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 = $\frac{1}{2}bh$

                                 $=\frac{1}{2}\times 31\times 20$

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 = $\frac{1}{2}\times \text{Sum of the parallel sides}\times \text{Height}$

                                     $=\frac{1}{2}\times {(16+31)}\times{15}$

                                     $=\frac{1}{2}\times {47}\times{15}$

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