Difference between revisions of "2017 AMC 8 Problems/Problem 18"
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<math>\textbf{(A) }12 \qquad \textbf{(B) }24 \qquad \textbf{(C) }26 \qquad \textbf{(D) }30 \qquad \textbf{(E) }36</math> | <math>\textbf{(A) }12 \qquad \textbf{(B) }24 \qquad \textbf{(C) }26 \qquad \textbf{(D) }30 \qquad \textbf{(E) }36</math> | ||
| − | ==Solution== | + | ==Solution 1== |
We first connect point <math>B</math> with point <math>D</math>. | We first connect point <math>B</math> with point <math>D</math>. | ||
| Line 12: | Line 12: | ||
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> | 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> | ||
| + | |||
| + | ==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, s={\frac {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> | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 13:26, 17 June 2021
Problem
In the non-convex quadrilateral 
shown below, 
is a right angle, 
, 
, 
, and 
. What is the area of quadrilateral ?
![[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]](http://latex.artofproblemsolving.com/b/e/3/be32bf473ea4532b8cc3eb844835e13b70bbfba9.png)
Solution 1
We first connect point 
with point 
.
![[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]](http://latex.artofproblemsolving.com/d/6/6/d66f241a2e30d6733c30bc1e3c4236af99f628ce.png)
We can see that 
is a 3-4-5 right triangle. We can also see that 
is a right triangle, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of is 
, and the area of the smaller 3-4-5 triangle is 
. Thus, the area of quadrialteral is 
Solution 2
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, s={\frac {5+12+13}{2}}= 15 
. is
Video Solution
https://youtu.be/tJm9KqYG4fU?t=2812
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. 
