Difference between revisions of "2017 AMC 8 Problems/Problem 18"
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==Solution 2== | ==Solution 2== | ||
− | <math>\triangle BCD</math> is a 3-4-5 right triangle. Then we can use Heron's formula to resolve the problem. The area of <math>\triangle ABD</math> whose sides have lengths 5-12-13 is { | + | <math>\triangle BCD</math> is a 3-4-5 right triangle. Then we can use Heron's formula to resolve the problem. The area of <math>\triangle ABD</math> whose sides have lengths 5-12-13 is {A={\sqrt {s(s-a)(s-b)(s-c)}},} A = \sqrt{s(s-a)(s-b)(s-c)},. S is the semi-perimeter of the triangle, {s={\frac {a+b+c}{2}}.} s=\frac{a+b+c}{2}=30<math>. So </math>ABCD<math> is </math>30-6 = \boxed{\textbf{(B)}\ 24}.$ |
+ | |||
+ | Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is | ||
+ | |||
+ | {\displaystyle A={\sqrt {s(s-a)(s-b)(s-c)}},} A = \sqrt{s(s-a)(s-b)(s-c)}, | ||
+ | where s is the semi-perimeter of the triangle; that is, | ||
+ | |||
+ | {\displaystyle s={\frac {a+b+c}{2}}.} s=\frac{a+b+c}{2}. | ||
https://youtu.be/6yrwfRMV-5k | https://youtu.be/6yrwfRMV-5k |
Revision as of 14:36, 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. Then we can use Heron's formula to resolve the problem. The area of whose sides have lengths 5-12-13 is {A={\sqrt {s(s-a)(s-b)(s-c)}},} A = \sqrt{s(s-a)(s-b)(s-c)},. S is the semi-perimeter of the triangle, {s={\frac {a+b+c}{2}}.} s=\frac{a+b+c}{2}=30ABCD30-6 = \boxed{\textbf{(B)}\ 24}.$
Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is
{\displaystyle A={\sqrt {s(s-a)(s-b)(s-c)}},} A = \sqrt{s(s-a)(s-b)(s-c)}, where s is the semi-perimeter of the triangle; that is,
{\displaystyle s={\frac {a+b+c}{2}}.} s=\frac{a+b+c}{2}.
https://youtu.be/tJm9KqYG4fU?t=2812
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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