Question

Today's interesting but hard geometric Question.
find the area of blue region.
it is a square of side length of 1cm.
don't scam for 5 points.​

Answer 1

Answer:

0.215cm²

if it wrong please tell me

Answer 2

$\large\underline{\sf{Solution-}}$

Let the first square is represented as ABCD with AB = 1 cm.

So, Diameter of circle, d = 1 cm

So, Radius of circle, r = 1/2 cm

Hence, the area of first region is

$\rm :\longmapsto\:A_1 = A_{square} \: - \: A_{circle}$

$\rm \:  =  \:  \: {(1)}^{2} - \pi {\bigg(\dfrac{1}{2} \bigg) }^{2}$

$\rm \:  =  \:  \: 1 - \dfrac{\pi}{4} \:$

$\boxed{ \bf{ A_1 \:  =  \:  \: 1 - \dfrac{\pi}{4} \: {cm}^{2} }}$

Now,

Consider Region 2,

Let the square of this region is represented as EFGH such that diagonal HF = 1 cm

Let assume that side of square EFGH is 'x' cm

So, In right triangle FGH

By using Pythagoras Theorem,

$\rm :\longmapsto\: {x}^{2} + {x}^{2} = {1}^{2}$

$\rm :\longmapsto\: 2{x}^{2} = {1}^{}$

$\rm :\longmapsto\: {x}^{2} = \dfrac{1}{2}$

$\rm :\longmapsto\: {x}^{} = \dfrac{1}{ \sqrt{2} } \: cm$

So,

$\rm :\longmapsto\:Side \: of \: square \: = \dfrac{1}{ \sqrt{2} }$

and

$\rm :\longmapsto\:Radius \: of \: circle \: = \dfrac{1}{ 2\sqrt{2} }$

Hence, Area of second region is

$\rm :\longmapsto\:A_2 = A_{square} \: - \: A_{circle}$

$\rm \:  =  \:  \: {\bigg(\dfrac{1}{ \sqrt{2} } \bigg) }^{2} - \pi {\bigg(\dfrac{1}{2 \sqrt{2} } \bigg) }^{2}$

$\rm \:  =  \:  \: \dfrac{1}{2} - \dfrac{\pi}{8} \:$

$\boxed{ \rm{A_2 \:  =  \:  \: \dfrac{1}{2} - \dfrac{\pi}{8} \: {cm}^{2} }}$

Consider, Region 3,

Let the square of this region is represented as IJKL such that diagonal JL = 1 / sqrt(2) cm

Let assume that side of square IJKL is 'y' cm

So, In right triangle JKL

By using Pythagoras Theorem,

$\rm :\longmapsto\: {y}^{2} + {y}^{2} = {\bigg(\dfrac{1}{ \sqrt{2} } \bigg) }^{2}$

$\rm :\longmapsto\:2 {y}^{2} = \dfrac{1}{ 2 }$

$\rm :\longmapsto\: {y}^{2} = \dfrac{1}{ 4 }$

$\rm :\longmapsto\: {y}^{} = \dfrac{1}{ 2 }$

So,

$\rm :\longmapsto\:Side \: of \: square \: = \dfrac{1}{2 }$

and

$\rm :\longmapsto\:Radius \: of \: circle \: = \dfrac{1}{4}$

Hence, Area of third region is

$\rm :\longmapsto\:A_3 = A_{square} \: - \: A_{circle}$

$\rm \:  =  \:  \: {\bigg(\dfrac{1}{ 2 } \bigg) }^{2} - \pi {\bigg(\dfrac{1}{4 } \bigg) }^{2}$

$\rm \:  =  \:  \: \dfrac{1}{4} - \dfrac{\pi}{16} \:$

$\boxed{ \rm{A_3 \:  =  \:  \: \dfrac{1}{4} - \dfrac{\pi}{16} \: {cm}^{2} }}$

So, goes on like this

The total area of blue region is

$\rm \:  =  \:  \: A_1 + A_2 + A_3 + - - -$

$\rm \:  =  \:  \: \bigg(1 - \dfrac{\pi}{4} \bigg) + \bigg(\dfrac{1}{2} - \dfrac{\pi}{8} \bigg) + \bigg(\dfrac{1}{4} - \dfrac{\pi}{16}\bigg) + - - -$

can be re-arranged as

$\rm \:  =  \bigg(1 + \dfrac{1}{2} + \dfrac{1}{4} + - - - \bigg) - \bigg(\dfrac{\pi}{4} + \dfrac{\pi}{8} + \dfrac{\pi}{16} + - - - \bigg)$

So, these two forms an infinite GP series.

We know,

Sum of infinite geometric sequence is given by

$\boxed{ \rm{ S_ \infty \: = \: \frac{a}{1 - r} , \: provided \: that \: |r| < 1}}$

where,

a is first term of GP series

r is the common ratio of GP series.

So, using this formula, we get

$\rm \:  =  \:  \: \dfrac{1}{1 - \dfrac{1}{2} } - \pi \: \dfrac{\dfrac{1}{4} }{1 - \dfrac{1}{2} }$

$\rm \:  =  \:  \: \dfrac{1}{ \dfrac{1}{2} } - \pi \: \dfrac{\dfrac{1}{4} }{\dfrac{1}{2} }$

$\rm \:  =  \:  \: 2 - \dfrac{\pi}{2}$

$\rm \:  =  \:  \: 2 - \dfrac{22}{14}$

$\rm \:  =  \:  \: 2 - \dfrac{11}{7}$

$\rm \:  =  \:  \: \dfrac{14 - 11}{7}$

$\rm \:  =  \:  \: \dfrac{3}{7} \: {cm}^{2}$