Question

sum of the areas of two squares is 468 m square if the difference of their perimeter is 24 metre then find the side of the square ​

Answer 1

Step-by-step explanation:

Let , A1&A2 be the side of the two squares respectively.

Given A1^2+A2^2=468m------(1)

4(A1-A2)=24m

(A1-A2)=6-----(1)

A1=6+A2, substituting in eq.1

We get, A2=approx.32m and

A1=38m

Answer 2

Answer:-

★ Point to remember:-

→ Let the sides of the two squares be x m and ym.

→ 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, wkt:-

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