Question

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

Answer 1

$\bold{GIVEN}$

Sum of squares of two squares = 157

Sum of there perimeter= 68 m

Let the side of first square be S

And of other square be = s

Area of first square = $S^{2}$

Area of second square = $s^{2}$

Sum of there areas is 157

$\boxed{ S^{2} + s^{2} = 157}$----(1)

Perimeter of first square = 4S

Perimeter of second square = 4s

Sum of these is 68

$\boxed{ 4S + 4s = 68}$----(2)

On dividing whole equation by 4

S + s = 17

S= 17 - s

on substitution value of S in eq (2)

$(17 -s) ^{2} + s^{2} = 157$

$289 + s^{2} - 34s + s^{2} = 157$

$2s^{2} - 34s + 289 - 157 = 0$

$2s^{2} - 34s + 132= 0$

since the equation has a multiple 2 , on dividing it by 2

$s^{2} - 17s + 66= 0$

On solving the following quadratic equation we will get two values,

$s^{2} - 6s - 11x + 66= 0$

$s(s - 6)- 11(s - 6)= 0$

$(s - 6)(s - 11)= 0$

On taking separately,

s - 6 =0

$\boxed{s= 6}$

or

$\boxed{s= 11}$-

On placing its value in equation (2)

If s= 11

4S + 4 ×11 = 68

4S + 44 = 68

4S = 68 -44

4S = 24

S = $\frac{24}{4}$

$\boxed{S= 6}$

If s= 6

4S + 4 ×6 = 68

4S + 24 = 68

4S = 68 -24

4S = 44

S = $\frac{44}{4}$

$\boxed{S= 11}$

So, the side of first square can be 6 or 11.

The side of square can be 11 or can be 6.