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The Inside Perimeter of a Running Track (Shown in the Following Figure) is 400 M. the Length of Each of the Straight Portion is 90 M and the Ends Are Semi-circles. If the Track is Everywhe - Mathematics

Sum

The inside perimeter of a running track (shown in the following figure) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If the track is everywhere 14 m wide. find the area of the track. Also find the length of the outer running track.

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Solution

It is given that, `"length of each straight portion"`and `"width of track=14m"` 

We know that the circumference C of semicircle of radius be r is \[C = \pi r\] 

The inside perimeter of running track is the sum of twice the length of straight portion and circumferences of semicircles. So,
inside perimeter of running track = 400 

\[2l + 2\pi r = 400 m\] 

\[\Rightarrow 2 \times 90 + 2 \times \frac{22}{7} \times r = 400 m\]

\[\Rightarrow 2 \times 90 + 2 \times \frac{22}{7} \times r = 400 m\]

Thus, radius of inner semicircle is 35 m.

Now,
radius of outer semi circle r' = 35 + 14 = 49 m

Area of running track =\[2 \times \text{ Area of rectangle } + 2 \times \text{ Area of outer semi circle } - 2 \times \text{ Area of inner semicircle }\]

\[= 2 \times 90 \times 14 + 2 \times \frac{\pi(49 )^2}{2} - 2 \times \frac{\pi(35 )^2}{2}\]

\[= 2520 + \pi \times \left( 49 + 35 \right)\left( 49 - 35 \right)\]

\[= 2520 + \pi \times \left( 49 + 35 \right)\left( 49 - 35 \right)\]

\[= 2520 + 3696 = 6216 m^2\] 

Hence, the area of running track = 6216 m2

Now, length L of outer running track isL = 2 × l + 2 \[\pi\]

\[= 2 \times 90 + 2\pi \times 49\]

\[= 180 + 2 \times \frac{22}{7} \times 49\]

\[= 180 + 308 = 488 m\]

Hence, the length L of outer running track is 488 m.

 

 

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 13 Areas Related to Circles
Exercise 13.4 | Q 11 | Page 57
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