Question:

A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, .... as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take \(\pi = \frac {22}{7}\))
A spiral is made up of successive semicircles
Fig. 5.4

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Length of successive semicircles is \(l_1, l_2, l_3, l_4, ......\) with centres at \(A, B, A, B, ....\), respectively.

Updated On: Jun 15, 2024
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Solution and Explanation

Semi-perimeter of circle \(= \pi𝑟\)
\(l_1 = \pi (0.5) = \frac {\pi}{2} \ cm\)
\(l_2 = \pi (1) = \pi\ cm\)
\(l_3 = \pi (1.5) = \frac {3\pi}{2} cm\)
Therefore, \(l_1, l_2, l_3\), i.e. the lengths of the semi-circles are in an A.P.,
\(\frac \pi2, π, \frac {3\pi}{2}, 2\pi, ......\)
\(a = \frac \pi2\)
\(d = \pi - \frac \pi2 = \frac \pi2\)
\(S_{13} =?\)
We know that the sum of n terms of an a A.P. is given by
\(S_n = \frac n2[2a + (n-1)d]\)

\(S_{13}\) \(=\) \(\frac {13}{2}[2(\frac \pi2) + (13-1)(\frac \pi2)]\)

\(S_{13}\) \(= \frac {13}{2}[\pi + \frac {12\pi}{2}]\)

\(S_{13}\) \(= (\frac{13}{2})(7\pi)\)

\(S_{13}\) \(= \frac {91π}{2}\)
\(S_{13}\) \(= 91 \times \frac {22}{2} \times 7\)
\(S_{13}\) \(= 13 \times 11\)
\(S_{13}\) \(=143\)

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

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