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 14:46, 22 November 2017
Problem 18
In the non-convex quadrilateral 
shown below, 
is a right angle, 
, 
, 
, and 
. ![[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)
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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