Question

A spiral is made up of successive semicircles, with centres alternately at $A$ and $B,$ starting with centre at $A$. If the radii are in sequence, $0.5\text{cm},1.0\text{cm},1.5\text{cm},2.0\text{cm},\dots$ as shown in figure below.


Then the total length of the spiral made by thirteen consecutive semicircles is
$(\text{Take}\pi =\frac{22}{7})$

• A. $133\text{cm}$
• B. $140\text{cm}$
• C. $143\text{cm}$
• D. $150\text{cm}$

Answer 1

The correct option is C $143\text{cm}$

We know, the perimeter of a semi-circle
$=\pi r$
Now,
$P_1=\pi \left(0.5\right)=\frac{\pi }{2}\text{cm}{P}_2=\pi \left(1\right)=\pi \text{cm}{P}_3=\pi \left(1.5\right)=\frac{3\pi }{2}\text{cm}$
Given sequence is an A.P. whose
$a=\frac{\pi }{2},d=\frac{\pi }{2}$

Therefore, the required length of the spiral
$S_{13}=\frac{13}{2}[2a+(13-1\left)d\right]\Rightarrow S_{13}=\frac{13}{2}[\pi +6\pi ]\Rightarrow S_{13}=\frac{13}{2}\left[7\pi \right]=\frac{13}{2}\times \frac{7\times 22}{7}\therefore S_{13}=13\times 11=143\text{cm}$