Question

Sum of the area of two squares is 400cm^ if the difference of their perimeter is 16cm. Find the sides of the two squares?

Answer 1

We assume that the side of the smaller square be x cm.
Perimeter of any square = (4 × side of the square) cm.

It is given that the difference of the perimeters of two squares is 16 cm.
Then side of the bigger square = 16+4x4=(4+x)16+4x4=4+x cm.

The difference between the perimeters is given by,

x2+(4+x)2=400
⇒x2+16+x2+8x=400
⇒2x2+8x−384=0
⇒x2+4x−192=0
⇒x2+16x−12x−192=0
⇒x(x+16)−12(x+16)=0
⇒(x−12)(x+16)=0
⇒x−12=0orx+16=0

⇒x=12orx=-16

Since, side of the square cannot be negative.

Thus, the side of the smaller square is 12 cm.

and the side of the bigger square is (4 + 12) = 16 cm.

Answer 2

Answer:

Step-by-step explanation:

let the side of square be x

and the side of another square be y

Condition : 1

SUM OF AREAS OF TWO SQUARE = 400

x²+y² = 400........(1)

Condition : 2

difference of perimeter is 16

4x - 4y = 16

x-y=4

SOLUTION:

x-y=4

x= y+4........(2)

SUB 2 in 1

(y+4)² + y² =400

y²+16+8y+y²=400

2y²+8y+16=400

2y²+8y-384=0

divide by 2

y²+4y-192

y²+16y-12y-192=0

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

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

y=12 (or) y= -16

neglect negative term

so y=12

now sub in(2)

x = 12+4

x = 16

the side of the square is 16cm and 12cm