Difference between revisions of "2017 AMC 8 Problems/Problem 18"
Larryflora (talk | contribs) (→Solution 2) |
Larryflora (talk | contribs) (→Solution 2) |
||
(7 intermediate revisions by the same user not shown) | |||
Line 14: | Line 14: | ||
==Solution 2== | ==Solution 2== | ||
− | <math>\triangle BCD</math> is a 3-4-5 right triangle. Then we can use Heron's formula to | + | <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> =30, so the area of <math>ABCD</math> is <math>30-6 = \boxed{\textbf{(B)}\ 24}.</math> ---LarryFlora |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
==See Also== | ==See Also== |
Latest revision as of 14:48, 17 June 2021
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. 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 =30, so the area of is ---LarryFlora
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.