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Difference between revisions of "2017 AMC 8 Problems/Problem 18"
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==Problem 18==
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==Problem==
 
In the non-convex quadrilateral <math>ABCD</math> shown below, <math>\angle BCD</math> is a right angle, <math>AB=12</math>, <math>BC=4</math>, <math>CD=3</math>, and <math>AD=13</math>. What is the area of quadrilateral <math>ABCD</math>?
 
In the non-convex quadrilateral <math>ABCD</math> shown below, <math>\angle BCD</math> is a right angle, <math>AB=12</math>, <math>BC=4</math>, <math>CD=3</math>, and <math>AD=13</math>. What is the area of quadrilateral <math>ABCD</math>?
  

Revision as of 13:08, 18 January 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/tJm9KqYG4fU?t=2812

See Also

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

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