# The given figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:(i) the distance around the track along its inner edge.(ii) the area of the track.

AcademicMathematicsNCERTClass 10

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Given:

The given figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long.

The track is 10 m wide.

To do:

We have to find

(i) the distance around the track along its inner edge.

(ii) the area of the track.

Solution:

(i) Circumference of the semicircular ends at both sides $=\frac{2 \pi r}{2}$

$=\pi r$
Radius of the circular ends on both sides $r=\frac{60}{2}$

$=30 \mathrm{~cm}$
Circumference of the semicircular ends $=\frac{22}{7} \times 30$

$=\frac{660}{7} \mathrm{~cm}$

Total length of the semicircular ends $=\frac{2 \times 660}{7}$

$=\frac{1320}{7} \mathrm{~cm}$

Total length of the inner racing track $=106+106+\frac{1320}{7}$

$=212+\frac{1320}{7}$

$=\frac{1484+1320}{7}$

$=\frac{2804}{7} \mathrm{~cm}$

The distance around the track along its inner edge is $\frac{2804}{7} \mathrm{~cm}$.

(ii) Radius of the outer semicircular end $=30+10$

$=40 \mathrm{~cm}$

Area of the outer semicircular ends $=\frac{1}{2}(\pi r^{2})$

$=\frac{1}{2} \times \frac{22}{7} \times 40 \times 40$

$=\frac{11 \times 40 \times 40}{7}$

$=\frac{17600}{7} \mathrm{~cm}^{2}$

Area of the inner semicircular ends $=\frac{1}{2} \times \frac{22}{7} \times 30 \times 30$

$=\frac{11 \times 900}{7}$

$=\frac{9900}{7} \mathrm{~cm}^{2}$

Area of the track between semicircular ends $=\frac{17600}{7}-\frac{9900}{7}$

$=\frac{7700}{7} \mathrm{~cm}^{2}$

$=1100 \mathrm{~cm}^{2}$

Area of the tracks at both semicircular ends $= 2 \times 1100$

$= 2200\ cm^2$

Area of the two rectangular portions $= 2 l \times h$

$= 2 \times 106 \times 10$

$= 2120\ cm^2$

Total area of the track $=$ Area of the track at semicircular ends $+$ Area of the rectangular portions

$= 2200 + 2120$

$= 4320\ cm^2$

Updated on 10-Oct-2022 13:24:10

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