Chapter 11.6, Problem 88AYU

To determine

To find: The length of the side of the square and the radius of the circle.

Answer to Problem 88AYU

Solution:

With a 60 ft long wire it is not possible to make square and a circle with a total area of 100 sq ft.

Explanation of Solution

Given:

• Total length of the wire is 60 feet.

• The wire is cut and one piece bend to a square and another is bend to a circle.

• The total area enclosed by the two pieces is 100 sq feet.

Formula Used:

$\text{Area} \text{of} \text{a} \text{circle} A = \pi r^2$

$\text{Perimeter} \text{of} \text{a} \text{circle} = 2\pi r$

$\text{Area} \text{of} \text{a} \text{square} = side.side$

$\text{Perimeter} \text{of} \text{a} \text{square} = 4.side$

Calculation:

Let,

$l$ be the side of the square

$r$ be the radius of the circle

The perimeters and area can be written as the following equations

$4l + 2\pi r = 60$   -----Eq(1)

$l^2 + \pi r^2 = 100$   -----Eq(2)

Substituting $l = \frac{60 - 2\pi r}{4}$ in Eq(2)

${\left(\frac{60 - 2\pi r}{4}\right)}^2 + \pi r^2 = 100$

rewritting the above equation as quadratic equation of $r$

$\left({\pi }^2 + 4\pi \right)r^2 - 60\pi r + 500 = 0$

The solution of the quadratic equation is $r = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}$

Real solutions exists if $b^2 - 4ac > 0$

Here $a = {\pi }^2 + 4\pi , b = - 60\pi$ and $c = 500$.

$\begin{array}{lll}{b}^2-4ac\hfill & =\hfill & {(-60\pi )}^2-4({\pi }^2+4\pi )(500)\hfill \\ \hfill & =\hfill & -9341.4\hfill \end{array}$

Since, $b^2 - 4ac < 0$ real solution does not exist. Hence it is not possible to make square and a circle with a total area of 100 sq ft, with a wire of 60 ft length.