Chapter 12.6, Problem 26PPS

To determine

Find the surface area of the fish when it is not puffed up.

Answer to Problem 26PPS

The surface area of the fish is $52.36i{n}^2$ .

Explanation of Solution

Given: It is given in the question that the puffer fish is like the sphere with a diameter of $5inch$ and its surface area when inflated is about $1.5$ times its normal surface area.

Concept Used:

In this, use the concept of surface area of a sphere i.e. $A=4\pi r^2$ .

Calculation:

Here, $d=5inch$ then $r=\frac{5}{2}inch=2.5inch$ .

Now,

  $\begin{array}{l}A=4\pi r^2\\ $ A=4\pi {(2.5)}^2$A=4\pi \times 6.25\\ A=25\pi \\ A\approx 78.54i{n}^2\end{array}$

According to question,

Surface area when puffed up = $1.5$ (normal surface area)

  $\begin{array}{l}78.54=1.5(n)\\ n=\frac{78.54}{1.5}=52.36\end{array}$

Conclusion:

The surface area is $52.36i{n}^2$ .