Question

Sum of the areas of two squares is 468 m2 . if the difference of their perimeter is 24 m. Find the sides of the squares?

Answer 1

■ $\blue {\rm \huge \underline{S \large OLUTION}}$ ■


★ Sum of their areas
⇒ a² + b²
⇒ 468

★ Difference of their perimeters
⇒ 4a - 4b = 24
⇒ a - b = 6
⇒ a = b + 6

★ So, we get the equation

⇒ (b + 6)^2 + b^2 = 468
⇒ 2b^2 + 12b + 36 = 468
⇒ b^2 + 6b - 216 = 0
⇒ b = 12
⇒ a = 18

∴ The sides of the two squares are 12 and 18.

Answer 2

Step-by-step explanation:

Answer:

→ 18m and 12 m .

Step-by-step explanation:

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

Case 1 .

→ Sum of the areas of two squares is 468 m² .

A/Q,

∵ x² + y² = 468 . ...........(1) .

[ ∵ area of square = side² . ]

Case 2 .

→ The difference of their perimeters is 24 m .

A/Q,

∵ 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) , we get

∵ 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), we get

∵ y = x - 6 .

⇒ y = 18 - 6 .

∴ y = 12 m . ......

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