Difference between revisions of "2017 AMC 8 Problems/Problem 18"
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<math>\triangle BCD</math> is a 3-4-5 right triangle. So the area of <math>\triangle BCD</math> is 6. Then we can use Heron's formula to compute the area of <math>\triangle ABD</math> whose sides have lengths 5,12,and 13. The area of <math>\triangle ABD</math> = <math>\sqrt{s(s-5)(s-12)(s-13)}</math> , where s is the semi-perimeter of the triangle, that is <math>s=(5+12+13)/2=15.</math> Thus, the area of <math>\triangle ABD</math> is 30, so the area of <math>ABCD</math> is <math>30-6 = \boxed{\textbf{(B)}\ 24}.</math> ---LarryFlora | <math>\triangle BCD</math> is a 3-4-5 right triangle. So the area of <math>\triangle BCD</math> is 6. Then we can use Heron's formula to compute the area of <math>\triangle ABD</math> whose sides have lengths 5,12,and 13. The area of <math>\triangle ABD</math> = <math>\sqrt{s(s-5)(s-12)(s-13)}</math> , where s is the semi-perimeter of the triangle, that is <math>s=(5+12+13)/2=15.</math> Thus, the area of <math>\triangle ABD</math> is 30, so the area of <math>ABCD</math> is <math>30-6 = \boxed{\textbf{(B)}\ 24}.</math> ---LarryFlora | ||
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| − | + | https://youtu.be/aH4GrC5Nfwk?si=tFdEHOoFbkp4q7tD | |
| − | + | ~ savannahsolver | |
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==See Also== | ==See Also== | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
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| + | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 16:41, 8 June 2025
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 is 
. Thus, the area of quadrilateral is 
~CHECKMATE2021
Solution 2
is a 3-4-5 right triangle. So the area of is 6. Then we can use Heron's formula to compute the area of 
whose sides have lengths 5,12,and 13. The area of = 
, where s is the semi-perimeter of the triangle, that is 
Thus, the area of is 30, so the area of is ---LarryFlora
Video Solution by Pi Academy
https://youtu.be/aH4GrC5Nfwk?si=tFdEHOoFbkp4q7tD
~ savannahsolver
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
