Question

The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of the distances travelled their tip s in 2 days (48 hours).

Answer 1

Solution:-
Short hand is the hour hand and it completes 1 revolution in 12 hours.
In 12 hours, distance = 2πr = 2*22/7*4
And, in 2 days or 48 hours, the distance covered = 2*4*22/7*4
= 704/7
= 100.57 cm
Long hand is the minute hand and it completes 1 revolution in 1 hour.
So, in 48 hours, it will take 48 revolutions.
= 2*22/7*6*48
= 12672/7
= 1810.285
Total distance covered by the two hands = 100.57 + 1810.285
= 1910.855 cm
Answer.

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Answer 2

Ans..

No. of times shorter hand moves in one day= 2

''          ''             ''                                2 days= 4

''        ''           larger               ''        2days= 48

Distance covered by shorter hand in 2 days =4×2πr = 704/7

''             ''                      larger ''             ''            = 48×2πr = 12676/7

sum of distances = 13376/7= 1910.85 cm (ans)

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