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A chord of a circle of radius 30 cm makes an angle of 60° at the centre of the circle. Find the areas of the minor and major segments. [Take π = 3.14 and √3 = 1.732.]
Area of minor segment = Area of sector – Area of triangle → eqn1
Where R = radius of the circle and θ = central angle of the sector
∴ Area of sector = 471 cm2
Where a = side of the triangle
Area of triangle = √3/4×30×30
Area of triangle = √3/4×900
Area of triangle = (900×√(3 ))/4
(put √3 = 1.732)
Area of triangle = (1.732×900)/4
∴ Area of triangle = 389.7 cm2
Put the values of area of triangle and area of sector in equation 1
Area of minor segment = 471 – 389.7
⇒ Area of minor segment = 81.3 cm2
Area of major segment = πR2 – Area of minor segment
Put the value of area of minor segment and R in above equation
⇒ Area of major segment = π×(302) – 81.3 (put π = 3.14)
⇒ Area of major segment = 3.14×30×30 – 81.3
⇒ Area of major segment = 2826 – 81.3
= 2744.7 cm2
Area of major segment is 2744.7cm2 and of minor segment is 81.3 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.]