Question

In the figure shown, square 2 is formed by joining the mid-points of square 1, square 3 is formed by joining the mid-points of square 2 and so on. In this way total five squares are drawn. The side of the square 1 is 'a' cm. What is the sum of perimeters of all the five squares ?

Answer 1

'A' is the midpoint of BD

$\therefore AD=\frac{a}{2}=AB$

$\angle ABC={90}^o-$Angle of square

Applying Pythagoras theorem,

$(AC{)}^2=(a/2{)}^2+(a/2{)}^2$

$AC=\sqrt{a^2}/2=a/\sqrt{2}$

Similarly length of side of square $3$ would be

$(s_3{)}^2=(a/2\sqrt{2}{)}^2+(a/2\sqrt{2}{)}^2$

$(s_3{)}^2=a^2/4$

$s_3=a/2$

Similarly length of side of square $4$ would be

$(s_4{)}^2=(a/4{)}^2+(a/4{)}^2$

$(s_4{)}^2=a^2/2$

$s_4=a/\sqrt{2}$

$\therefore$ Length of side of square $5$ would be

$s_5=a/2$

$\therefore$ Total perimeter$=P\left(square1\right)+P\left(square2\right)+P\left(square3\right)+P\left(square4\right)+P\left(square5\right)$

$=\left(4a\right)+\left(4a/\sqrt{2}\right)+\left(2a\right)+\left(4a/\sqrt{2}\right)+\left(2\right(a\left)\right)$

$=8a+8a/\sqrt{2}$ sq.cm.