Question

sum of areas of two squares is 157 metre square. If the sum of their perimetersis 68m.Find the sides of the two square

Answer 1

Answer:

a² = x

b²= y

x+y = 157

4a + 4b = 68

a²+b²= x+y

4[a+b] = 68

a+b = 68/4 = 17

a+b = 17

a²+b² =157

a = 17-b

a²=157-b²

Answer 2

The sides of squares are 11 meters and 6 meters

Solution:

Let the sides of the two squares be x and y respectively

Then,

$Area\ of\ first\ square = x^2 \\\\Area\ of\ second\ square = y^2$

Given, sum of areas of two squares is 157 metre square

$x^2+y^2 = 157 ----- eqn\ 1$

$Perimeter\ of\ first\ square = 4x \\\\Perimeter\ of\ second\ square = 4y$

Given,  sum of their perimeters is 68 m

$4x+4y = 68\\\\Divide\ by\ 4 \\\\x + y = 17\\\\x = 17 - y ----- eqn\ 2$

$Substitute\ eqn\ 2\ in\ eqn\ 1 \\\\(17-y)^2+y^2 = 157 \\\\2y^2 - 34y + 132 = 0$

$Divide\ by\ 2 \\\\y^2 - 17y + 66 = 0 \\\\\mathrm{Break\:the\:expression\:into\:groups} \\\\\left(y^2-6y\right)+\left(-11y+66\right) = 0 \\\\ \mathrm{Factor\:out\:}y\mathrm{\:from\:}y^2-6y\mathrm{:\quad }\\\\\mathrm{Factor\:out\:}-11\mathrm{\:from\:}-11y+66\mathrm{:\quad }$

$y\left(y-6\right)-11\left(y-6\right) = 0 \\\\\mathrm{Factor\:out\:common\:term\:}y-6\\\\\left(y-6\right)\left(y-11\right) \\\\y = 6\\\\y = 11$

Thus, from eqn 2

x = 17 - 6

x = 11

x = 17 - 11

x = 6

So the values can either be x = 6 and y = 11 or x = 11 and y = 6

The sides are : 11 meters and 6 meters

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