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The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of distances travelled by their tips in 2 days. [Take π = 3.14.]
In an hour the minute hand completes one rotation therefore in 24 hours the minute hand will complete 24 rotations similarly the hour hand completes one rotation in 12 hours therefore in 24 hours it will complete 2 rotations. Now we have to just calculate the perimeter of the circle traced by minute hand and hour hand and multiply it with the number of rotations of minute hand and hour hand in 2 days respectively.
Length of short/hour hand = r = 4 cm
Length of long/minute hand = R = 6 cm
∴ The perimeter of circle traced by short hand = p = 2πr → eqn1
∴ The perimeter of circle traced by Long hand = P = 2πR → eqn2
Now put the value of ‘r’ and ‘R’ in the equation 1 and 2 respectively.
⇒ p = 2π(4) & P = 2π(6) (put π = 3.14)
⇒ p = 2×3.14×4 & P = 2×3.14×6
∴ p = 25.12 cm & P = 37.68 cm
Therefore, distance covered by short hand in one rotation = 25.12 cm
Distance covered by long hand in one rotation = 37.68 cm
Number of rotation of short hand in one day = 2
Number of rotation of long hand in one day = 24
Therefore number of rotation of small hand in two days = 4
Number of rotation of long hand in two days = 48
Total distance covered by long hand in 2 days = P × no. of rotations in 2 days
⇒ Total distance covered by long hand in 2 days = 37.68×48
⇒ Total distance covered by long hand in 2 days = 1808.64 cm → eqn3
Total distance covered by short hand in 2 days = p × no. of rotations in 2 days
⇒ Total distance covered by short hand in 2 days = 25.12×24
⇒ Total distance covered by short hand in 2 days = 100.48 cm → eqn4
Now total distance covered by tip of both hands in 2 days = eqn3 + eqn4
⇒ Total distance covered by both hands in 2 days = 1808.64 + 100.48
⇒ Total distance covered by both hands in 2 days = 1909.12 cm
The distance covered by both hands tip in 2 days is 1909.12 cm
ABCD is a field in the shape of a trapezium, AD || BC, ∠ABC = 90° and ∠ADC = 60°. Four sectors are formed with centres A, B, C and D, as shown in the figure. The radius of each sector is 14 m.
Find the following:
(i) total area of the four sectors,
(ii) area of the remaining portion, given that AD = 55 m, BC = 45 m and AB = 30 m.
A child draws the figure of an aeroplane as shown. Here, the wings ABCD and FGHI are parallelograms, the tail DEF is an isosceles triangle, the cockpit CKI is a semicircle and CDFI is a square. In the given figure, BP ⊥ CD, HQ ⊥ FI and EL ⊥ DF. If CD = 8 cm, BP = HQ = 4 cm and DE = EF = 5 cm, find the area of the whole figure. [Take π = 3.14.]