Question

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

Answer 1

Let the sides of the two squares be x m and y m.

Therefore, their perimeter will be 4x and 4y respectively

And area of the squares will be x² and y² respectively.

Given,

➪ 4x – 4y = 24

➪ x – y = 6

➪ x = y + 6

Also, x² + y² = 468

➪ (6 + y²) + y² = 468

➪ 36 + y² + 12y + y² = 468

➪ 2y² + 12y + 432 = 0

➪ y² + 6y – 216 = 0

➪ y² + 18y – 12y – 216 = 0

➪ y(y +18) –12(y + 18) = 0

➪ (y + 18)(y – 12) = 0

➪ y = –18, 12

As we know, the side of a square cannot be negative.

Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.

Answer 2

Answer:

A²+a²= 468 (where A and a are the sides of square)

4A-4a=24 --- A-a=6

___A²+a²-2Aa= 36

___468-2Aa=36

____Aa=216

A=216/a

a²+6a-216=0

a=12

A=18

Sides of square are 12m and 18m