Question

The sides of a given square is 10 cm. The mid points of its sides
are joined to form a new square. Again, the mid points of the
sides of this new square are joined to form another square.
This process is continued indefinitely. Find the sum of the
areas and the sum of the perimeters of the squares formed
through this process. 10cm 10cm 10cm 10cm

Pls help me​

Answer 1

Question :

The sides of a given square is 10 cm. The mid points of its sides are joined to form a new square. Again, the mid points of the sides of this new square are joined to form another square.This process is continued indefinitely. Find the sum of the areas and the sum of the perimeters of the squares formed through this process.

Solution :

Let ABCD be a square of side 10 cm.Now it's mid- points of its side are joined together to form a new square PQRS , again the mid points of sides are joined to form another square EFGH ,and process is continuously continued indefinitely.

From the diagram :

In ∆ PCQ

<PCQ = 90° ,

Then PQ ²= PC²+QC²

Since , P and Q are mid - points of square ABCD

thus , PC = BP = QC = DQ = 5

Thus , PQ = √ 25+25 =

⇒PQ = √50= 5√2

Since; E,F,G and H are mid-points of square PQRS

Thus ,$\sf\:EP= PF =FQ = QG =\frac{PQ}{2}=\frac{5\sqrt{2}}{2}$

Now , In ∆ FQG

<FQG = 90°

Then ,FG²= FQ²+GQ²

$\sf \: FG {}^{2} = \sqrt{( \frac{5 \sqrt{2} }{2}) {}^{2} + ( \frac{5 \sqrt{2} }{2} ) {}^{2} }$

$\implies\sf \: {FG}^{2} = \sqrt{25 } = 5$

Similarly , The process the continued indefinitely .

1) Sum of the areas

= Area of square ABCD+ Area of square PQRS +Area of EFGH + ...... +∞

= 10× 10+ 5√2× 5√2+ 5× 5+.....+∞

= 100+50+25+........+∞

Clearly it is a Geometric progression series

Where a = 100 and

r = 1/2

We know that ,

Sum of infinite GP = a/1-r

$\sf\:S_{n}=\dfrac{100}{1-\frac{1}{2}}$

$\sf\:S_{n}= 200$

Therefore , The sum of the areas is 200cm²

2)sum of the perimeters of the squares formed

= perimeter of square ABCD+ Perimeter of square PQRS + Perimeter of square PQRS+....+∞

= (4×10)+ (4×5√2)+(4×5)+......+∞

= 40+20√2+20+......+∞

In this Infinte geometric series :

a = 40 and r = √2/2

Sum of infinite GP = a/1-r

$\sf\:S_{n} = \dfrac{40}{1-\frac{ \sqrt{2} }{2}}$

$\implies\sf\:S_{n}=\dfrac{40 \sqrt{2} }{ \sqrt{2} - 1 }$

$\implies\sf\:S_{n}=\dfrac{40 \sqrt{2} }{ \sqrt{2} - 1} \times \dfrac{ \sqrt{2} + 1 }{ \sqrt{2} + 1 }$

$\sf\:S_{n} = \dfrac{40 \sqrt{2}( \sqrt{2} + 1) }{( \sqrt{2}) {}^{2} - 1 { }^{2} }$

$\implies\sf\:S_{n}=80 + 40 \sqrt{2}$

Therefore , The sum of the perimeters of the squares formed through this process is 80+40√2.

Answer 2

Let the side of the larger (given) square be $\sf{a_1,}$ so the area of the square is $\sf{(a_1)^2}$ and perimeter is $\sf{4a_1.}$

Let the side length of the square formed by joining the midpoints be $\sf{a_1}$ whose area will be,

$\longrightarrow\sf{(a_2)^2=\left(\dfrac{a_1}{2}\right)^2+\left(\dfrac{a_1}{2}\right)^2}$

$\longrightarrow\sf{(a_2)^2=\dfrac{(a_1)^2}{4}+\dfrac{(a_1)^2}{4}}$

$\longrightarrow\sf{(a_2)^2=\dfrac{(a_1)^2}{2}}$

And the perimeter is,

$\longrightarrow\sf{4a_2=\dfrac{4a_1}{\sqrt2}}$

The area of the new square formed by joining the midpoints of sides of this square will be,

$\longrightarrow\sf{(a_3)^2=\dfrac{(a_2)^2}{2}}$

$\longrightarrow\sf{(a_3)^2=\dfrac{(a_1)^2}{2^2}}$

And perimeter is,

$\longrightarrow\sf{4a_3=\dfrac{4a_2}{\sqrt2}}$

$\longrightarrow\sf{4a_3=\dfrac{4a_1}{(\sqrt2)^2}}$

And that of the new square formed by joining midpoints of sides of this square will be,

$\longrightarrow\sf{(a_4)^2=\dfrac{(a_3)^2}{2}}$

$\longrightarrow\sf{(a_4)^2=\dfrac{(a_1)^2}{2^3}}$

And perimeter is,

$\longrightarrow\sf{4a_4=\dfrac{4a_3}{\sqrt2}}$

$\longrightarrow\sf{4a_4=\dfrac{4a_1}{(\sqrt2)^3}}$

Hence the area of $\sf{n^{th}}$ square formed (including given square) is given by,

$\longrightarrow\sf{(a_n)^2=\dfrac{(a_1)^2}{2^{n-1}}}$

And perimeter is,

$\longrightarrow\sf{4a_n=\dfrac{4a_1}{(\sqrt2)^{n-1}}}$

So the sum of areas of the infinite squares formed (excluding the given square) is,

$\displaystyle\longrightarrow\sf{A=\sum_{i=2}^{\infty}(a_i)^2}$

$\displaystyle\longrightarrow\sf{A=\sum_{i=2}^{\infty}\dfrac{(a_1)^2}{2^{i-1}}}$

$\displaystyle\longrightarrow\sf{A=(a_1)^2\sum_{i=2}^{\infty}\dfrac{1}{2^{i-1}}\quad\quad\dots(1)}$

The sum $\displaystyle\sf{\sum_{i=2}^{\infty}\dfrac{1}{2^{i-1}}}$ is an infinite geometric series with first term $\sf{\dfrac{1}{2^{2-1}}=\dfrac{1}{2}}$ and common ratio $\sf{\dfrac{1}{2},}$ which is given by,

$\longrightarrow\displaystyle\sf{\sum_{i=2}^{\infty}\dfrac{1}{2^{i-1}}=\dfrac{\dfrac{1}{2}}{\left(1-\dfrac{1}{2}\right)}}$

$\longrightarrow\displaystyle\sf{\sum_{i=1}^{\infty}\dfrac{1}{2^{i-1}}=1}$

since $\displaystyle\sf{\sum_{i=1}^{\infty}ar^{i-1}=\dfrac{a}{1-r}}$ for $\sf{|r|\ \textless\ 1.}$

Hence (1) becomes,

$\displaystyle\longrightarrow\sf{A=(a_1)^2}$

Since $\sf{a_1=10\ cm}$ as per the question,

$\displaystyle\longrightarrow\sf{A=10^2\ cm^2}$

$\displaystyle\longrightarrow\sf{\underline{\underline{A=100\ cm^2}}}$

And sum of perimeters of the infinite squares formed (excluding the given one) is,

$\displaystyle\longrightarrow\sf{P=\sum_{i=2}^{\infty}4a_i}$

$\displaystyle\longrightarrow\sf{P=\sum_{i=2}^{\infty}\dfrac{4a_1}{\left(\sqrt2\right)^{i-1}}}$

$\displaystyle\longrightarrow\sf{P=4a_1\sum_{i=2}^{\infty}\dfrac{1}{\left(\sqrt2\right)^{i-1}}\quad\quad\dots(2)}$

Here the sum $\displaystyle\sf{\sum_{i=2}^\infty\dfrac{1}{\left(\sqrt2\right)^{i-1}}}$ is also an infinite geometric series with first term $\sf{\dfrac{1}{\left(\sqrt2\right)^{2-1}}=\dfrac{1}{\sqrt2}}$ and common ratio $\sf{\dfrac{1}{\sqrt2},}$ which is given by,

$\longrightarrow\displaystyle\sf{\sum_{i=2}^\infty\dfrac{1}{\left(\sqrt2\right)^{i-1}}=\dfrac{\dfrac{1}{\sqrt2}}{\left(1-\dfrac{1}{\sqrt2}\right)}}$

$\longrightarrow\displaystyle\sf{\sum_{i=2}^\infty\dfrac{1}{\left(\sqrt2\right)^{i-1}}=\dfrac{1}{\sqrt2-1}}$

$\longrightarrow\displaystyle\sf{\sum_{i=2}^\infty\dfrac{1}{\left(\sqrt2\right)^{i-1}}=\dfrac{\sqrt2+1}{(\sqrt2-1)(\sqrt2+1)}}$

$\longrightarrow\displaystyle\sf{\sum_{i=2}^\infty\dfrac{1}{\left(\sqrt2\right)^{i-1}}=1+\sqrt2}$

Hence (2) becomes,

$\displaystyle\longrightarrow\sf{P=4a_1\left(1+\sqrt2\right)}$

Taking $\sf{a_1=10\ cm,}$

$\displaystyle\longrightarrow\sf{\underline{\underline{P=40\left(1+\sqrt2\right)\ cm}}}$

Hence the sum of areas is $\bf{100\ cm^2}$ and that of perimeters is $\bf{40\left(1+\sqrt2\right)\ cm.}$