2017 AMC 8 Problems/Problem 18

Problem 18

In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB=12$ , $BC=4$ , $CD=3$ , and $AD=13$ . $[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N);[/asy]$ What is the area of quadrilateral $ABCD$ ?

$\textbf{(A) }12\qquad\textbf{(B) }24\qquad\textbf{(C) }26\qquad\textbf{(D) }30\qquad\textbf{(E) }36$

Solution

We can see a Pythagorean triple's two longer lengths: 12, 13. So BD should be 5. This is certainly the case because $3^2 + 4^2 = 5^2$ , which is $BD$ . Thus the area of triangle ABD is $30$ . So $30 - 6 = 24$ , or $\text{B)}$ $24$ .

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2017 AMC 8 Problems/Problem 18

Problem 18

Solution

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