Simple circle problem

WELL let us take the centre to be O . Let the perpendicular from X on WZ be A and the perpendicular from Y on WZ be B . THEN the upper triangle is AOX and the lower triangle is BOY .

So as per the question $\angle XOZ = 135 ^ \circ.$

Which follows that $\angle XOW = 45 ^ \circ.$ tHIS makes the triangle isoceles .

SO the sides of $AOX, AO=AX.$ Using Pythagorean Theorem We get...

$AO^2+AX^2=XO^2.$ .

$\Rightarrow$ $AO^2+AO^2=(6)^2.$

This is because XY is the diameter and is bisected at the centre O ., also $AX=AO=6.$ $\Rightarrow$ $2 \times AO^2=36.$

THerefore, $AO^2=18.$

$\Rightarrow$ $Area Of AOX=1/2 \times AO \times AO.$

$\Rightarrow$ $Area Of AOX=1/2 \times 18.$

$\Rightarrow$ $Area Of AOX=9.$

$\Rightarrow$ $Area Of Shaded Region=2 \cdot Area Of AOX=18.$

As both the shaded regions are equal.

$Quite ?asy know. :)$

2(x^2)=6^2 x=4.24

Area = (1/2)(x^2) Area = 9 Area of two triangles = 9 x 2 = 18