Question

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

Answer 1

Answer:

Answer

Let the sides of the two squares be

x cm andy cm, where x> y

Then, their areas are x2 and y2 and

their perimeters are 4x and 4y.

By given condition,

x+y2 = 400

-(1)

And 4x - 4y = 16

4(x - y) = 16

x-y = 4

x = y +4

Substituting the value of x from (2) in (1

), we get,

(y+4)2 + y2 = 400

y+16+8y +y2 = 400

2y2+ 16 +8y = 400

y+ 4y 192=0

y+16y 12y -192 = 0

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

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

y=-16 or 12

Since, y cannot be negative, y = 12

x=y +4= 12 + 4 16

.'. Sides of the two squares are 16 cm and

12 cm.

Required difference = 16 12 = 4 cm

Step-by-step explanation:

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