Question

The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of the distances travelled by their tips in 2 days. (Take π=22/7)
Areas related to circles

Answer 1

2....days, the short hand will complete 4 rounds.


Therefore, distance moved = 4 (circumference of a circle of radius 4cm)
= 4 (2 * 22/7 * 4) cm = 707/7 cm
In 2 days the long hand will complete 48 rounds.
Therefore, distance moved by its tips = 48 (circumference of a circle of radius 6cm)
= 48 (2 * 22/7 * 6) cm = 12672/7 cm
Hence, sum of the distances moved by the tips of two hands of the clock
= 707/7 cm + 12672/7 cm
=1910.86 cm..

Answer 2

Answer:

Given :

Length of short hand = 4 cm

Length of long hand = 6 cm

We know short hand make 1 revolution in 12 hours while long hand in 1 hour.

Distance travelled by short hand in 48 hours = 4 × 2 π × 4 = 32 π cm

Distance travelled by long hand in 48 hours = 48 × 2 π × 6 = 576 π cm

Total distance travelled = 32 π cm +  576 π cm = 608 π cm

Therefore ,  sum of distances travelled by their tips in two day is 608 π cm.