Chapter 15, Problem 38CR

a.

To determine

To find: the diameter of the pipe that Peter must put in to the pool.

Answer to Problem 38CR

  $RS=16$ .

Explanation of Solution

Given:

The diameter of both the pipes are $12\text{ and }16$ .

Calculation:

According to the question, add the area of both the pipes. Then,

  $\begin{array}{l}A=\pi r_1^2+\pi r_2^2\\ A=\pi \left(r_1^2+r_2^2\right)\\ A=\pi \left({\left(\frac{12}{2}\right)}^2+{\left(\frac{16}{2}\right)}^2\right)\\ A=\pi \left(6^2+8^2\right)\\ A=\pi \left(36+64\right)\\ A=100\pi \end{array}$

  $\begin{array}{l}A=100\pi \\ \pi r^2=100\pi \{\because A=\pi r^2\}\\ r^2=100\\ r=\sqrt{100}=10\end{array}$

Thus, the diameter of the pipe is $2\times 10=20$ .

Hence, the diameter of the pipe is $20$ .

b.

To determine

To find: the general relationship exists between both the three diameters.

Answer to Problem 38CR

Pythagoras theorem.

Explanation of Solution

Given:

The diameter of both the pipes are $12\text{ and }16$ .

Calculation:

According to the question, to find the general relationship exists between both the three diameters.

From part a,

The three diameters are $12,16,20$

The ratio of the diameters is $3,4,5$

Therefore, by Pythagoras theorem, the general relationship exists between both the three diameters.