Question

What is the area of trapezoid ABCD?

Enter your answer as a decimal or whole number in the box. Do not round at any steps.

Answer 1

Check the picture below.

let's firstly find the values for a, b and h.

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{-3}~,~\stackrel{y_1}{2})\qquad D(\stackrel{x_2}{0}~,~\stackrel{y_2}{-2})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ AD=√((~~0 - (-3)~~)^2 + (~~-2 - 2~~)^2) \implies AD=√((0 +3)^2 + (-2 -2)^2) \\\\\\ AD=√(( 3 )^2 + ( -4 )^2) \implies AD=√( 9 + 16 ) \implies \boxed{AD=5=a} \\\\[-0.35em] ~\dotfill`

`B(\stackrel{x_1}{1}~,~\stackrel{y_1}{5})\qquad C(\stackrel{x_2}{7}~,~\stackrel{y_2}{-3})\qquad \qquad \\\\\\ BC=√((~~7 - 1~~)^2 + (~~-3 - 5~~)^2) \implies BC=√((7 -1)^2 + (-3 -5)^2) \\\\\\ BC=√(( 6 )^2 + ( -8 )^2) \implies BC=√( 36 + 64 ) \implies \boxed{BC=10=b} \\\\[-0.35em] ~\dotfill`

`A(\stackrel{x_1}{-3}~,~\stackrel{y_1}{2})\qquad B(\stackrel{x_2}{1}~,~\stackrel{y_2}{5})\qquad \qquad \\\\\\ AB=√((~~1 - (-3)~~)^2 + (~~5 - 2~~)^2) \implies AB=√((1 +3)^2 + (5 -2)^2) \\\\\\ AB=√(( 4 )^2 + ( 3 )^2) \implies AB=√( 16 + 9 ) \implies \boxed{AB=5=h} \\\\[-0.35em] ~\dotfill`

`\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h~~=height\\ a,b=\stackrel{parallel~sides}{bases~\hfill }\\[-0.5em] \hrulefill\\ a=5\\ b=10\\ h=5 \end{cases}\implies A=\cfrac{5(5+10)}{2}\implies {\Large \begin{array}{llll} A=37.5 \end{array}}`