Question

Sum of the areas of two squares is 305 m2 . If the difference of their perimeters is 36

cm then find the sides of the two squares.​

Answer 1

Answer:

Let x be the length of the side of First square and y be the length of side of the second square.

Then, x 2 +y 2=305...(i)

Let x be the length of the side of the bigger square

4x−4y=36

⇒ x−y=9 or x=y+9...(ii)

Putting the value of x in terms of y from equation (ii), in equation (i), we get

(y+9) 2+y 2 =305

⇒ y 2 +18y+81+y 2=305

⇒ 2y 2+18y−224=0

Y2+9Y-112=0

⇒ y(y+14)8−(y+14)=0

⇒ (y+14)(y−8)=0

Either y+14=0 or y−8=0

⇒ y=−14 or y=8

But sides cannot be negative, so y=8

Therefore, x=8+6=14

Hence, sides of two square are 14m and 8m

Answer 2

Answer:

Let x be the length of the side of First square and y be the length of side of the second square.

Then, x

2

+y

2

=468...(i)

Let x be the length of the side of the bigger square

4x−4y=24

⇒ x−y=6 or x=y+6...(ii)

Putting the value of x in terms of y from equation (ii), in equation (i), we get

(y+6)

2

+y

2

=468

⇒ y

2

+12y+36+y

2

=468

⇒ y

2

+18y−12y−216=0

⇒ y(y+18)12−(y+18)=0

⇒ (y+18)(y−12)=0

Either y+18=0 or y−12=0

⇒ y=−18 or y=12

But sides cannot be negative, so y=12

Therefore, x=12+6=18

Hence, sides of two square are 18m and 12m

Step-by-step explanation:

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