Question

sum of areas of two squares is 400 cm 2 .If the difference in their perimeters is 16cm. find the sides of the two squares

Answer 1

Let side of first square be x and the side of another square be y.
difference in perimeter of two squares:-
4x-4y=16
4(x-y)=16
x-y=16/4
therefore, x-y=4
and x=4+y---take this(1)
 
sum of areas of two squares:-
 x²+y²=400
Put (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 (here we have taken 2 common and take to another side. then it becomes:-)
y²+4y-192=0
we have done with factorisation method:-
y²-12y+16y-192=0
y(y-12)+16(y-12)=0
(y+16) (y-12)=0
either:-                             |             or:-
y+16=0                            |                   y-12=0
y= -16                              |                   y=12

we will take y=12 because y being side cannot be negative.
So, y=12
put y=12 in (1)
x=4+y
x=4+12
therefore, x=16
and hence, Side of first square= x = 16
and Side of another square= y = 12.

AND DONE.

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let the sides of the square be x and y.

According to the Question,

⇒ x² + y² = 400

⇒ 4x - 4y = 16

⇒ x - y = 4

⇒ x = 4 + y

⇒ (4 + y)² + y² = 400

⇒ 2y² + 8y + 16 = 400

⇒ y² + 4y – 192 = 0

⇒ y² + 16y - 12y - 192 = 0

⇒ (y + 16)(y - 12 ) = 0

⇒ y = - 16, 12 (y can't be negative)

⇒ y = 12

And x = 16

Hence, the two sides of square  are 12 and 16.