Question

Sum of the areas of two squares is 157 m². If the sum of their perimeters is 68 m, find the sides of the two squares.

Answer 1

Let the side of one square be x
And a side of other square be y
Sum of an area of two square is 157
Equation becomes

`x^2 + y^2 = 157`    ....(1)   (∵ area of square is side²)

Now, sum of their perimeters is 68
Equation becomes

`4x + 4y = 68`    (∵ perimeter of square is 4 × side)

solving the two-equation by substitution method

`4x + 4y = 68`

`x + y = 17`

⇒ x = 17 - y ....(2)

Substitute (2) in (1)

`(17 - y)^2 + y^2 = 157`

`289 + y^2 - 34y + y^2 = 157`

`2y^2 - 34y + 132 = 0`

`y^2 - 17y + 66 = 0`

Using y = `(-b ± sqrt(b^2 - 4ac))/(2a)`

Plugging the values in the formula we get

`y = (17 ± sqrt(289 - 4(66)))/2`

`y = (17 ± sqrt(25))/2`

`y = (17 ± 5)/2`

`y = 12/2,22/2`

y = 6,11

when y = 6 then x = 11
And when y = 11 then x = 6
Therefore, the sides of square are 6 m and 11 m.