle + Area of a triangle  

= 16 + 8

= 24 cm²

Problem 5 :

Solution :

By observing the figure,

Area of a rectangle A = l × w

Length of a rectangle (l) = 10 cm

Width of a rectangle (w) = 4 cm

Area of a rectangle A = 10 × 4

= 40 cm²

Area of a triangle A = 1/2 (b × h)

Base of a triangle (b) = 10 cm

Height of a triangle (h) = 3 cm

= 1/2 (10 × 3)

= 1/2 (30)

= 15 cm²

Area of a triangle A = 1/2 (b × h)

Base of a triangle (b) = 10 cm

Height of a triangle (h) = 4 cm

= 1/2 (10 × 4)

= 1/2 (40)

= 20 cm²

Area of Compound Shapes = Area of a rectangle + Area of a triangle + Area of a triangle  

= 40 + 15 + 20

= 75 cm²

Problem 6 :

Solution :

By observing the figure,

Area of a rectangle A = l × w

Length of a rectangle (l) = 10 cm

Width of a rectangle (w) = 4 cm

Area of a rectangle A = 10 × 4

= 40 cm²

Area of a triangle A = 1/2 (b × h)

Base of a triangle (b) = 7 cm

Height of a triangle (h) = 6 cm

= 1/2 (7 × 6)

= 1/2 (42)

= 21 cm²

Area of Compound Shapes = Area of a rectangle + Area of a triangle  

= 40 + 21

= 61 cm²

Problem 7 :

Find the area of the portion of the basketball court shown.

Solution :

The half of the circle is merged in the rectangle and finding out half of the circle is enough.

Area of portion of the basketball = area of rectangle + (1/2) πr²

Length = 19 ft, width = 12 ft and diameter of circle = 12 ft, radius = 6 ft

= 12 x 19 + (1/2) x 3.14 x 6²

= 228 + 1.57 x 36

= 228 + 56.52

= 284.52 square ft.

Problem 8 :

Solution :

Area of compound shape = area of triangle + area of rectangle

= (1/2) x base x height + length x width

= (1/2) x 9 x 6 + 7 x 9

= 27 + 63

= 100 square meter.

Problem 9 :

Solution :

Area of compound shape = area of square + 4 (area of semicircle)

= side x side + 4 x  (1/2) πr²

= 2 x 2 + 4 x (1/2) x 3.14 x 1²

= 4 + 6.28

= 10.58 square ft.

Problem 10 :

The figure is made up of a square and a rectangle. Find the area of the shaded region.

Solution :

base of the triangle to the left = 3 m, since it is the shape of square, height of the triangle = 7 m.

Base of the triangle to the right = 16 - 7 ==> 9 m

height = 3 m

Area of shaded region = 1/2 x 3 x 7 + 1/2 x 9 x 3

= 21/2 + 27/2

= 10.5 + 13.5

= 24 square meter