2007 AMC 8 Problems/Problem 14: Difference between revisions
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== Solution == | == Solution == | ||
The area of a triangle is shown by 1 | 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>). | ||
a = 12 b = 5 | <math>a = 12</math>, <math>b = 5</math>, | ||
c = 13 | <math>c = 13</math>. | ||
The answer is C | The answer is <math>\boxed{\textbf{(C)}\ 13}</math> | ||
==Video Solution by WhyMath== | ==Video Solution by WhyMath== | ||
Revision as of 11:22, 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
