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A path of 8 m width runs around the outside of a circular park whose radius is 17 m. Find the area of the path.
Given radius of circular park = R = 17 m
Width of the circular path outside the park = d = 8 m
Therefore, the radius of the outer circle = R’ = R + d
Outer radius = R’ = 17 + 8
R’ = 25 m
Area of inner circle = πR2 and,
Area of outer circle = πR’2
Area of path = Area of outer circle – Area of inner circle
= πR’2 – πR2 (put values of R’ & R)
= π(252) – π(172)
= π(252 – 172) (taking π common from R.H.S)
= π(625 – 289)
(put π = 22/7)
= 1056 m2
The area of the path is 1056 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.]