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

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<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>  
 
<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>  
  
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We can see that <math>\triangle BCD</math> is a 3-4-5 right triangle. We can also see that <math>\triangle BDA</math> is a right triangle, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of <math>\triangle BDA</math> is <math>\frac{5\cdot 12}{2}</math>, and the area of the smaller 3-4-5 triangle is <math>\frac{3\cdot 4}{2}</math>. Thus, the area of quadrialteral <math>ABCD</math> is <math>30-6 = \boxed{\textbf{(B)}\ 24}.</math>
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We can see that <math>\triangle BCD</math> is a 3-4-5 right triangle. We can also see that <math>\triangle BDA</math> is a right triangle, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of <math>\triangle BDA</math> is <math>\frac{5\cdot 12}{2}</math>, and the area of <math>\triangle BCD</math> is <math>\frac{3\cdot 4}{2}</math>. Thus, the area of quadrilateral <math>ABCD</math> is <math>30-6 = \boxed{\textbf{(B)}\ 24}.</math>
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~CHECKMATE2021
  
 
==Solution 2==
 
==Solution 2==
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<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>
+
<math>\triangle BCD</math> is a 3-4-5 right triangle. So the area of <math>\triangle BCD</math> 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 of <math>\triangle ABD</math> = <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=15.</math> Thus, the area of <math>\triangle ABD</math> is 30, so the area of <math>ABCD</math> is <math>30-6 = \boxed{\textbf{(B)}\ 24}.</math> ---LarryFlora
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== Video Solution by Pi Academy ==
 +
 
 +
https://youtu.be/aH4GrC5Nfwk?si=tFdEHOoFbkp4q7tD
 +
 
 +
~ savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 15:02, 27 December 2024

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 $\triangle BCD$ is $\frac{3\cdot 4}{2}$. Thus, the area of quadrilateral $ABCD$ is $30-6 = \boxed{\textbf{(B)}\ 24}.$

~CHECKMATE2021

Solution 2

$\triangle BCD$ is a 3-4-5 right triangle. So the area of $\triangle BCD$ 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 of $\triangle ABD$ = $\sqrt{s(s-5)(s-12)(s-13)}$ , where s is the semi-perimeter of the triangle, that is $s=(5+12+13)/2=15.$ Thus, the area of $\triangle ABD$ is 30, so the area of $ABCD$ is $30-6 = \boxed{\textbf{(B)}\ 24}.$ ---LarryFlora

Video Solution by Pi Academy

https://youtu.be/aH4GrC5Nfwk?si=tFdEHOoFbkp4q7tD

~ savannahsolver

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

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

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