Question

From a piece of tin in the shape of a square 6 inches on a side, the largest possible circle is cut out. What is the ratio of the area of the circle to the original square?

Answer 1

Answer:

$\sf \dfrac{1}{4} \pi \quad or \quad \dfrac{7}{9}$

Step-by-step explanation:

The width of a square is its side length.

The width of a circle is its diameter.

Therefore, the largest possible circle that can be cut out from a square is a circle whose diameter is equal in length to the side length of the square.

Formulas

$\sf \textsf{Area of a square}=s^2 \quad \textsf{(where s is the side length)}$

$\sf \textsf{Area of a circle}=\pi r^2 \quad \textsf{(where r is the radius)}$

$\sf \textsf{Radius of a circle}=\dfrac{1}{2}d \quad \textsf{(where d is the diameter)}$

If the diameter is equal to the side length of the square, then:
$\implies \sf r=\dfrac{1}{2}s$

Therefore:

$\begin{aligned}\implies \sf Area\:of\:circle & = \sf \pi \left(\dfrac{s}{2}\right)^2\\& = \sf \pi \left(\dfrac{s^2}{4}\right)\\& = \sf \dfrac{1}{4}\pi s^2 \end{aligned}$

So the ratio of the area of the circle to the original square is:

$\begin{aligned}\textsf{area of circle} & :\textsf{area of square}\\\sf \dfrac{1}{4}\pi s^2 & : \sf s^2\\\sf \dfrac{1}{4}\pi & : 1\end{aligned}$

Given:

• side length (s) = 6 in
• radius (r) = 6 ÷ 2 = 3 in

$\implies \sf \textsf{Area of square}=6^2=36\:in^2$

$\implies \sf \textsf{Area of circle}=\pi \cdot 3^2=28\:in^2\:\:(nearest\:whole\:number)$

Ratio of circle to square:

$\implies \dfrac{28}{36}=\dfrac{7}{9}$