Difference between revisions of "2007 AMC 8 Problems/Problem 14"
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=13|num-a=15}} | {{AMC8 box|year=2007|num-b=13|num-a=15}} |
Revision as of 13:20, 9 December 2012
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
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 |