Q. 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.

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: sum of the areas of two squares is 468 m²

…… (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 18² = 324 m² and 12² = 144 m²

If we add these two values, we get 324 + 144 = 468 m² 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.

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