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

Revision as of 13:33, 17 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 1

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

Solution 2

$\triangle BCD$ is a 3-4-5 right triangle. Then we can use Heron's formula to resolve the problem. The area of $\triangle ABD$ whose sides have lengths 5-12-13 is \sqrt {s(s-5)(s-12)(s-13)}. S is the semi-perimeter of the triangle, $s=(5+12+13)/2= 15$ . Then the area is \sqrt {15(15-5)(15-12)(15-13)}= 30 $. So$ 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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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 14: / Line 14: @@
 ==Solution 2==
-<math>\triangle BCD</math> is a 3-4-5 right triangle. Then we can use Heron's formula to resolve the problem. The area of a triangle whose sides have lengths 5, 12, and 13 is \sqrt {s(s-5)(s-12)(s-13)}. S is the semi-perimeter of the triangle, <math>s=(5+12+13)/2= 15 </math>. Then the area is \sqrt {15(15-5)(15-12)(15-13)}= 30<math>. </math>ABCD<math> is </math>30-6 = \boxed{\textbf{(B)}\ 24}.$
+<math>\triangle BCD</math> is a 3-4-5 right triangle. Then we can use Heron's formula to resolve the problem. The area of <math>\triangle ABD</math> whose sides have lengths 5-12-13 is \sqrt {s(s-5)(s-12)(s-13)}. S is the semi-perimeter of the triangle, <math>s=(5+12+13)/2= 15</math>. Then the area is \sqrt {15(15-5)(15-12)(15-13)}= 30<math>. So </math>ABCD<math> is </math>30-6 = \boxed{\textbf{(B)}\ 24}.$
 ==Video Solution==

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