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A racetrack is in the form of a ring whose inner circumference is 352 m and outer circumference is 396 m. Find the width and the area of the track.
Consider the race track as shown below,
The inner and outer radius of track is ‘r’ cm and ‘R’ cm respectively.
Let inner and outer circumference be ‘C1’ and C2’ respectively.
C1 = 352 m and C2 = 396 m.
Circumference of circle = 2πr
Where r = radius of the circle
C1 = 2πr and C2 = 2πR
⇒ 2πr = 352 and 2πR = 396
⇒ r = 56 m and R = 63 m
So, the width of the race track = R – r,
⇒ Width of the race track = 63 – 56
⇒ Width of the race track = 7 m
Area of race track = area of outer circle – area of inner circle
⇒ Area of track = πR2 – πr2 (put values of r and R)
⇒ Area of track = π(632) – π(562)
⇒ Area of track = π(632 – 562) (taking π common from R.H.S)
⇒ Area of track = π(3969 – 3136)
⇒ Area of track = π×833
= 2618 m2
The width of tack is 7 m and area of track is 2618 m2.
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.]