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2017 AMC 8 Problems/Problem 18: Difference between revisions

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==Problem 18==
==Problem 18==
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>. [asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label("<math>B</math>", (0, 0), SW); label("<math>A</math>", (12, 0), ESE); label("<math>C</math>", (2.4, 3.6), SE); label("<math>D</math>", (0, 5), N);[/asy] 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>. <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> What is the area of quadrilateral <math>ABCD</math>?


<math>\textbf{(A) }12\qquad\textbf{(B) }24\qquad\textbf{(C) }26\qquad\textbf{(D) }30\qquad\textbf{(E) }36</math>  
<math>\textbf{(A) }12\qquad\textbf{(B) }24\qquad\textbf{(C) }26\qquad\textbf{(D) }30\qquad\textbf{(E) }36</math>


==Solution==
==Solution==

Revision as of 14:29, 22 November 2017

Problem 18

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

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

Solution

We can see a Pythagorean triple's two longer lengths: 12, 13. So BD should be 5. This is certainly the case because $3^2 + 4^2 = 5^2$, which is $BD$. Thus the area of triangle ABD is $30$. So $30 - 6 = 24$, or $B$.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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