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

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==Solution==
 
==Solution==
We can see a Pythagorean triple's two  longer lengths: 12, 13. So BD should be 5. This is certainly the case because <math>3^2 + 4^2 = 5^2</math>, which is <math>BD</math>. Thus the area of triangle ABD is <math>30</math>. So <math>30 - 6 = 24</math>, or <math>\text{B)}</math> <math>24</math>.
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We can see a Pythagorean triple's two  longer lengths: <math>12</math>, <math>13</math>. So <math>BD</math> should be <math>5</math>. This is certainly the case because <math>3^2 + 4^2 = 5^2</math>, which is <math>BD</math>. Thus the area of quadrialteral ABCD is <math>30-6 = \boxed{\textbf{(B)}\ 24}.</math>
  
 
==See Also==
 
==See Also==

Revision as of 15:46, 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 quadrialteral ABCD is $30-6 = \boxed{\textbf{(B)}\ 24}.$

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
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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