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Difference between revisions of "2007 AMC 8 Problems/Problem 14"

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<math>\mathrm{(A)}\ 5 \qquad \mathrm{(B)}\ 8 \qquad \mathrm{(C)}\ 13 \qquad \mathrm{(D)}\ 14 \qquad \mathrm{(E)}\ 18</math>
 
<math>\mathrm{(A)}\ 5 \qquad \mathrm{(B)}\ 8 \qquad \mathrm{(C)}\ 13 \qquad \mathrm{(D)}\ 14 \qquad \mathrm{(E)}\ 18</math>
  
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== Solution ==
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==Solution 1==
  
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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 triangle's height, or altitude, 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 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>
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 +
==Solution 2 (Heron's Formula)==
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 +
According to [[Heron's Formula]], setting side <math>a</math> as <math>24</math>, we have <cmath>\sqrt{s(s-24)(s-b)(s-c)}=60</cmath>
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where <math>s</math> is the triangle's semiperimeter (i.e. <math>\frac{a+b+c}{2}</math>). Since the triangle is isosceles, <math>b=c</math>, so we can rewrite <math>s</math> as <math>\frac{24+2b}{2}=12+b</math>. Substituting and solving the equation and taking the positive solution for <math>b</math>, <cmath>\sqrt{(12+b)(-12+b)(12)(12)}=60</cmath> <cmath>\sqrt{144(144-b^2)}=60</cmath> <cmath>144(144-b^2)=3600</cmath> <cmath>-b^2=-169</cmath> <cmath>b=\boxed{\textbf{(C)}\ 13}</cmath>
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 +
~megaboy6679
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/9sVdsKcpJ9U
 +
 +
~savannahsolver
 +
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=omFpSGMWhFc
 +
 +
==Video Solution by AliceWang==
 +
https://youtu.be/U8v4XVPXr18
  
 
==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 22:58, 28 October 2024

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 1

The area of a triangle is shown by $\frac{1}{2}bh$. We set the base equal to $24$, and the area equal to $60$, and we get the triangle's height, or altitude, to be $5$. In this isosceles triangle, the height bisects the base, so by using the Pythagorean Theorem, $a^2+b^2=c^2$, we can solve for one of the legs of the triangle (it will be the hypotenuse, $c$). $a = 12$, $b = 5$, $c = 13$. The answer is $\boxed{\textbf{(C)}\ 13}$

Solution 2 (Heron's Formula)

According to Heron's Formula, setting side $a$ as $24$, we have \[\sqrt{s(s-24)(s-b)(s-c)}=60\] where $s$ is the triangle's semiperimeter (i.e. $\frac{a+b+c}{2}$). Since the triangle is isosceles, $b=c$, so we can rewrite $s$ as $\frac{24+2b}{2}=12+b$. Substituting and solving the equation and taking the positive solution for $b$, \[\sqrt{(12+b)(-12+b)(12)(12)}=60\] \[\sqrt{144(144-b^2)}=60\] \[144(144-b^2)=3600\] \[-b^2=-169\] \[b=\boxed{\textbf{(C)}\ 13}\]

~megaboy6679

Video Solution by WhyMath

https://youtu.be/9sVdsKcpJ9U

~savannahsolver

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=omFpSGMWhFc

Video Solution by AliceWang

https://youtu.be/U8v4XVPXr18

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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