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> | ||
− | == Solution == | + | ==Solution 1== |
− | The area of a triangle is shown by <math>\frac{1}{2}bh</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>c = 13</math>. | ||
+ | The answer is <math>\boxed{\textbf{(C)}\ 13}</math> | ||
− | + | ==Solution 2 (Heron's Formula)== | |
− | + | 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> | ||
− | + | ~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
Contents
[hide]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 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 ,
~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.