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.

 Q.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.

Solution:

Sum of the areas of two squares is 468 m².

∵ x² + y² = 468 . ………..(1) [ ∵ area of square = side²] → The difference of their perimeters is 24 m.

∵ 4x – 4y = 24 .[ ∵ Perimeter of square = 4 × side] ⇒ 4( x – y ) = 24 .
⇒ x – y = 24/4 .
⇒ x – y = 6 .
∴ y = x – 6 ……….(2)

From equation (1) and (2),

∵ x² + ( x – 6 )² = 468

⇒ x² + x² – 12x + 36 = 468

⇒ 2x² – 12x + 36 – 468 = 0

⇒ 2x² – 12x – 432 = 0

⇒ 2( x² – 6x – 216 ) = 0

⇒ x² – 6x – 216 = 0

⇒ x² – 18x + 12x – 216 = 0

⇒ x( x – 18 ) + 12( x – 18 ) = 0

⇒ ( x + 12 ) ( x – 18 ) = 0

⇒ x + 12 = 0 and x – 18 = 0

⇒ x = – 12m [ rejected ] and x = 18m

∴ x = 18 m

Put the value of ‘x’ in equation (2),

∵ y = x – 6

⇒ y = 18 – 6

∴ y = 12 m

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