Question

Sum of areas of two squares is 468 m² . If the difference of their perimeters is 24 m. Find the sides of the two square.​

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) .

[ ∵ 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

Sum of areas of 2 squares is 468

Let sides of 2 squares be a and b

a^2 + b^2 = 468

Difference of their perimeters is 24

4a- 4b = 24

a- b= 6 .............(1)

a^2 + b^2 = 468

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

36 + 2ab = 468

ab = 432/2 = 216

a^2 + b^2 = 468

( a+ b)^2 = 468 + 2ab = 468 +2(216)

= 468 + 432 = 900

( a+b) = 30 ( -30 neglected as both a,b positive)

So, a+ b= 30....(2)

Also, a- b= 6. ( from eq1)

(1) +(2)

2a = 36

a= 18

a- b= 6

b= a- 6 = 18-6 = 12

So, Sides of 2 squares are 18 and 12