Question

The sum of areas of two squares is 640 m\(^2\). If the difference in perimeters is 64 m, find the sides.

Hint:

In speed-distance problems, the equation is usually written as: \(\text{Time}_{\text{slow}} - \text{Time}_{\text{fast}} = \text{Difference}\).

Answer 1

Step 1: Let the Required Sides be:
Let side of first square = a metres
Let side of second square = b metres

Given:
4a − 4b = 64

Factor 4:
4(a − b) = 64
a − b = 16 → (1)

From (1):
a = b + 16

Step 2: Using Sum of Areas Condition:
Area of first square = a²
Area of second square = b²

Given sum of areas = 640

a² + b² = 640

Substitute a = b + 16:
(b + 16)² + b² = 640

Expand:
b² + 32b + 256 + b² = 640

2b² + 32b + 256 = 640
2b² + 32b − 384 = 0

Divide entire equation by 2:
b² + 16b − 192 = 0

Step 3: Solving the Quadratic Equation:
b² + 16b − 192 = 0

Factorising:
(b + 24)(b − 8) = 0

So,
b = −24 (not possible, side cannot be negative)
b = 8

Now substitute in (1):
a = b + 16
a = 8 + 16
a = 24

Final Answer:
The sides of the two squares are:
24 m and 8 m