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Mensuration | Mensuration Exercise 20E

Question 3

A racetrack is in the form of a ring whose inner circumference is 528 m and the outer circumference is 616 m. Find the width of the track.

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Let the inner and outer radii of the track be r meters and R meters respectively.

Then,

= 2πr = 528 m

= 2 × (22/7) × r = 528

= r = (528/ (2 × (22/7)))

= r = (528/ 2) × (7/22)

= r = (264/2) × (7/11)

= r = 132 × 0.6364

= r = 84 m

And,

= 2πR = 616 m

= 2 × (22/7) × R = 616

= R = (616/ (2 × (22/7)))

= R = (616/ 2) × (7/22)

= R = (308/2) × (7/11)

= R = 154 × 0.6364

= R = 98 m

Now,

(R – r) = (98 – 84)

= 14 m

Hence, the width of the track is 14 m

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Our top 5% students will be awarded a special scholarship to Lido.

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Question 3

A racetrack is in the form of a ring whose inner circumference is 528 m and the outer circumference is 616 m. Find the width of the track.

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

Let the inner and outer radii of the track be r meters and R meters respectively.

Then,

= 2πr = 528 m

= 2 × (22/7) × r = 528

= r = (528/ (2 × (22/7)))

= r = (528/ 2) × (7/22)

= r = (264/2) × (7/11)

= r = 132 × 0.6364

= r = 84 m

And,

= 2πR = 616 m

= 2 × (22/7) × R = 616

= R = (616/ (2 × (22/7)))

= R = (616/ 2) × (7/22)

= R = (308/2) × (7/11)

= R = 154 × 0.6364

= R = 98 m

Now,

(R – r) = (98 – 84)

= 14 m

Hence, the width of the track is 14 m

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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