Area of a square inscribed in a circle ABCD is a square inscribed in a circle whose diameter is L cm.
If P and Q are mid points of BC and CD, respectively, find the shaded area MDCNT
Thanks

I tried this
If I knew the M value I could solve

 A: The shaded region is the union of two triangles that overlap, so the area is the sum of areas of CND and DQM, minus the area of DQT. All three of these triangles lie on segment CD, which we will take as base. These three triangles have base lengths of $L/\sqrt{2}$, $L/2\sqrt 2$, and $L/2\sqrt 2$, respectively, so we just need their heights (horizontal width, in the image).
By similar triangles, area of DQM is the same as the are of CPN, so the sum of areas of CND and DQM is the area of CPD, which is $\frac{1}{2}L/\sqrt{2} \times L/2\sqrt 2 = L^2/8$.
Notice that DQT is a right triangle since ADQ is 90 degrees rotated from DCP, to which it is congruent. The ratio of length TQ to TD is $\frac{1}{2}$ (from P being the midpoint of BC). Let $\theta = \angle TDQ$, then the area is $\frac{1}{2}QD^2 \sin\theta \cos\theta$ with $\tan\theta = \frac{1}{2}$. This gives $\sin\theta\cos\theta = \frac{2}{5}$, so area of DQT is $\frac{1}{5}L^2/8$
Thus the area of the shaded part is $\frac{4}{5}L^2/8 = L^2/10$.
