Question

sum of area of two square is 468 m2. if the difference of their perimeter is 24 m2.find the Sides of square​

Answer 1

Answer:

Let the two squares be A and B. A has sides of a, while B has sides of b,

Now for the two equations.

a^2 + b^2 = 468 …(1) [Sum of the areas]

4a - 4b = 24 …(2) (perimeter of the two squares), which can be written as

a - b = 6, or

a = b+6. Put that value in (1) to get

(b+6)^2 + b^2 = 468, or

b^2 + 12b + 36 + b^2 = 468, or

2b^2 + 12b + 36 = 468, or

b^2 + 6b + 18 = 234, or

b^2 + 6b + 18 - 234 = 0, or

b^2 + 6b -216 = 0, or

b^2 + 18b - 12b - 216 = 0, or

b(b + 18) - 12(b + 18) = 0, or

(b-12)(b+18) = 0

b = 12 m or -18m (inadmissible)

Then a = b+6 = 12 + 6 = 18m

So A is a square of 18 m and B is a square of 12 m

Check: Area of A = 18^2 = 324 sq m. area of B = 12^2 = 144 sq m and their sum is 324 + 144 = 468 sq m