Squares and Square Roots | FIO

Question 17

Construct a square whose area is equal to the difference of the areas of squares of sidelengths 5 units and 7 units.

Solution

Understand the Question

The area of a square of side length $s$ is given by $\text{Area} = s^2$ .
The difference between the areas of squares with side lengths $7\text{ units}$ and $5\text{ units}$ is $7^2 - 5^2 = 49 - 25 = 24\text{ square units}$ .
The required square must have an area of $24\text{ square units}$ , which corresponds to a side length of $\sqrt{24}\text{ units}$ .
We can geometrically construct the side length $\sqrt{24}$ using the Pythagoras theorem in a right-angled triangle where $\text{Hypotenuse} = 5\text{ units}$ and one leg $= 1\text{ unit}$ since $5^2 - 1^2 = 24$ , and then construct the square on this segment.

Construct a square whose area is equal to the difference of the areas of squares of sidelengths 5 units and 7 units.

Step 1 · Calculate the Required Area and Side Length

Calculate the areas of the two given squares

\begin{aligned} \text{Area}_1 &= 7 \times 7 = 49 \text{ square units} \\ \text{Area}_2 &= 5 \times 5 = 25 \text{ square units} \end{aligned}

Find the difference between the two areas

\begin{aligned} \text{Required Area} &= \text{Area}_1 - \text{Area}_2 \\ &= 49 - 25 \\ &= 24 \text{ square units} \end{aligned}

Let the side length of the required square be $x\text{ units}$ $x^2 = 24 \implies x = \sqrt{24} \text{ units}$

Using the Pythagorean theorem $a^2 = c^2 - b^2$ , choose hypotenuse $c = 5$ and leg $b = 1$ $a^2 = 5^2 - 1^2 = 25 - 1 = 24 \implies a = \sqrt{24} \text{ units}$

Step 2 · Construct the Side Length $\sqrt{24}\text{ units}$

Draw a line segment and mark point $\text{A}$ on it.
At point $\text{A}$ , construct a perpendicular line segment $\text{AB} = 1\text{ unit}$ .
With $\text{B}$ as center and radius $5\text{ units}$ , draw an arc cutting the initial line at point $\text{C}$ .

In right-angled triangle $\text{ABC}$

\begin{aligned} \text{AC}^2 + \text{AB}^2 &= \text{BC}^2 \\ \text{AC}^2 + 1^2 &= 5^2 \\ \text{AC}^2 + 1 &= 25 \\ \text{AC}^2 &= 25 - 1 = 24 \\ \text{AC} &= \sqrt{24} \text{ units} \end{aligned}

Step 3 · Construct the Square

At point $\text{C}$ , draw a line perpendicular to $\text{AC}$ and mark point $\text{D}$ such that $\text{CD} = \text{AC} = \sqrt{24}\text{ units}$ .
With center $\text{D}$ and radius $\text{AC}$ , draw an arc.
With center $\text{A}$ and radius $\text{AC}$ , draw an arc intersecting the previous arc at point $\text{E}$ .
Join $\text{D}$ to $\text{E}$ and $\text{A}$ to $\text{E}$ .

$\text{Area of square ACDE} = (\sqrt{24})^2 = 24 \text{ square units}$

Answer

A square $\text{ACDE}$ of side length $\sqrt{24}\text{ units}$ and area $24\text{ square units}$ .

Common Mistakes

Subtracting Side Lengths Instead of Areas: Incorrectly setting the side length to $7 - 5 = 2\text{ units}$ instead of finding the difference in area ( $49 - 25 = 24$ ) and side length $\sqrt{24}\text{ units}$ .
Pythagoras Theorem Setup: Using addition instead of subtraction for finding the leg length; ensure the hypotenuse is the longest side ( $c^2 - b^2 = a^2$ ).

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