**Q. Sum of the areas of two squares is 468 m ^{2}. If the difference of their perimeters is 24 m, find the sides of the two squares.**

**Ans: **Let consider the sides of the two squares be x m and y m.

We know that the area of a square is given by:

Hence, area of 1st square will be

and area of 2nd square will be

By given 1^{st} condition: s**um of the areas of two squares is 468 m ^{2}**

…… (i)

By given 2^{nd} condition, we have: **difference of their perimeters is 24 m**

perimeter of a square is given by: P = 4 x Side,

Therefore and

…….. (ii)

By substituting value of x from equation (ii) in equation (i), we get:

Therefore, y = – 18 and y = 12

Here, we reject y = – 18 because side can not be negative, and we accept y = 12

By substituting the value of y in equation (ii), we get:

x = y + 6

x = 12 + 6 = 18

**Hence, sides of the squares are 18 m and 12 m.**

**Check:**

*If sides of the two squares are 18 m and 12m, there squares will be 18 ^{2} = 324 m^{2} and 12^{2} = 144 m^{2}*

*If we add these two values, we get 324 + 144 = 468 m ^{2 }*

*and it meets our 1*

^{st}given condition.*The perimeters of the two squares will be 18 x 4 = 72 m and 12 x 4 = 48 m*

*The difference of these two perimeters is 72 – 48 = 24 m*

*Since our 2 ^{nd} condition is also met. *

**hence our answer is correct.****Please do press “Heart” button if you liked the solution. **