Let the sides of the two squares be a and b. The area of the squares can be expressed as:
- Area of the first square: a²
- Area of the second square: b²
According to the problem, we have the first equation:
a² + b² = 468 (1)
The perimeter of a square is given by the formula P = 4s (where s is the length of a side). Thus, the perimeter of the first square is 4a and the perimeter of the second square is 4b. The problem states that the difference of their perimeters is 24 m:
|4a – 4b| = 24 (2)
This simplifies to:
|a – b| = 6 (3)
From equation (3), we can have two cases:
- a – b = 6 (i)
- b – a = 6 (ii)
Let’s consider case (i):
From (i), we can express a in terms of b:
a = b + 6
Now substitute this expression for a into equation (1):
(b + 6)² + b² = 468
This expands to:
b² + 12b + 36 + b² = 468
Combining like terms gives:
2b² + 12b – 432 = 0
To simplify, divide the entire equation by 2:
b² + 6b – 216 = 0
Now, we can factor this quadratic equation:
(b + 18)(b – 12) = 0
This gives us two solutions for b:
b + 18 = 0 → b = -18 (not valid since a side length cannot be negative)
b – 12 = 0 → b = 12
Now substitute b = 12 back into our equation for a:
a = 12 + 6 = 18
Thus, we have found the sides of the two squares:
Side of the first square (a): 18 m
Side of the second square (b): 12 m
Now, let’s quickly verify:
- Area of the first square: 18² = 324 m²
- Area of the second square: 12² = 144 m²
- Sum of areas: 324 + 144 = 468 m² (check)
- Perimeter of the first square: 4 * 18 = 72 m
- Perimeter of the second square: 4 * 12 = 48 m
- Difference in perimeters: 72 – 48 = 24 m (check)
Therefore, the sides of the squares are indeed 18 m and 12 m.