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Three equal circles, each of radius 6 cm, touch one another as shown in the figure. Find the area enclosed between them. [Take π = 3.14 and √3 = 1.732.]
Consider the above figure,
Here, first we join the center of all adjacent circles then the distance between the center of circles touching each other is equal to the side of an equilateral triangle formed by joining the center of adjacent circles. Therefore, we can say that the side of the equilateral triangle is equal to the twice of the radius of circle. Now by simply calculating the area of the 3 sectors and then subtracting it from the area of the equilateral triangle we can easily calculate the area of the enclosed region.
Given radius of each circle = r = 6 cm
Central angle of each sector = θ = 60° (∵ ∆ABC is equilateral)
Side of equilateral ∆ABC = a = 2×r = 2×6
∴ Side of equilateral ∆ABC = a = 12 cm
⇒ Area of one sector = 6π cm2→ eqn1
Area of all the 3 sector = 3×Area of one sector
= 3×6π (from eqn1)
= 18π cm2→ eqn2
Area of enclosed region = Area of equilateral ∆ABC – Area of all 3 sectors
⟹ Area of enclosed region = 36√3-18π (from eqn 3 and eqn 2)
⟹ Area of enclosed region = (36×1.732)-(18×3.14)
(put π = 3.14 &√3 = 1.732)
= 62.352 - 56.52
= 5.832 cm2
The area of enclosed region is 5.832 cm2.
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.]