Question

Plz plz help me so i can get a good grade

Answer 1

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 =
`(1)/(2)bh`

`$=(1)/(2)* 31* 20`

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 =
`(1)/(2)* \text{Sum of the parallel sides}* \text{Height}`

`$=(1)/(2)* {(16+31)}*{15}`

`$=(1)/(2)* {47}*{15}`

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