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A boy is cycling in such a way that the wheels of his bicycle are making 140 revolutions per minute. If the diameter of a wheel is 60 cm, calculate the speed (in km/h) at which the boy is cycling.
Given diameter of wheel = d = 60 cm
Number of revolutions in one minute = 140
Number of revolutions in one hour = 140×60
Number of revolutions in one hour = 8400
Distance covered in one revolution = circumference of wheel
⇒ Distance covered in one revolution = πd
= 188.57 cm
Distance covered in one hour = Distance in 1 revolution × no. of revolutions
⇒ Total distance covered in one hour = 188.57× 8400
⇒ Total distance covered in one hour = 1583988 cm = 15.839 km
∴ speed with which boy is cycling = 15.839 km/hr
The speed with which boy is cycling is 15.839 km/hr
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.]