Difference between revisions of "2007 AMC 8 Problems/Problem 14"
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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> | 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> | ||
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> | 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> | ||
| + | |||
| + | ~megaboy6679 | ||
==Video Solution by WhyMath== | ==Video Solution by WhyMath== | ||
Latest revision as of 22:58, 28 October 2024
Contents
Problem
The base of isosceles 
is 
and its area is 
. What is the length of one
of the congruent sides?
Solution 1
The area of a triangle is shown by 
. We set the base equal to , and the area equal to , and we get the triangle's height, or altitude, 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 hypotenuse, 
).
, 
,
.
The answer is 
Solution 2 (Heron's Formula)
According to Heron's Formula, setting side 
as , we have ![\[\sqrt{s(s-24)(s-b)(s-c)}=60\]](http://latex.artofproblemsolving.com/d/6/4/d64d36b1785366960c41603f5dc5f65ef11b3a09.png)
where 
is the triangle's semiperimeter (i.e. 
). Since the triangle is isosceles, 
, so we can rewrite as 
. Substituting and solving the equation and taking the positive solution for 
, ![\[\sqrt{(12+b)(-12+b)(12)(12)}=60\]](http://latex.artofproblemsolving.com/d/c/6/dc6712389e4a9df732b4da6fe8eea2e044d2976f.png)
![\[\sqrt{144(144-b^2)}=60\]](http://latex.artofproblemsolving.com/d/6/3/d63353540fbddd35a64d0a7bc8527b239a7397a1.png)
![\[144(144-b^2)=3600\]](http://latex.artofproblemsolving.com/c/f/e/cfefacf5a3e3fb613a42836fcb9652ea8b26195f.png)
![\[-b^2=-169\]](http://latex.artofproblemsolving.com/c/a/a/caa081d12325431e64e4eedac47e2e04405bd95e.png)
![\[b=\boxed{\textbf{(C)}\ 13}\]](http://latex.artofproblemsolving.com/7/7/d/77dd75a7d588e0cc5ea54d12012682c0da83e624.png)
~megaboy6679
Video Solution by WhyMath
~savannahsolver
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=omFpSGMWhFc
Video Solution by AliceWang
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. 
