Geometry Problems on Squares

Problem Statement: When the sides of a square are each increased by 2 feet, its area increases by \(44 \, \text{ft}^2\). Find the original side length \(s\).

Solution:

Let \(s\) be the original side length.

Original area: \(A_1 = s^2\)

New side length: \(s + 2\)

New area: \(A_2 = (s + 2)^2\)

Given: \(A_2 = A_1 + 44\)

So: \(s^2 + 44 = (s + 2)^2\)

Expand: \(s^2 + 44 = s^2 + 4s + 4\)

Simplify: \(44 = 4s + 4\)

Solve: \(4s = 40\)

Thus: \(s = 10 \, \text{feet}\)

Check: Original area = \(10^2 = 100 \, \text{ft}^2\), New area = \(12^2 = 144 \, \text{ft}^2\), Difference = \(44 \, \text{ft}^2\) ✓

Problem Statement: Find the area and perimeter of a square with a diagonal of 200 meters.

Solution:

Using the diagonal formula: \(d = s\sqrt{2}\)

Given \(d = 200 \, \text{m}\), so \(s\sqrt{2} = 200\)

Thus: \(s = \dfrac{200}{\sqrt{2}} = 100\sqrt{2} \, \text{m}\)

Area: \(A = s^2 = (100\sqrt{2})^2 = 20000 \, \text{m}^2\)

Perimeter: \(P = 4s = 4 \times 100\sqrt{2} = 400\sqrt{2} \, \text{m}\)

Problem Statement: What happens to the area of a square if we double its side length?

Solution:

Let original side = \(s\), area = \(s^2\)

New side = \(2s\), new area = \((2s)^2 = 4s^2\)

Ratio: \(\dfrac{4s^2}{s^2} = 4\)

Answer: The area becomes 4 times larger.

Problem Statement: A square garden of \(400 \, \text{m}^2\) is surrounded by a walkway of constant width \(x\). The total area of the walkway is \(500 \, \text{m}^2\). Find the width \(x\).

Solution:

Garden area = \(400 \, \text{m}^2\), so side length = \(\sqrt{400} = 20 \, \text{m}\)

Total area (garden + walkway) = \(400 + 500 = 900 \, \text{m}^2\)

Let walkway width = \(x\)

Total side length = \(20 + 2x\)

Equation: \((20 + 2x)^2 = 900\)

Take square root: \(20 + 2x = 30\) (positive root since length > 0)

Solve: \(2x = 10\)

Thus: \(x = 5 \, \text{m}\)

Check: Total side = 30 m, total area = \(900 \, \text{m}^2\), walkway area = \(900 - 400 = 500 \, \text{m}^2\) ✓