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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 basketbal = 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
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May 21, 24 08:51 PM
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