Ex.5.3 Q18 Arithmetic Progressions Solution - NCERT Maths Class 10

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A spiral is made up of successive semicircles, with centres alternately at \(A\) and \(B,\) starting with centre at \(A\) of radii \(0.5,\) \(1.0 \,\rm{cm}, 1.5 \,\rm{cm}, 2.0 \,\rm{cm}, ......... \)as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles?

\(\begin{align}\left[ {{\rm{Take}}\,\,\pi = \frac{{22}}{7}} \right] \end{align}\)

 Video Solution
Arithmetic Progressions
Ex 5.3 | Question 18

Text Solution

What is Known?

Radii of the \(13\) semicircles.

What is Unknown?

Length of the spiral made.


Sum of the first \(n\)terms of an AP is given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\)

Where \(a\) is the first term, \(d\)is the common difference and \(n\) is the number of terms.


Semi-perimeter of circle, \(l = \pi r\)

\[\begin{align}{{l}_{1}}& =\pi \times \left( 0.5\,\,\text{cm} \right)=0.5\pi \,\,\text{cm } \\  {{l}_{2}}&=\pi \times \left( 1\,\,\text{cm} \right)=\pi \,\,\text{cm} \\ {{l}_{3}} &=\pi \times \left( 1.5\,\,\text{cm} \right)=1.5\pi \,\text{cm} \\\end{align}\]

Therefore, \({l_1},{\rm{ }}{l_2},{\rm{ }}{l_3},\) i.e. the lengths of the semi-circles are in an A.P.,

\[\begin{align}&0.5\pi ,\pi ,1.5\pi ,2\pi\dots\dots\dots\\&a = 0.5\pi \\&d = \pi - 0.5\pi = 0.5\pi\end{align}\]

We know that the sum of \(n\) terms of an A.P. is given by

\[\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{13}} &\!=\!\frac{{13}}{2}\left[ {2 \!\times\! \left( {0.5\pi } \right)\!+\!\left( {13\!-\!1} \right)\left( {0.5\pi } \right)}\!\right]\\ &= \frac{{13}}{2}\left[ {\pi + 6\pi } \right]\\ &= \frac{{13}}{2} \times 7\pi \\ &= \frac{{13}}{2} \times 7 \times \frac{{22}}{7}\\ &= 143\end{align}\]

Therefore, the length of such spiral of thirteen consecutive semi-circles will be \(143\, \rm{cm.}\)