SOLUTION: A circle O is tangent to the hypotenuse BC of isosceles right ABC. AB and AC are extended and are tangent to circle O at E and F, respectively. The area of the triangle is X^2. Fin

Algebra ->  Circles -> SOLUTION: A circle O is tangent to the hypotenuse BC of isosceles right ABC. AB and AC are extended and are tangent to circle O at E and F, respectively. The area of the triangle is X^2. Fin      Log On


   

Question 911784: A circle O is tangent to the hypotenuse BC of isosceles right ABC. AB and AC are extended and are tangent to circle O at E and F, respectively. The area of the triangle is X^2. Find the area of the circle. Answer and solution please.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The area of the triangle is A%5Bt%5D=X%5E2
Since it's an isosceles right triangle,
b=h
%281%2F2%29bh=X%5E2
%281%2F2%29b%5E2=X%5E2
b%5E2=2X%5E2
b=sqrt%282%29X
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Calculate Y.
Since the bisector cuts the triangle in half, the area is also half.
Y is one leg.
Since it's still an isosceles right triangle Y is the other leg.
%281%2F2%29Y%2AY=%281%2F2%29X%5E2
Y=X
So then, use the Pythagorean theorem,
%28R%2BY%29%5E2=R%5E2%2BR%5E2
%28R%2BX%29%5E2=2R%5E2
R%2BX=sqrt%282%29R
R-sqrt%282%29R=-X
R%281-sqrt%282%29%29=-X
R=-X%2F%281-sqrt%282%29%29
R=X%2F%28sqrt%282%29-1%29
So then the area becomes,
A%5Bc%5D=pi%2AR%5E2
A%5Bc%5D=%28pi%2A%28X%29%5E2%29%2F%28sqrt%282%29-1%29%5E2
A%5Bc%5D=%28pi%2AA%5Bt%5D%29%2F%28sqrt%282%29-1%29%5E2