Difference between revisions of "2007 AMC 8 Problems/Problem 14"
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The area of a triangle is shown by <math>\frac{1}{2}bh</math>. We set the base equal to <math>24</math>, and the area equal to <math>60</math>, and we get the height, or altitude, of the triangle to be <math>5</math>. In this isosceles triangle, the height bisects the base, so by using the Pythagorean Theorem, <math>a^2+b^2=c^2</math>, we can solve for one of the legs of the triangle (it will be the the hypotenuse, <math>c</math>). | The area of a triangle is shown by <math>\frac{1}{2}bh</math>. We set the base equal to <math>24</math>, and the area equal to <math>60</math>, and we get the height, or altitude, of the triangle to be <math>5</math>. In this isosceles triangle, the height bisects the base, so by using the Pythagorean Theorem, <math>a^2+b^2=c^2</math>, we can solve for one of the legs of the triangle (it will be the the hypotenuse, <math>c</math>). | ||
<math>a = 12</math>, <math>b = 5</math>, | <math>a = 12</math>, <math>b = 5</math>, | ||
| − | <math>c = 13</math> | + | <math>c = 13</math>. |
The answer is <math>\boxed{\textbf{(C)}\ 13}</math> | The answer is <math>\boxed{\textbf{(C)}\ 13}</math> | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/9sVdsKcpJ9U | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=13|num-a=15}} | {{AMC8 box|year=2007|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 12:22, 16 August 2021
Problem
The base of isosceles 
is 
and its area is 
. What is the length of one
of the congruent sides?
Solution
The area of a triangle is shown by 
. We set the base equal to , and the area equal to , and we get the height, or altitude, of the triangle to be 
. In this isosceles triangle, the height bisects the base, so by using the Pythagorean Theorem, 
, we can solve for one of the legs of the triangle (it will be the the hypotenuse, 
).
, 
,
.
The answer is 
Video Solution by WhyMath
~savannahsolver
See Also
| 2007 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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. 
