We have a triangle $(\triangle POB)$ within a semicircle. $OP$ and $BP$ are extended to $OA$ and $BQ$. $AP = 5$ and $PQ = 7$. What is the area of the triangle?
It's a problem I stumbled upon on chance. I haven't been able to solve it since I am a little weak in geometry.
The best approach I have tried is connecting $Q$ to the diameter of semi circle and then approaching the two triangles as similar triangles as they both are right angled triangles.
I am still new to the site and can't quite use MathJax to round up the problem. I hope you all will see my mistakes lightly.
Thank You
Best Answer
It is known that if two chords intersect, then product of their segments is the same (see Intersecting chords theorem). In the picture two chords intersect at point $P$, if you continue the vertical one. $PB$ is found from the Pythagorean theorem. Hence
$$5\cdot (2r-5)=7\cdot \sqrt{r^2+(r-5)^2}$$
$$25\cdot(4r^2-20r+25)=49\cdot(2r^2-10r+25)$$
$$2r^2-10r-600=0$$
$$r=-15,\quad r=20$$
Since $2r-5$ is positive, $r=20$. Then the area is $$\frac{r(r-5)}2=\frac{20\cdot15}2=150.$$