# 10 Swimming pool in the shape of two intersecting circles

**Situation**

A swimming pool is shaped from two intersecting circles 9 m in radius with their centers 9 m apart.

Part 1: What is the area common to the two circles?

A. 85.2 m^{2}

B. 63.7 m^{2}

C. 128.7 m^{2}

D. 99.5 m^{2}

Part 2: What is the total water surface area?

A. 409.4 m^{2}

B. 524.3 m^{2}

C. 387.3 m^{2}

D. 427.5 m^{2}

Part 3: What is the perimeter of the pool, in meters?

A. 63.5 m

B. 75.4 m

C. 82.4 m

D. 96.3 m

**Solution**

$\theta = 120^\circ$

$\beta = 360^\circ - \alpha = 360^\circ - 120^\circ$

$\beta = 360^\circ - 120^\circ$

$\beta = 240^\circ$

$A_1 = \frac{1}{2}r^2 (\theta_{rad} - \sin \theta_{deg})$

$A_1 = \frac{1}{2}(9^2) \left[ 120^\circ \left( \dfrac{\pi}{180^\circ} \right) - \sin 120^\circ \right]$

$A_1 = 49.749 \, \text{ m}^2$

$\text{Common area} = 2A_1 = 2(49.749)$

$\text{Common area} = 99.498 \, \text{ m}^2$ Part 1: [ D ]

$\text{Area of water surface} = 2(\pi r^2 – A_1)$

$\text{Area of water surface} = 2[ \, \pi(9^2) – 49.749 \, ]$

$\text{Area of water surface} = 409.44 \, \text{ m}^2$ Part 2: [ A ]

$\text{Perimeter} = 2 \times \dfrac{\pi r \beta_{deg}}{180^\circ} = 2 \times \dfrac{\pi(9)(240^\circ)}{180^\circ}$

$\text{Perimeter} = 75.398 \, \text{ m}$ Part 3: [ B ]