Multiple choice

The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, The side length of one of these square is _____.

  1. 10 m

  2. 11 m

  3. 12 m

  4. 13 m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let sides be x and y. x^2 + y^2 = 468 and 4x - 4y = 24 (so x - y = 6). Substituting y = x - 6 into the first equation gives x^2 + (x-6)^2 = 468, which simplifies to 2x^2 - 12x - 432 = 0, or x^2 - 6x - 216 = 0. Factoring gives (x-18)(x+12) = 0, so x = 18 and y = 12.

AI explanation

Let the sides of the two squares be a and b, where a is greater than b. The difference of their perimeters gives 4a minus 4b equals 24, meaning a minus b equals 6. The sum of their areas gives a squared plus b squared equals 468; substituting a equals b plus 6 yields b squared plus 6b minus 216 equals 0. Solving this quadratic equation gives the side lengths as 18 m and 12 m. The side length of one of the squares is 12 m.

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