Question

The length of a side of a square is ‘a’ meter. A second square is formed by joining the middle points of this square. Then a third square is formed by joining the middle points of the second square and so on. Then the sum of the area of the squire which carried upto infinity is
A) $a^{2}$
B) $2a^{2}$
C) $3a^{2}$
D) $4a^{2}$

Answer 1

Hint: In this question it is given that the length of a side of a square is ‘a’ meter. A second square is formed by joining the middle points of this square. Then a third square is formed by joining the middle points of the second square and so on. Then we have to find the sum of the area of the squire. So to understand it in better way, we have to draw the diagram,

So we can easily find the area of the outer circle, i.e, Side$\times$Side, so to find the area of ${1}^{st}$ inner circle, we first need to find its side, as we can see from the diagram $\triangle GCF$ forming a right angle triangle, where GF is the hypotenuse, so we can easily obtain the length GF by pythagorean theorem which states that,$$h^{2}=p^{2}_{{}}+b^{2}$$,
Where h= hypotenuse, p= perpendicular, b= base of a right angle triangle.
So in this way we are able to find the side of the ${2}^{nd}$ inner circle and so on.
And by adding their area upto infinity we will get our required solution.

Complete step-by-step solution:
Let us consider ABCD to be a square whose side is ‘a’, and E, F, G, H are the midpoint of side AB, BC, CD, DA respectively. By joining the mid points we will get another square EFGH.
Now area of square ABCD = Side $\times$ Side = a$\times$a=$a^{2}$........(1)

Now since F is the midpoint of BC , then, BF=FC=$$\dfrac{a}{2}$$.
Now from $\triangle GCF$, by using pythagoras theorem, we get,
$$h^{2}=p^{2}_{{}}+b^{2}$$
$$\Rightarrow GF^{2}=CF^{2}+GC^{2}$$
$$\Rightarrow GF^{2}=\left( \dfrac{a}{2} \right)^{2} +\left( \dfrac{a}{2} \right)^{2} $$
$$\Rightarrow GF^{2}=2\cdot \left( \dfrac{a}{2} \right)^{2} $$
$$\Rightarrow GF^{2}=\dfrac{a^{2}}{2}$$
$$\Rightarrow GF=\dfrac{a}{\sqrt{2} }$$
So the area of first inner square EFGH = Side $\times$ Side = $$\dfrac{a}{\sqrt{2} } \times \dfrac{a}{\sqrt{2} }$$=$$\dfrac{a^{2}}{2}$$.....(2)

Now by following a similar process for the ${2}^{nd}$ inner square, we can find the sides are $$\dfrac{a}{2}$$.
Therefore area= $$\dfrac{a}{2} \times \dfrac{a}{2} =\dfrac{a^{2}}{4}$$.....(3)
Similarly, For the ${3}^{rd}$ inner square the sides will be $$\dfrac{a}{2\sqrt{2} }$$ and the area $$\dfrac{a^{2}}{8}$$.....(4)
and so on……..,

Therefore the sum of the area of the squares
A= $$a^{2}+\dfrac{a^{2}}{2} +\dfrac{a^{2}}{4} +\dfrac{a^{2}}{8} +\cdots \cdots$$............(5)
Now the above series is in the form of G.P(Geometric Progression), whose first term is $$a^{2}$$ and common ratio is $$\dfrac{1}{2}$$
Therefore the sum of the above series is,
A= $$a^{2}+\dfrac{a^{2}}{2} +\dfrac{a^{2}}{4} +\dfrac{a^{2}}{8} +\cdots \cdots$$
 =$$\dfrac{a^{2}}{\left( 1-\dfrac{1}{2} \right) }$$
 =$2a^{2}$.
Hence the correct option is option B.

Note: If any series follows infinite geometric progression i.e, $$a+ar+ar^{2}+\cdots$$, whose first term is ‘a’ and common ratio ‘r’, then the summation of the series, $$S_{\infty }=\dfrac{a}{\left( 1-r\right) }$$.
Also in the above diagram we only draw upto fourth square, since it is not possible to draw the squares upto infinity and the length of sides of each inner square is dependent upon its outer square, like if you know the side of outermost square, then its first inner square side is $$\dfrac{a}{\sqrt{2} }$$, i.e, dividing side by $\sqrt{2}$, similarly the ${2}^{nd}$ inner square side will be $$\dfrac{\left( \dfrac{a}{\sqrt{2} } \right) }{\sqrt{2} } =\dfrac{a}{2}$$ i.e, dividing the first inner square side by $\sqrt{2}$, and so on.