Area of a trapezium inscribed in a circle?

Question

A circle, having center at $(2, 3)$ and radius $6$, crosses $y$-axis at the points $P$ and $Q$. The straight line with equation $x= 1$ intersects the radii $CP$ and $CQ$ at points $R$ and $S$ respectively. Find the area of the trapezium $PQSR$.

I am getting stuck to find the length of parallel sides? As area of a trapezium is $A=\frac{a+b}{2} \cdot h$

here $a$ = $SR$ , b = $PQ $

How to find $h$?

Answer 1

Using the Pythagorean theorem you can find $PQ$. $$PQ = 2\cdot\sqrt{r^2 - (2h)^2},$$ where $r$ is the radius of the circle, and $h$ is the altitude of the trapezoid $PQSR$. $$RS = \frac{PQ}{2},$$ because the triangles $PCQ$ and $RCS$ are similar and $PC = 2RC.$ So you can plug in the values and find the area of the trapezoid by using $A = h\cdot\frac{PQ + RS}{2}$. $$r = 6, h =\frac{HC}{2} = 1, PH = 4\sqrt 2, PQ = 8\sqrt 2, RS = 4\sqrt 2,$$ $$A = 6\sqrt 2$$

Answer 2

$h = 1$ since the distance between the $y$-axis ($x=0$) and $x=1$ is $1$. The tricky part is finding $a$ and $b$.

Answer 3

Hint:

The equation of the circle is $$(x-2)^2 +(y-3)^2 =36$$ To find $P$ and $Q$, set $x=0$ and solve the quadratic in $y$. Now, find the equations of lines $CP$ and $CQ$, and finally set $x=1$ in these equations to find $R$ and $S$. You have everything you need now.