# 2005 AMC 8 Problems/Problem 19

## Problem

What is the perimeter of trapezoid $ABCD$?

$[asy]size(3inch, 1.5inch); pair a=(0,0), b=(18,24), c=(68,24), d=(75,0), f=(68,0), e=(18,0); draw(a--b--c--d--cycle); draw(b--e); draw(shift(0,2)*e--shift(2,2)*e--shift(2,0)*e); label("30", (9,12), W); label("50", (43,24), N); label("25", (71.5, 12), E); label("24", (18, 12), E); label("A", a, SW); label("B", b, N); label("C", c, N); label("D", d, SE); label("E", e, S);[/asy]$

$\textbf{(A)}\ 180\qquad\textbf{(B)}\ 188\qquad\textbf{(C)}\ 196\qquad\textbf{(D)}\ 200\qquad\textbf{(E)}\ 204$

## Solution

Draw altitudes from $B$ and $C$ to base $AD$ to create a rectangle and two right triangles. The side opposite $BC$ is equal to $50$. The bases of the right triangles can be found using Pythagorean or special triangles to be $18$ and $7$. Add it together to get $AD=18+50+7=75$. The perimeter is $75+30+50+25=\boxed{\textbf{(A)}\ 180}$.

## Video Solution

https://youtu.be/h0V8qcEo0U8 Soo, DRMS, NM

## Video Solution by OmegaLearn

~ pi_is_3.14

 2005 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 20 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions