
Concentric circles of radii 1,2,3….,100cm are drawn. The interior of the smallest circle coloured red and the angular regions are coloured alternately green and red, so that no two adjacent regions are the same colour. The total area of the green region in sq.cm is equal to
A. $1000\pi $
B. $5050\pi $
C. $4950\pi $
D. $5151\pi $
Answer
617.4k+ views
Hint: In order to solve this question, first convert the areas of concentric circles in the form of sum series of an AP. Then we calculate the total no. of terms. After that we will get the total area by using the formula of sum of n terms of an AP.
Complete step-by-step answer:
We should also know that concentric circles are those circles who have the same centre.
Area of the smallest circle( radius = 1cm ) is red coloured so the green coloured area will be for the second circle, fourth and so on.
Area of circle coloured green=$\pi \left[ {{2^2} - {1^2}} \right]$
$\therefore $ Total area of green coloured regions
$
= \pi \left[ {{2^2} - {1^2}} \right] + \pi \left[ {{4^2} - {3^2}} \right] + \pi \left[ {{6^2} - {5^2}} \right] + ..... + \pi \left[ {{{100}^2} - {{99}^2}} \right] \\
= \pi \left[ {\left( {2 - 1} \right)\left( {2 + 1} \right) + \left( {4 - 3} \right)\left( {4 + 3} \right) + \left( {6 - 5} \right)\left( {6 + 5} \right) + ...\left( {100 - 99} \right)\left( {100 + 99} \right)} \right] \\
= \pi \left[ {3 + 7 + 11.... + 199} \right] \\
$
Here we can clearly see that it forms an AP where
a = 3, ${a_n} = 199$, d = 4
Now, we know that the formula for ${n^{th}}$ term of an AP is
${a_n} = a + \left( {n - 1} \right) \times d$.
Or $n = \dfrac{{{a_n} - a}}{d} + 1$ which is the simplified form of this.
No. of terms in 3,7,11….199= $\dfrac{{199 - 3}}{4} + 1$
$
= \dfrac{{196}}{4} + 1 \\
= 50 \\
$
Now we know that sum of n terms of an AP is
${s_n} = \dfrac{n}{2} \times \left( {a + {a_n}} \right)$
Total area= $\pi \times 50 \times \left[ {\dfrac{{3 + 199}}{2}} \right]$
$
= \pi \times 50 \times 101 \\
= 5050\pi \\
$
Hence, option (B) is correct.
Note: We should know that during the formation of an AP one can mention the wrong way of making the areas of concentric circles in series. Also one must remember the formula of sum of terms and no. of terms of an AP as it can lead to making mistakes in the solution.
Complete step-by-step answer:
We should also know that concentric circles are those circles who have the same centre.
Area of the smallest circle( radius = 1cm ) is red coloured so the green coloured area will be for the second circle, fourth and so on.
Area of circle coloured green=$\pi \left[ {{2^2} - {1^2}} \right]$
$\therefore $ Total area of green coloured regions
$
= \pi \left[ {{2^2} - {1^2}} \right] + \pi \left[ {{4^2} - {3^2}} \right] + \pi \left[ {{6^2} - {5^2}} \right] + ..... + \pi \left[ {{{100}^2} - {{99}^2}} \right] \\
= \pi \left[ {\left( {2 - 1} \right)\left( {2 + 1} \right) + \left( {4 - 3} \right)\left( {4 + 3} \right) + \left( {6 - 5} \right)\left( {6 + 5} \right) + ...\left( {100 - 99} \right)\left( {100 + 99} \right)} \right] \\
= \pi \left[ {3 + 7 + 11.... + 199} \right] \\
$
Here we can clearly see that it forms an AP where
a = 3, ${a_n} = 199$, d = 4
Now, we know that the formula for ${n^{th}}$ term of an AP is
${a_n} = a + \left( {n - 1} \right) \times d$.
Or $n = \dfrac{{{a_n} - a}}{d} + 1$ which is the simplified form of this.
No. of terms in 3,7,11….199= $\dfrac{{199 - 3}}{4} + 1$
$
= \dfrac{{196}}{4} + 1 \\
= 50 \\
$
Now we know that sum of n terms of an AP is
${s_n} = \dfrac{n}{2} \times \left( {a + {a_n}} \right)$
Total area= $\pi \times 50 \times \left[ {\dfrac{{3 + 199}}{2}} \right]$
$
= \pi \times 50 \times 101 \\
= 5050\pi \\
$
Hence, option (B) is correct.
Note: We should know that during the formation of an AP one can mention the wrong way of making the areas of concentric circles in series. Also one must remember the formula of sum of terms and no. of terms of an AP as it can lead to making mistakes in the solution.
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