Chapter 11.6, Problem 3P

To determine

To calculate: The sum of the perimeters of the square.

Answer to Problem 3P

The sums of perimeter of the squares are $\frac{48\sqrt{2}}{\sqrt{2}-1}\text{ cm}$ .

Explanation of Solution

Given information: The side of 1^st square is 12 cm

Figure:

  

According to the question, 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.

Perimeter of first square = 4 ×12 = 48 cm

Therefore, side of the second square:

P is midpoint of AB

PB = BQ = 6cm

Apply the Pythagoras Theorem:

Hence, the sides of second square $6\sqrt{2}$

  $\therefore PQ=\sqrt{{\left(PB\right)}^2+{\left(BQ\right)}^2}$

  $\begin{array}{l}=\sqrt{{\left(6\right)}^2+{\left(6\right)}^2}\\ =\sqrt{36+36}\\ =\sqrt{72}\\ =6\sqrt{2}\text{ cm}\\ \end{array}$

Therefore, perimeter = 4× $6\sqrt{2}$ = $24\sqrt{2}$ cm

Similarly,

Side of Third Square:

  $\therefore KL=\sqrt{{\left(KP\right)}^2+{\left(PL\right)}^2}$

  $\begin{array}{l}=\sqrt{{\left(3\sqrt{2}\right)}^2+{\left(3\sqrt{2}\right)}^2}\\ =\sqrt{18+18}\\ =\sqrt{36}\\ =6\text{ cm}\end{array}$

Perimeter = 4×6 = 24 cm

Hence, the perimeters of all square are in GP

Therefore, $48+ 24\sqrt{2}+24+\mathrm{...........}\infty$

The first term, $a=48$

Second terms, $a_1=24\sqrt{2}$

The common ratio, $r=\frac{a_1}{a}=\frac{24\sqrt{2}}{48}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

Formula used: As we know sum of infinity term of a G.P is $S=\frac{a}{1-r}$ .

Calculation:

  $\begin{array}{l}\therefore S=\frac{a}{1-r}\\ =\frac{48}{1-\frac{1}{\sqrt{2}}}\\ =\frac{48}{\frac{\sqrt{2}-1}{\sqrt{2}}}\\ =\frac{48\sqrt{2}}{\sqrt{2}-1}\text{ cm}\end{array}$

Hence, the sums of perimeter of the squares are $\frac{48\sqrt{2}}{\sqrt{2}-1}\text{ cm}$ .