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Difference between revisions of "2007 AMC 8 Problems/Problem 14"
(→‎Solution)
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== Solution ==
 
== Solution ==
  
−
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 1 half bh We set the base equal to 24 and the area equal to 60 and we get the height, or altitude, of the triangle to be 5 In this isosceles triangle, the height bisects the base a^2+b^2=c^2 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>,
+
a = 12 b = 5
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<math>c = 13</math>.
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c = 13
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The answer is <math>\boxed{\textbf{(C)}\ 13}</math>
+
The answer is C:13
  
 
==Video Solution by WhyMath==
 
==Video Solution by WhyMath==

Revision as of 18:22, 1 August 2021

Problem

The base of isosceles $\triangle ABC$ is $24$ and its area is $60$. What is the length of one of the congruent sides?

$\mathrm{(A)}\ 5 \qquad \mathrm{(B)}\ 8 \qquad \mathrm{(C)}\ 13 \qquad \mathrm{(D)}\ 14 \qquad \mathrm{(E)}\ 18$

Solution

The area of a triangle is shown by 1 half bh We set the base equal to 24 and the area equal to 60 and we get the height, or altitude, of the triangle to be 5 In this isosceles triangle, the height bisects the base a^2+b^2=c^2 we can solve for one of the legs of the triangle (it will be the the hypotenuse, $c$). a = 12 b = 5 c = 13 The answer is C:13

Video Solution by WhyMath

https://youtu.be/9sVdsKcpJ9U

~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

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