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2017 AMC 8 Problems/Problem 18: Difference between revisions

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==Solution 2==
==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 <math>\triangle ABD</math> whose sides have lengths 5-12-13 is {A={\sqrt {s(s-a)(s-b)(s-c)}},} A = \sqrt{s(s-a)(s-b)(s-c)},. S is the semi-perimeter of the triangle, {s={\frac {a+b+c}{2}}.} s=\frac{a+b+c}{2}=30<math>. So </math>ABCD<math> is </math>30-6 = \boxed{\textbf{(B)}\ 24}.$
<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 5-12-13. Area = \sqrt{s(s-5)(s-12)(s-13)}. 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>
 
Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is
 
{\displaystyle A={\sqrt {s(s-a)(s-b)(s-c)}},} A = \sqrt{s(s-a)(s-b)(s-c)},
where s is the semi-perimeter of the triangle; that is,
 
{\displaystyle s={\frac {a+b+c}{2}}.} s=\frac{a+b+c}{2}.
 
https://youtu.be/6yrwfRMV-5k
 
https://youtu.be/tJm9KqYG4fU?t=2812


==See Also==
==See Also==

Revision as of 13:39, 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 5-12-13. Area = \sqrt{s(s-5)(s-12)(s-13)}. The s is the semi-perimeter of the triangle, $s=(5+12+13)/2=30$. So $ABCD$ is $30-6 = \boxed{\textbf{(B)}\ 24}.$

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

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