Finding the Side Lengths of Two Squares
To solve the problem, let’s denote the side lengths of the two squares as s1 and s2.
We know from the problem:
- The sum of their areas: s12 + s22 = 468
- The difference of their perimeters: 4s1 – 4s2 = 24
From the perimeter equation, we can simplify:
s1 – s2 = 6
Now we can express s1 in terms of s2:
s1 = s2 + 6
Next, we’ll substitute this expression into the area equation:
(s2 + 6)2 + s22 = 468
Expanding the first term:
(s22 + 12s2 + 36) + s22 = 468
Simplifying gives us:
2s22 + 12s2 + 36 = 468
Next, subtract 468 from both sides:
2s22 + 12s2 – 432 = 0
Now, we can simplify by dividing everything by 2:
s22 + 6s2 – 216 = 0
We can solve this quadratic equation using the quadratic formula, where a = 1, b = 6, and c = -216:
s2 = (-b ± √(b2 – 4ac)) / 2a
Substituting the values:
s2 = (-6 ± √(62 – 4 * 1 * -216)) / (2 * 1)
s2 = (-6 ± √(36 + 864)) / 2
s2 = (-6 ± √900) / 2
s2 = (-6 ± 30) / 2
This yields two potential solutions:
- s2 = 12 m (choosing the positive root for a side length)
- s2 = -18 m (not a valid solution as lengths cannot be negative)
So we find s2 = 12 m.
Next, we can find s1:
s1 = s2 + 6 = 12 + 6 = 18 m
Final Result:
- The side length of the first square is: 18 m
- The side length of the second square is: 12 m
In conclusion, the side lengths of the two squares are 18 m and 12 m, respectively.