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 $75+30+50+25=\boxed{\textbf{(A)}\ 180}$.

See Also

2005 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AJHSME/AMC 8 Problems and Solutions

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