Question

The sum of the areas of two squares is 400 sq.m. If the difference of their perimeters is 16 m,find the sides of 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

Step-by-step explanation:

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