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Sum of the areas of two squares is 468m². If the difference of their perimeters is 24 m, find the sides of the two squares.

User Natxet
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Final answer:

The sides of the two squares are found by setting up a system of equations based on the given area sum and perimeter difference, which are solved simultaneously to find that one square has a side length of 18 m and the other has a side length of 12 m.

Step-by-step explanation:

We are given the sum of the areas of two squares to be 468 m² and the difference of their perimeters to be 24 m. To find the sides of the squares, let's denote the side of the first square as a and the side of the second square as b.

The area of a square is calculated as side × side, which is a² for the first square and b² for the second square.

Accordingly, the sum of their areas is a² + b² = 468.

The perimeter of a square is the sum of all its sides, so it's 4a for the first square and 4b for the second square. Thus, their difference is 4a - 4b = 24. Simplifying the equation gives a - b = 6.

Now we have two equations:

  1. a² + b² = 468
  2. a - b = 6

By solving these two equations simultaneously, we can find the values of a and b. First, let's solve the second equation for a: a = b + 6.

Substituting a = b + 6 into the first equation:

(b + 6)² + b² = 468
b² + 12b + 36 + b² = 468
2b² + 12b - 432 = 0
b² + 6b - 216 = 0
(b + 18)(b - 12) = 0

From the above, we get two possible values for b, either b = -18 or b = 12. Since a side length can't be negative, we discard b = -18 and keep b = 12. Now we can find a: a = b + 6 = 12 + 6 = 18.

Therefore, the sides of the two squares are 18 m and 12 m, respectively.

User Mahmoud Gamal
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