Difference between revisions of "2017 AMC 8 Problems/Problem 18"

Revision as of 19:51, 14 June 2021

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

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 $30-6 = \boxed{\textbf{(B)}\ 24}.$

Video Solution

https://youtu.be/6yrwfRMV-5k

https://youtu.be/tJm9KqYG4fU?t=2812

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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Difference between revisions of "2017 AMC 8 Problems/Problem 18"

Revision as of 19:51, 14 June 2021

Contents

Problem

Solution

Video Solution

See Also

@@ Line 14: / Line 14: @@
 ==Video Solution==
+https://youtu.be/6yrwfRMV-5k
 https://youtu.be/tJm9KqYG4fU?t=2812