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 $\pi = \frac {22}{7}$)Fig. 5.4

Answer 1

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 $143 \ cm$.