SOLUTION: I keep coming up with a solution that is different from the answer key. Here is the problem: Find the area of an isosceles triangle where the base is 3 and the sides are the square

Algebra ->  Triangles -> SOLUTION: I keep coming up with a solution that is different from the answer key. Here is the problem: Find the area of an isosceles triangle where the base is 3 and the sides are the square      Log On


   

Question 122278: I keep coming up with a solution that is different from the answer key. Here is the problem: Find the area of an isosceles triangle where the base is 3 and the sides are the square root of 11.
Thanks
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Triangle is ABC.
AB side is sqrt%2811%29.
AC side is sqrt%2811%29.
BC side is 3.
Split BC in half at point Z to form two equal line segments BZ=ZC=3/2.
You can now form two right triangles, ABZ and AZC.
Call the length of AZ side, H.
Using the Pythagorean theorem, you can find the height, H, because,
H%5E2%2B%283%2F2%29%5E2=%28sqrt%2811%29%29%5E2
H%5E2%2B%289%2F4%29=11
H%5E2=11-%289%2F4%29
H%5E2=44%2F4-9%2F4
H%5E2=36%2F4
H%5E2=9
H=3
For the area of the original isoceles triangle,
A=%281%2F2%29bH
A=%281%2F2%29%283%2F2%29%283%29
A=9%2F4
A is 2.25 square units.
Does that answer match the answer key?