Question

the short and long hands of a clock are 4 cm and 6 cm long. Find the sum of distances travelled by thier tips in 48 hours

Answer 1

Answer:

IN 48 HOURS THE SHORT HAND WILL COMPLETE 4 ROUNDS

SO , DISTANCE MOVED BY ITS TIP = 4(CIRCUMFERENCE OF CIRCLE MADE BY SHORT HAND)

                                                              OR , 4(2 X 22/7 X 4) CM

                                                              OR , 4(176/7) CM

                                                              OR , 4 X 176/7 CM

                                                              OR , 704/7 CM

IN 48 HOURS THE LONG HAND WILL COMPLETE 48 ROUNDS

SO , DISTANCE MOVED BY ITS TIP = 48(CIRCUMFERENCE OF CIRCLE MADE BY SHORT HAND)

                                                              OR , 48(2 X 22/7 X 6) CM

                                                              OR , 48(264/7) CM

                                                              OR , 48 X 264 /7 CM

                                                              OR , 12672/7 CM

HENCE , SUM THE DISTANCES MOVED BY THE TIPS OF TWO HANDS OF THE CLOCK = (704/7 + 12672/7) CM

                      OR , (704 + 12672/7) CM

                      OR , (13376/7) CM

                      ANS =  1910.85 CM

HOPE U WILL UNDERSTAND IT

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.