Find the area of the triangle $\triangle ABC$ embedded in a circle

Question

Find the area $ABC$ of the triangle.

Are these similar triangles? Can I just use a ratio of the sides to find the area?

I solved for all sides of the large triangle and got $5,7,\sqrt(74)$. Then used similar triangles to get the 2 missing sides of the smaller triangle. My final area was $3.87 $

Answer 1

We know that: $$\angle{DAE}=\arctan(5/7)\rightarrow \sin(\angle{DAE}=5/\sqrt{74}$$ Now, because $ABC$ is a right triangle, we have: $$A=\frac{1}{2}AB\cdot AC \sin(\angle{DAE})=\frac{1}{2}AB^2\sin(\angle{DAE})\cos(\angle{DAE})$$ Substituing, we have: $$A=\frac{1}{2}\cdot4^2\cdot\frac{5}{\sqrt{74}}\frac{7}{\sqrt{74}}=\frac{140}{37}$$

Answer 2

Yes, they are similar triangles.

Thus, $$\frac{S_{\Delta ABC}}{S_{\Delta ADE}}=\left(\frac{AB}{AE}\right)^2,$$ which gives $$S_{\Delta ABC}=\frac{7\cdot 5}{2}\cdot\left(\frac{4}{\sqrt{7^2+5^2}}\right)^2=\frac{140}{37}=3.78....$$