Q. 80 - Precalculus Enhanced with Graphing Utilities

Question

An area of $52 fee{t}^2$ is to be enclosed by two squares whose sides are in the ratio of $2:3$. Find the sides of the squares.

Step-by-Step Solution

Answer

The sides of the squares are $6 feet and 4 feet$.

Step 1. Given information.

Assume the sides of the two squares are x,y.

From the question,

$$\begin{gathered} x^2+y^2=52 ...... \left(i\right) \\ \frac{x}{y}=\frac{2}{3} ...... \left(ii\right) \end{gathered}$$

Solve equation (ii) for x,

$x=\frac{2}{3}y$

Substitute the value of x in equation (i),

$$\begin{gathered} {\left(\frac{2}{3}y\right)}^2+y^2=52 \\ \frac{4}{9}{y}^2+y^2=52 \\ \frac{4y^2+9y^2}{9}=52 \end{gathered}$$

Step 2. Find the value of x.

Continuing the above equation,

$$\begin{gathered} \frac{13y^2}{9}=52 \\ {y}^2=\frac{52}{13}\times 9 \\ {y}^2=36 \\ y=\pm 6 \end{gathered}$$

Negative value of the side of the square is not considered as length cannot be negative.

Substitute $y=6$ in equation (ii),

$$\begin{gathered} \frac{x}{6}=\frac{2}{3} \\ x=\frac{2}{3}\times 6 \\ x=4 \end{gathered}$$