Question

plz help guys plz ...........HELP

Answer 1

Let the side of first square be x
and that of the second be y

So perimeter of first square = 4x
Similarly, perimeter of second square = 4y


Given that, the difference of their perimeter is 24 cm

=> 4x - 4y = 24

=> 4(x - y) = 24

=> (x - y) = 24/4

=> (x - y) = 6 cm.............( 1 )

Now the area of first square would be x²

And that of the second square be y²

Given that their sum is 818

=> x² + y² = 818

Now square both the sides in equation ( 1 )

=> (x - y)² = (6)²

=> x² + y² - 2xy = 36

But x² + y² = 818

=> 818 - 2xy = 36

=> -2xy = 36 - 818

=> -2xy = -782

=> 2xy = 782

=> xy = 782/2

=> xy = 391.


Now, in x² + y² = 818, add 2xy on both sides

=> x² + y² + 2xy = 818 + 2xy

=> (x + y)² = 818 + 782

=> (x + y)² = 1600

=> (x + y)² = (40)²

=> x + y = 40..............( 2 )


Now add ( 1 ) and ( 2 )

=> x - y + x + y = 6 + 40

=> 2x = 46

=> x = 46/2

=> x = 23 cm.

Now out value of x in any of the equation

x - y = 6

=> 23 - y = 6

=> - y = 6 - 23

=> -y = - 17

=> y = 17 cm.

So the side of first square is 23 cm and side of second square is 17 cm


Hope it helps dear friend ☺️✌️

Answer 2

Let the sides of the squares be x and y.

(i)

Given : Sum of areas of squares is 818 m^2.

⇒ x^2 + y^2 = 818

(ii)

Given : Difference of their perimeters is 24 m.

⇒ 4x - 4y = 24  

⇒ 4x = 24 + 4y

⇒ 4x = 4(6 + y)

⇒ x = 6 + y

Now,

Substitute (ii) in (i), we get

⇒ (6 + y)^2 + y^2 = 818

⇒ 36 + y^2 + 12y + y^2 = 818

⇒ 2y^2 + 12y = 782

⇒ 2y^2 + 12y - 782 = 0

⇒ y^2 + 6y - 391 = 0

⇒ y + 23y - 17y - 391 = 0

⇒ y(y + 23) - 17(y + 23) = 0

⇒ (y - 17)(y + 23) = 0

⇒ y = 17,-23{It cannot be negative}

⇒ y = 17.

Substitute y = 17 in (2), we get

⇒ x = 6 + y

⇒ x = 6 + 17

⇒ x = 23

Therefore, the sides of the squares are 23 and 17 respectively.

Hope it helps!