Question

The difference of perimeters of two squares is 32 meters and the difference of their areas is 208 square meters. What is the difference between the lengths of their sides.? Find the length of their sides? ​

Answer 1

Question -

The difference of perimeters of two squares is 32 m and the difference of their areas is 208 m². What is the difference between the lengths of their sides? Find the length of their sides?

Solution -

Let the side of first square be x and second square be y

According to the question,

x²- y²=208

(x+y) (x-y) =208.......1)

And,

$4x - 4y = 32$

$4(x - y) = 32$

$x - y = \frac{32}{4}$

$x - y = 8$

$x = 8 + y$

Substituting the value of x in equation 1) we will get -

$(8 + y + y)(8 + y - y) = 208$

$(8 + 2y)8 = 208$

$64 + 16y = 208$

$16y = 144$

$y = \frac{144}{16}$

$y = 9$

Substituting the value of y in equation 1) we will get -

$(x + 9)(x - 9) = 208$

${x}^{2} - 81 = 208$

${x}^{2} = 289$

$x = \frac{ + }{ - } \sqrt{289}$

$x = \frac{ + }{ - } 17$

We can't have a negative side so x =17 is chosen

The difference between their sides =17-9

=8m

Therefore.

The length of first square = 17 m

And,

The length of second square = 9 m