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 \[\frac{3\times4}{2} + \frac{5\times12}{2} = 6 + 30 = \boxed{\textbf{(B)}\ 36}\]

See also

2014 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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 AMC 12 Problems and Solutions

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