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The wheels of a car make 2500 revolutions in covering a distance of 4.95 km. Find the diameter of a wheel.
Let the diameter of the wheel be ‘d’ cm
Total distance covered in 250 revolutions = 49.5 km = 495000 m
⇒ Distance covered in one revolution = 198 cm → eqn1
Also, Distance covered in one revolution = circumference of wheel
∴ Distance covered in one revolution = πD where d = diameter of wheel
Equate equation 1 and 2 we get,
⟹ d = 9×7
⟹ d = 63 cm
The diameter of the wheel is 63 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.]