A swimming po is shaped from two intersecting circles 9m

Step 1:
a) To find the area common to the two circles, we need to calculate the intersection area between the two circles. Let's consider the centers of the circles as points A and B, and the intersection area as the sector formed by the two radii and the enclosed region of the two circles.
The formula to find the area of a sector is given by:
$\text{{Area of sector}\}=\frac{\text{Central angle}}{{360}^{\circ }}\times \pi r^2$
Since the centers of the circles are 9m apart, the distance between points A and B is the diameter of each circle. Therefore, the distance AB is 18m.
To calculate the central angle, we can use the cosine rule. The cosine of an angle in a triangle can be found using the formula:
$\cos \theta =\frac{a^2+b^2-c^2}{2ab}$
Here, a = 9m (radius of each circle), b = 9m (distance between the centers), and c = 18m (distance between the points A and B).
Substituting the values into the formula, we have:
$\cos \theta =\frac{9^2+9^2-{18}^2}{2\times 9\times 9}$
Simplifying the equation gives:
$\cos \theta =\frac{81+81-324}{162}=\frac{-162}{162}=-1$
Since the cosine of an angle cannot be greater than 1 or less than -1, we conclude that the angle between the radii is 180°.
Using the formula for the area of a sector, with the radius (r) as 9m and the central angle as 180°, we get:
$\text{{Area of common region}\}=\frac{{180}^{\circ }}{{360}^{\circ }}\times \pi \times (9\text{{m}\}{)}^2$
Simplifying the equation gives:
$\text{{Area of common region}\}=\frac{1}{2}\times \pi \times 81\text{{m}{\}}^2$
Therefore, the area common to the two circles is $\frac{1}{2}\times \pi \times 81\text{{m}{\}}^2$.
Step 2:
b) The total water surface area of the pool includes the area of both circles and the area of the common region. To find this, we sum the areas of the two circles and subtract the area of the common region:
$\text{{Total water surface area}\}=(\pi \times (9\text{{m}\}{)}^2)+(\pi \times (9\text{{m}\}{)}^2)-\left(\frac{1}{2}\times \pi \times 81\text{{m}{\}}^2\right)$
Simplifying the equation gives:
$\text{{Total water surface area}\}=2\times \pi \times 81\text{{m}{\}}^2-\frac{1}{2}\times \pi \times 81\text{{m}{\}}^2$
Combining like terms, we get:
$\text{{Total water surface area}\}=\frac{3}{2}\times \pi \times 81\text{{m}{\}}^2$
Therefore, the total water surface area of the pool is $\frac{3}{2}\times \pi \times 81\text{{m}{\}}^2$.
Step 3:
c) The perimeter of the pool is the sum of the circumferences of both circles minus the length of the common region between them. The circumference of a circle can be calculated using the formula:
$\text{{Circumference}\}=2\times \pi \times \text{{radius}\}$
Therefore, the perimeter of the pool is given by:
$\text{{Perimeter}\}=(2\times \pi \times 9\text{{m}\})+(2\times \pi \times 9\text{{m}\})-(2\times 9\text{{m}\})$
Simplifying the equation gives:
$\text{{Perimeter}\}=4\times \pi \times 9\text{{m}\}-18\text{{m}\}$
Therefore, the perimeter of the pool is $4\times \pi \times 9\text{{m}\}-18\text{{m}\}$.

a) We must determine the area of the intersection region before we can determine the area shared by the two circles. The following is the formula to determine the area of the intersection of two circles:
$A=r^2{\cos }^{-1}\left(\frac{d^2+r^2-R^2}{2dr}\right)+R^2{\cos }^{-1}\left(\frac{d^2+R^2-r^2}{2dR}\right)-\frac{1}{2}\sqrt{(-d+r+R)(d+r-R)(d-r+R)(d+r+R)}$ where $r$ is the radius of one circle, $R$ is the radius of the other circle, and $d$ is the distance between the centers of the circles.
For this problem, $r=R=9\text{m}$ and $d=9\text{m}$. Plugging in these values into the formula, we get:
$A=9^2{\cos }^{-1}\left(\frac{9^2+9^2-9^2}{2·9·9}\right)+9^2{\cos }^{-1}\left(\frac{9^2+9^2-9^2}{2·9·9}\right)-\frac{1}{2}\sqrt{(-9+9+9)(9+9-9)(9-9+9)(9+9+9)}$
Simplifying the expression, we find:
$A=81{\cos }^{-1}(1)+81{\cos }^{-1}(1)-\frac{1}{2}\sqrt{(9)(18)(18)(27)}$
Using the values ${\cos }^{-1}(1)=0$ and $\sqrt{(9)(18)(18)(27)}=972$, the equation becomes:
$A=81(0)+81(0)-\frac{1}{2}(972)$
Therefore, the area common to the two circles is $A=-486{\text{m}}^2$.
b) The total water surface area is the sum of the areas of the two circles plus the area of the intersection region. We can calculate it by using the formula:
$\text{Total Area}=2(\pi r^2)+A$ where $r$ is the radius of one circle and $A$ is the area common to the two circles.
For this problem, $r=9\text{m}$ and $A=-486{\text{m}}^2$. Plugging in these values, we get:
$\text{Total Area}=2(\pi ·9^2)+(-486)$
Simplifying the expression, we find:
$\text{Total Area}=162\pi -486{\text{m}}^2$
Therefore, the total water surface area is $\text{Total Area}=162\pi -486{\text{m}}^2$.
c) The perimeter of the pool is the sum of the perimeters of the two circles minus the length of the common boundary between them. The formula to calculate the perimeter of a circle is given by:
$\text{Perimeter}=2\pi r$
For this problem, $r=9\text{m}$. Plugging in this value, we get:
$\text{Perimeter}=2\pi ·9$
Simplifying the expression, we find:
$\text{Perimeter}=18\pi \text{m}$
Therefore, the perimeter of the pool is $\text{Perimeter}=18\pi \text{m}$.