Difference between revisions of "2017 AMC 8 Problems/Problem 18"
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==Solution== | ==Solution== | ||
− | We first connect point <math>B</math> with point <math>D</math>. We can see that it is a 3-4-5 right triangle. | + | We first connect point <math>B</math> with point <math>D</math>. We can see that it is a 3-4-5 right triangle. We can see that <math>\triangle BDA</math> is a right triangle [I NEED SOMEBODY TO PROVE THIS] , therefore <math>\overline{DA}</math> is <math>13</math>, by the 5-12-13 Pathagorean 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> |
==See Also== | ==See Also== |
Revision as of 13:03, 5 January 2018
Problem 18
In the non-convex quadrilateral shown below,
is a right angle,
,
,
, and
.
What is the area of quadrilateral
?
Solution
We first connect point with point
. We can see that it is a 3-4-5 right triangle. We can see that
is a right triangle [I NEED SOMEBODY TO PROVE THIS] , therefore
is
, by the 5-12-13 Pathagorean 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
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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