Question

A piece of wire of length 52 m is cut into two parts. Each part is then bent to form a square. It is found that the combined area of two squares is 109 m^2. Find the measures of the sides of the two squares.

Answer 1

Answer on Question #64032 – Math – Geometry

Question

A piece of wire of length $52\,\mathrm{m}$ is cut into two parts. Each part is then bent to form a square. It is found that the combined area of two squares is $109\,\mathrm{m}^2$. Find the measures of the sides of the two squares.

Solution

Let the side of the first square be $x$, and $y$ be the side of the second square.

The combined area of two squares is

$$x^2 + y^2 = 10. \quad (1)$$

On the other hand, the perimeter of the first square is $4x$, and the perimeter of the second square is $4y$.

The sum of these perimeters is equal to the length of the wire:

$$4x + 4y = 52. \quad (2)$$

It follows from equation (2) that $y = \frac{52 - 4x}{4}$, that is, $y = 13 - x$ and substitute it into equation (1):

$$x^2 + (13 - x)^2 = 109;$$ $$x^2 + 169 - 26x + x^2 = 109;$$ $$2x^2 - 26x + 60 = 0;$$ $$x^2 - 13x + 30 = 0;$$ $$x = \frac{13 - \sqrt{(-13)^2 - 4 \cdot 1 \cdot 30}}{2} \quad \text{or} \quad x = \frac{13 + \sqrt{(-13)^2 - 4 \cdot 1 \cdot 30}}{2};$$ $$x = \frac{13 - 7}{2} \quad \text{or} \quad x = \frac{13 + 7}{2}.$$ $$x = 3 \quad \text{or} \quad x = 10.$$

If $x = 3$, then $y = 10$, and vice versa. The measures of the sides of the two squares are $3\,\mathrm{m}$ and $10\,\mathrm{m}$.

Answer:

$3\,\mathrm{m}$ and $10\,\mathrm{m}$

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