Calculating the perimeter of triangle inside of a circle

Question

In triangle $DCB$, $BC = 10$ and is also the diameter. If the area of triangle $DCB = 11$, then determine the perimeter of the triangle.

I am a little stuck on this problem. I tried using the sine rule with angle $D$ seeing that it equals 90. Can someone assist me or give me a hint please?

Answer 1

Hint:

Let $x ,y $ be other two sides then we have $$\frac{xy}{2}=11$$ $${xy}=22$$ and $$x^2+y^2=100$$

$$(x+y)^2=(x^2+y^2)+2xy=100+44 \implies x+y=12$$

$$\text{ Perimeter }=x+y+10=22$$

Answer 2

HINT:

So, we can write $BD=10\cos y,CD=10\sin y$

We have $\dfrac12 BD\cdot CD=11$

We need to find $BD+CD+BC$