Area of the figure = 806.5 in²
Solution:
Length of the rectangle = 16 in
Breadth of the rectangle = 9 in
Area of the rectangle = length × breadth
= 16 × 9
Area of the rectangle = 144 in²
Base of the triangle = 31 in
Height of the triangle = 20 in
Area of the triangle = [tex]\frac{1}{2}bh[/tex]
[tex]$=\frac{1}{2}\times 31\times 20[/tex]
Area of the triangle = 310 in²
Parallel sides of the trapezium = 16 in and 31 in
Height of the trapezium = 35 – 20 = 15 in
Area of the trapezium = [tex]\frac{1}{2}\times \text{Sum of the parallel sides}\times \text{Height}[/tex]
[tex]$=\frac{1}{2}\times {(16+31)}\times{15}[/tex]
[tex]$=\frac{1}{2}\times {47}\times{15}[/tex]
Area of the trapezium = 352.5 in²
Area of the figure = Area of rectangle + Area of triangle + Area of trapezium
= 144 in² + 310 in² + 352.5 in²
Area of the figure = 806.5 in²
