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

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) .

[ $\because$ 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 .

Answer 2

Answer:

Step-by-step explanation:

let a^2 and b^2 be the area of the 2 squares and 4a and 4b be the perimeters

a^2+b^2=468-------------(1)

4a-4b=24

a-b=6

a=6+b ---------(2)                                                                (bringing 4 to rhs side)

substituting 2 in 1 we get

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

36+b^2+12b+b^2=468                                                   [    (a+b)^2 formula]

2b^2+12b-432=0

b^2+6b-216=0

solving the equation we get 18 and -12

so one side is 18 and the other one is 12 this happens when we substitute the value of 18 in eqn 2

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