2017 AMC 8 Problems/Problem 18

Problem

In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB=12$ , $BC=4$ , $CD=3$ , and $AD=13$ . What is the area of quadrilateral $ABCD$ ?

$[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N);[/asy]$

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

Solution 1

We first connect point $B$ with point $D$ .

$[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); draw((0,0)--(0,5)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N);[/asy]$

We can see that $\triangle BCD$ is a 3-4-5 right triangle. We can also see that $\triangle BDA$ is a right triangle, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of $\triangle BDA$ is $\frac{5\cdot 12}{2}$ , and the area of the smaller 3-4-5 triangle is $\frac{3\cdot 4}{2}$ . Thus, the area of quadrialteral $ABCD$ is $30-6 = \boxed{\textbf{(B)}\ 24}.$

Solution 2

$\triangle BCD$ is a 3-4-5 right triangle. So the area of $\triangle BCD$ is 6. Then we can use Heron's formula to compute the area of $\triangle ABD$ whose sides have lengths 5,12,and 13. The area of $\triangle ABD$ = $\sqrt{s(s-5)(s-12)(s-13)}$ , where s is the semi-perimeter of the triangle, that is $s=(5+12+13)/2=15.$ Thus, the area of $\triangle ABD$ =30, so the area of $ABCD$ is $30-6 = \boxed{\textbf{(B)}\ 24}.$ ---LarryFlora

Video Solution

http://youtube/DU95-maui9U

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/51K3uCzntWs?t=2010

~ pi_is_3.14

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AJHSME/AMC 8 Problems and Solutions

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