4.3.11. 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.

Question

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.

Sapling Academy · Accepted Answer

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: $A = Side^2$

Hence, area of 1st square will be $A_1 = x^2$

and area of 2nd square will be $A_2 = y^2$

By given 1st condition: sum of the areas of two squares is 468 m2

$	herefore x^2 + y^2 = 468$ ...... (i)

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

$	herefore P_1 - P_2 = 24$

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

Therefore $P_1 = 4x$ and $P_2 = 4y$

$	herefore 4 x - 4 y = 24$

$	herefore x - y = 6$

$	herefore x = y + 6$ ........ (ii)

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

$x^2 + y^2 = 468$

$	herefore (y + 6)^2 + y^2 = 468$

$	herefore y^2 + 12y + 36 + y^2 = 468$

$	herefore 2 y^2 + 12 y + 36 - 468 = 0$

$	herefore 2 y^2 + 12 y - 432 = 0$

$	herefore y^2 + 6 y - 216 = 0$

$ 	herefore y^2 + 18 y - 12 y - 216 = 0$

$ 	herefore y (y + 18) - 12 ( y + 18) = 0$

$ 	herefore (y + 18) ( y - 12) = 0 $

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 1st 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 2nd condition is also met. hence our answer is correct.

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