Question

PLEASE HELP ME!!!!

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

try to answer it with this formula
1/2 (b1+b2)(h)
that's the formula to find the area of a trapizoid I don't have a calculator sorry

Answer 2

`\bf \stackrel{\textit{area of a trapezoid}}{A=\cfrac{h(a+b)}{2}}~~ \begin{cases} h=height\\ a,b=bases \end{cases}\qquad therefore\qquad A=\cfrac{AB(AD+BC)}{2}`

so, let's find how long each of those segments are anyway.

`\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{-1}~,~\stackrel{y_1}{5})\qquad B(\stackrel{x_2}{3}~,~\stackrel{y_2}{2})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ AB=√([3-(-1)]^2+[2-5]^2)\implies AB=√((3+1)^2+(2-5)^2) \\\\\\ AB=√(16+9)\implies \boxed{AB=5}`

`\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_2}{3}~,~\stackrel{y_2}{2})\qquad C(\stackrel{x_1}{0}~,~\stackrel{y_1}{-2})\qquad \qquad \\\\\\ BC=√((0-3)^2+(-2-2)^2)\implies BC=√(9+16)\implies \boxed{BC=5}`

`\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ D(\stackrel{x_2}{-13}~,~\stackrel{y_2}{-11})\qquad A(\stackrel{x_1}{-1}~,~\stackrel{y_1}{5}) \\\\\\ AD=√([-1-(-13)]^2+[5-(-11)]^2)\implies AD=√((-1+13)^2+(5+11)^2) \\\\\\ AD=√(144+256)\implies AD=√(400)\implies \boxed{AD=20}`

`\bf \stackrel{\textit{area of a trapezoid}}{A=\cfrac{h(a+b)}{2}}~~ \begin{cases} h=height\\ a,b=bases \end{cases}\qquad therefore\qquad A=\cfrac{AB(AD+BC)}{2} \\\\\\ A=\cfrac{5(20+5)}{2}\implies A=\cfrac{5(25)}{2}\implies A=\cfrac{125}{2}\implies \blacktriangleright A=62.5 \blacktriangleleft`