Sum of the areas of two squares is 468 m square if the difference of their perimeter is 24 m find the sides of the two square .​

✎Δnsɯer࿐

Let the length of the side of the smaller square be x .

Perimeter of smaller square = 4x

Let the length of the side of the larger square be y m.

Perimeter of larger square = 4y

Now, 4y−4x=24

or, y−x=6 (Dividing both sides by 4)

or, y=x+6

Now, area of larger square = y 2=(x+6)2

= x2 +12x+36

Area of smaller square = x2

Now, x2+12x+36+x2=468 (Sum of areas)

or, 2 x2+12x+36=468

or, x2 +6 x +18=234 (Dividing both sides by 2)

or, x2 +6x+18−234=0

or, x2 +6x−216=0

or, (x+18)(x−12)=0 (Factoring the equation)

or, x=−18,12

Since x cannot be negative,therefore,

x=12m

Therefore, x + 6 = 18m

So the sides are 18m and 12m respectively.

Hope that helps.

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. Correct.