Question

sum of areas of two squares is 468m/square . if the difference of their perimeters is 24m , find the sides of the two squares.

Answer 1

Heya !!!

This is Your answer.

Let the side of one square be x and second be y.

We know that perimeter of square = 4 X side
= 4x  and 4y

It is Given that the difference of their perimeters is 24.
So, 
4x - 4y = 24
=> x - y = 6
=> x = 6 + y        ...........(i)

Also, area of two triangles 
= x² and y²                 {because area = side²}

It is given that the sum of area of the two squares is 468.
So, 
x² + y² = 468

From (i), we get x = y+6....
Putting,
(y+6)² + y² = 468
y² + 12y + 36 + y² = 468.
2y² + 12y - 432 = 0
y² + 6y - 216 = 0.

y²-12y+18y-192=0
y (y-12)+18  (y-12)=0
(y+18) (y-12)=0

y+18=0      or  y-12=0
y= -18  or        y=12.


As a side can't be negative. So, the side of the square is 12 cm.
And the Side of second one = y+6 = 18 cm.

Hope This Helps

Answer 2

hello...

the sides of the two squares are 'a' and 'b'

Sum of their areas = a² + b² = 468

Difference of perimeters = 4a - 4b = 24

= a - b = 6

= a = b + 6

we get the equation

(b + 6)² + b² = 468

=> 2b² + 12b + 36 = 468

=> b² + 6b - 216 = 0

= b=12

= a=18