Question 17

We need to find the area of a square. This square's area is the difference of two other squares' areas.

Step 1 — Calculate the required area

First, let us find the areas of the two given squares. One square has a side length of 7 units. Its area is side multiplied by side.

$\text{Area}_1 = 7 \times 7$

$= 49 \text{ square units}$

The other square has a side length of 5 units. Its area is also side multiplied by side.

$\text{Area}_2 = 5 \times 5$

$= 25 \text{ square units}$

Now, we find the difference between these two areas. This difference will be the area of our new square.

$\text{Required Area} = \text{Area}_1 - \text{Area}_2$

$= 49 - 25$

$\boxed{24 \text{ square units}}$

Step 2 — Find the side length of the new square

Let the side length of our new square be $x$ units. The area of this square is $x^2$. We know the area must be 24 square units.

$x^2 = 24$

So, the side length $x$ is the square root of 24.

$x = \sqrt{24} \text{ units}$

We will use the Pythagorean theorem to construct this length. The Pythagorean theorem helps us with right-angled triangles. It says: $a^2 + b^2 = c^2$. Here, $a$ and $b$ are the shorter sides (legs). And $c$ is the longest side (hypotenuse). We can also write it as $a^2 = c^2 - b^2$. Here, $a$ is one leg. $c$ is the hypotenuse. $b$ is the other leg. Let us find two numbers whose squares subtract to 24. We can choose a hypotenuse of 5 units. Its square is $5^2 = 25$. If one leg is 1 unit, its square is 1. Then the square of the other leg will be $25 - 1$. This means the square is 24. So, the other leg will be $\sqrt{24}$ units long.

Step 3 — Construct the side length $\sqrt{24}$ units

Let us draw a line segment. We will mark a point A on this line. At point A, we construct a perpendicular line. A perpendicular line forms a 90-degree angle. We mark a point B on the perpendicular line. The distance from A to B is 1 unit. Now, we use a compass. We place the compass point at B. We set the compass width to 5 units. We draw an arc that cuts the first line segment. Let this intersection point be C. Now, triangle ABC is a right-angled triangle. Angle A is 90 degrees. Side AB is 1 unit. Side BC is 5 units (this is the hypotenuse). Using the Pythagorean theorem:

$AC^2 + AB^2 = BC^2$

$AC^2 + 1^2 = 5^2$

$AC^2 + 1 = 25$

$AC^2 = 25 - 1$

$AC^2 = 24$

$AC = \sqrt{24} \text{ units}$

So, the length of AC is the side length we need.

Step 4 — Construct the square

We now have the side length AC, which is $\sqrt{24}$ units. Let us construct a square with this side length. We will use AC as one side of our square. At point C, we construct a line perpendicular to AC. On this perpendicular line, we mark a point D. The distance from C to D must be equal to AC. Now, we use a compass again. Place the compass point at D. Set the compass width to AC. Draw an arc. Place the compass point at A. Set the compass width to AC. Draw another arc. These two arcs will intersect at point E. Join points D to E and A to E. The figure ACDE is the required square. All its sides are equal to $\sqrt{24}$ units. All its angles are 90 degrees. The area of square ACDE is $(\sqrt{24})^2$.

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

$\boxed{24 \text{ square units}}$

Answer

(i) The area of the square with side 7 units is 49 square units. (ii) The area of the square with side 5 units is 25 square units. (iii) The area of the constructed square is 24 square units. (iv) The side length of the constructed square is $\sqrt{24}$ units.