2017 AMC 8 Problems/Problem 18
Contents
Problem
In the non-convex quadrilateral shown below,
is a right angle,
,
,
, and
. What is the area of quadrilateral
?
Solution 1
We first connect point with point
.
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 {\displaystyle 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, {\displaystyle s={\frac {a+b+c}{2}}.} s=\frac{a+b+c}{2}=30
ABCD
30-6 = \boxed{\textbf{(B)}\ 24}.$
https://youtu.be/tJm9KqYG4fU?t=2812
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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