# Give a composite shape of quarter of circle and rectangle with a total area of 570 " square feet" and and the diagonal angle of the rectangle equal to 18.43^0, calculate the radius?

Sep 1, 2016

$\text{Radius"~~22.92 "ft} .$

#### Explanation:

Let $r$ be the radius. Then, using the Figure given,

$I H = r \cos \theta , I A = r \sin \theta , \text{where, } \theta = {18.43}^{\circ}$.

$\therefore \text{ Area of the Rectangle } I H B A = I H \cdot I A = {r}^{2} \sin \theta \cos \theta$.

$= {r}^{2} / 2 \cdot \sin 2 \theta = {r}^{2} / 2 \cdot \frac{2 \tan \theta}{1 + {\tan}^{2} \theta}$

$= \frac{{r}^{2} \tan \theta}{1 + {\tan}^{2} \theta}$

For $\theta = {18.43}^{\circ} , \tan \theta = \frac{1}{3}$

$\text{Hence, Area of the Rectangle} = \frac{{r}^{2} / 3}{1 + \frac{1}{9}} = 3 {r}^{2} / 10$

$\text{Also, Area of the Quarter of the Circle} = \frac{1}{4} \cdot \pi {r}^{2}$

$\therefore \text{Total Area of the Composite Shape} = 3 {r}^{2} / 10 + \pi {r}^{2} / 4 = {r}^{2} / 20 \left(6 + 5 \pi\right)$, which is, $570 \text{ sq.ft.}$

$\therefore {r}^{2} / 20 \left(6 + 5 \pi\right) = 570$

$\therefore {r}^{2} = \frac{570 \cdot 20}{6 + 5 \pi} = \frac{11400}{6 + 15.7} = \frac{11400}{21.7} \approx 525.3456$

$\therefore r \approx 22.92 \text{ft.}$