Question

the difference of areas of two squares is 549 cm square. if the difference of their perimeters is 12cm, find the sides

Answer 1

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Answer 2

The sides are 15 cm and 18 cm

Step-by-step explanation:

Let the side of square 1 be x and side of square 2 be y

Area of square 1 =$Side^2 = x^2$

Perimeter of square 1 =$4 \times side = 4x$

Area of square 2 =$Side^2=y^2$

Perimeter of square 2 =$4 \times side = 4y$

The difference of areas of two squares is 549 cm square.

$x^2-y^2=549$ --- 1

The difference of their perimeters is 12 cm

$4x-4y=12$

$x-y=3$

$x= 3+y$

Substitute the value of x in 1

$(3+y)^2+y^2=549$

$9+y^2+6y+y^2=549$

$2y^2+6y+9=549$

$2y^2+6y-540=0$

(y+18)(y-15)=0

y= -18,15

Since side cannot be negative

So, side = y = 15 cm

x=3+y=3+15=18 cm

Hence the sides are 15 cm and 18 cm

#Learn more:

Find the perimeter of square of side 12cm

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