Difference between revisions of "2017 AMC 8 Problems/Problem 18"
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==Solution== | ==Solution== | ||
− | We can see a Pythagorean triple's two longer lengths: 12, 13. So BD should be 5. This is certainly the case because <math>3^2 + 4^2 = 5^2</math>, which is <math>BD</math>. Thus the area of | + | We can see a Pythagorean triple's two longer lengths: <math>12</math>, <math>13</math>. So <math>BD</math> should be <math>5</math>. This is certainly the case because <math>3^2 + 4^2 = 5^2</math>, which is <math>BD</math>. Thus the area of quadrialteral ABCD is <math>30-6 = \boxed{\textbf{(B)}\ 24}.</math> |
==See Also== | ==See Also== |
Revision as of 15:46, 22 November 2017
Problem 18
In the non-convex quadrilateral shown below, is a right angle, , , , and . What is the area of quadrilateral ?
Solution
We can see a Pythagorean triple's two longer lengths: , . So should be . This is certainly the case because , which is . Thus the area of quadrialteral ABCD 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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