2017 AMC 8 Problems/Problem 18
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 
whose sides have lengths 5-12-13 is \sqrt {s(s-5)(s-12)(s-13)}. S is the semi-perimeter of the triangle, 
. Then the area is \sqrt {15(15-5)(15-12)(15-13)}= 30
ABCD
30-6 = \boxed{\textbf{(B)}\ 24}.$
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. 
