2014 AMC 12B Problems/Problem 9

Problem

Convex quadrilateral $ABCD$ has $AB=3$ , $BC=4$ , $CD=13$ , $AD=12$ , and $\angle ABC=90^{\circ}$ , as shown. What is the area of the quadrilateral?

$[asy] pair A=(0,0), B=(-3,0), C=(-3,-4), D=(48/5,-36/5); draw(A--B--C--D--A); label("$A$",A,N); label("$B$",B,NW); label("$C$",C,SW); label("$D$",D,E); draw(rightanglemark(A,B,C,25)); [/asy]$

$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 58.5$

Solution

Note that by the pythagorean theorem, $AC=5$ . Also note that $\angle CAD$ is a right angle because $\triangle CAD$ is a right triangle. The area of the quadrilateral is the sum of the areas of $\triangle ABC$ and $\triangle CAD$ which is equal to $\frac{3\times4}{2} + \frac{5\times12}{2} = 6 + 30 = \boxed{\textbf{(B)}\ 36}$

Video Solution

https://youtu.be/4_x1sgcQCp4?t=923

~ pi_is_3.14

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AMC 12 Problems and Solutions

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2014 AMC 12B Problems/Problem 9

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Problem

Solution

Video Solution

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