Sum of the areas of two squares is 468 m2

Leave a Comment / NCERT book maths class 10th, quadratic / By Sapling Academy

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: $A = Side^2$

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

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

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

$\therefore x^2 + y^2 = 468$ …… (i)

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

$\therefore P_1 - P_2 = 24$

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

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

$\therefore 4 x - 4 y = 24$

$\therefore x - y = 6$

$\therefore x = y + 6$ …….. (ii)

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

$x^2 + y^2 = 468$

$\therefore (y + 6)^2 + y^2 = 468$

$\therefore y^2 + 12y + 36 + y^2 = 468$

$\therefore 2 y^2 + 12 y + 36 - 468 = 0$

$\therefore 2 y^2 + 12 y - 432 = 0$

$\therefore y^2 + 6 y - 216 = 0$

$\therefore y^2 + 18 y - 12 y - 216 = 0$

$\therefore y (y + 18) - 12 ( y + 18) = 0$

$\therefore (y + 18) ( y - 12) = 0$

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 m²and 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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