Let ABC be a rt angle isos triangle
<A = 90 , AD is the perpendicular bisector of BC
O is the centre of the circle
AO²= 2r² AD = r+r√2 = r(1+√2)
<C=45° AD = CD
Area of the triangle ABC = AD*CD = r²(1+√2)²
Ratio of circle to that of triangle ABC = πr²/r²(1+√2)²
IMO E

sanju09 wrote:A circle is inscribed inside a right angled isosceles triangle. What is the ratio of the area of the circle to that of the triangle?
(A) π (3 - 2 √2)
(B) π/(√2 - 1)
(C) π/(2 - √2)^2
(D) π/(√2 - 1)^2
(E) π/(√2 + 1)^2

i got A
Radius of the circle inscribed in the right triangle=(a+b-c)/2
where a, b legs and c-hypotenuse
in out case a=b so sides will be in 1:1:2^1/2 ratio
radius of the circle=(2-2^1/2)
and area
S=pi*(2-2^1/2)^2=pi*(3-2*2^1/2)/2
area of the triangle=1/2*1*1=1/2
so ratio of the area of the circle to that of triangle=pi*(3-2*2^1/2)

Actually both A & E options are same and are correct!!!