Question

the sum of areas of two squares is 400 cm square,if the difference of their perimeter is 16 then find the Side of each square

Answer 1

Hello...

Let side first square be x
the side of another square be y.

difference perimeter of two squares

4x-4y=16

4(x-y)=16

x-y=16/4

x-y=4

and x=4+y----- (1)

sum areas of two squares

x²+y²=400

Take (1) here

(4+y)²+y²=400

(4)²+(y)²+2×4×y+y²=400

16+y²+8y+y²=400

2y²+8y+16-400=0

2y²+8y-384=0

2(y²+4y-192)=0

we've taken 2 common and take at another side. then

y²+4y-192=0

Factorisation method

y²-12y+16y-192=0

y(y-12)+16(y-12)=0

(y+16) (y-12)=0

y+16=0

y= -16

y=12 because y being side cant be negative.

So, y=12

put y=12 in (1)

x=4+y

x=4+12

x=16

Side of first square= x = 16

Side of another square= y = 12.

Answer 2

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 16.
So, 
4x - 4y = 16
=> x - y = 4
=> x = 4 + 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 400.
So, 
x² + y² = 400

From (i), we get x = y+4....
Putting,
(y+4)² + y² = 400
y² + 8y + 16 + y² = 400.
2y² + 8y - 384 = 0
y² + 4y - 192 = 0.

y²-12y+16y-192=0
y(y-12)+16(y-12)=0
(y+16) (y-12)=0

y+16=0      or  y-12=0
y= -16  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+4 = 16 cm.

Hope it Helps