Question

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

Answer 1

$\huge\underline\mathsf{Question}:-$

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.

$\huge\underline\mathtt{Solution}:-$

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

Answer 2

Answer:

Step-by-step explanata

a = first square

b = sec square

difference= 4a -4b = 24

a - b= 6

a = 6+b

(6+b)sq = b square = 468

36 +B sq + 12 b + b sq = 468

2b sq+ 12b- 432 =0

on solving

b=12

a =b+6

a=18

hope it helps you