Question -
Answer -
┬аstarting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, тАжтАжтАж as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take┬а╧А = 22/7)
Solution
We know,
Perimeter of a semi-circle = ╧Аr
Therefore,
P1┬а= ╧А(0.5) = ╧А/2 cm
P2┬а= ╧А(1) = ╧А cm
P3┬а= ╧А(1.5) = 3╧А/2 cm
Where, P1,┬аP2,┬аP3┬аarethe lengths of the semi-circles.
Hence we got a series here, as,
╧А/2, ╧А, 3╧А/2, 2╧А, тАж.
P1┬а= ╧А/2 cm
P2┬а= ╧А cm
Common difference, d┬а=┬аP2┬атАУ P1┬а=┬а╧АтАУ ╧А/2 = ╧А/2
First term =┬аP1=┬аa┬а=┬а╧А/2cm
By the sum of n term formula, we know,
Sn┬а=┬аn/2┬а[2a┬а+ (n┬атАУ1)d]
Therefor, Sum of the length of 13 consecutivecircles is;
S13┬а=┬а13/2┬а[2(╧А/2)┬а+ (13┬атАУ 1)╧А/2]
= ┬а13/2┬а[╧А┬а+ 6╧А]
=13/2┬а(7╧А)
=┬а13/2 ├Ч 7 ├Ч┬а22/7
= 143 cm