1. ## Circles area proving

The diagram shows a rectangle abcd inscribed in a circle. Given $\displaystyle AB=4x$cm and $\displaystyle BC=8$ cm, show that the area, $\displaystyle P cm^2$ of the shaded region is given by $\displaystyle P = 4(\pi x^2+4 \pi -8x)$

2. [quote=Punch;527303]The diagram shows a rectangle abcd inscribed in a circle. Given $\displaystyle AB=4x$cm and $\displaystyle BC=8$ cm, show that the area, $\displaystyle P cm^2$ of the shaded region is given by $\displaystyle P = 4(\pi x^2+4 \pi -8x)$

You can work out the diameter using the Pythagorean Theorem.

$\displaystyle (4x)^2 + 8^2 = D^2$

$\displaystyle 16x^2 + 64 = D^2$

$\displaystyle D = \sqrt{16x^2 + 64}$.

Therefore the radius $\displaystyle r = \frac{\sqrt{16x^2 + 64}}{2}$.

So the area of the circle is

$\displaystyle A_{\textrm{Circle}} = \pi r^2$

$\displaystyle = \pi \left(\frac{\sqrt{16x^2 + 64}}{2}\right)^2$

$\displaystyle = \pi \left(\frac{16x^2 + 64}{4}\right)$

$\displaystyle = \pi (4x^2 + 16)$

$\displaystyle 4\pi x^2 + 16\pi$.

Now if we work out the area of the rectangle,

$\displaystyle A_{\textrm{Rectangle}} = 4x \cdot 8$

$\displaystyle = 32x$.

So the area of the shaded region is

$\displaystyle A_{\textrm{Shaded}} = A_{\textrm{Circle}} - A_{\textrm{Rectangle}}$

$\displaystyle = 4\pi x^2 + 16\pi - 32x$

$\displaystyle = 4(\pi x^2 + 4\pi - 8x)$.