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

(Solution 2)
(Solution 2)
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==Solution 2==
 
==Solution 2==
<math>\triangle BCD</math> is a 3-4-5 right triangle. The area of BCD is 6. Then we can use Heron's formula to compute the area of <math>\triangle ABD</math> whose sides have lengths 5,12, and 13. The Area = <math>\sqrt{s(s-5)(s-12)(s-13)}</math>. The s is the semi-perimeter of the triangle, <math>s=(5+12+13)/2=30</math>. So <math>ABCD</math> is <math>30-6 = \boxed{\textbf{(B)}\ 24}.</math>
+
<math>\triangle BCD</math> is a 3-4-5 right triangle. The area of BCD is 6. Then we can use Heron's formula to compute the area of <math>\triangle ABD</math> whose sides have lengths 5,12, and 13. The Area = <math>\sqrt{s(s-5)(s-12)(s-13)}</math> , where s is the semi-perimeter of the triangle, that is <math>s=(5+12+13)/2=30</math>. So <math>ABCD</math> is <math>30-6 = \boxed{\textbf{(B)}\ 24}.</math>
  
 
==See Also==
 
==See Also==

Revision as of 14:41, 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. The area of BCD is 6. Then we can use Heron's formula to compute the area of $\triangle ABD$ whose sides have lengths 5,12, and 13. The Area = $\sqrt{s(s-5)(s-12)(s-13)}$ , where s is the semi-perimeter of the triangle, that is $s=(5+12+13)/2=30$. So $ABCD$ 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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