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The area of an equilateral triangle is 49√3 cm2. Taking each angular point as centre, circles are drawn with radius equal to half the length of the side of the triangle. Find the area of the triangle not included in the circles. [Take √3 = 1.73.]
Each angle of triangle = θ = 60°
Area of triangle not included in circles = Area of triangle – Area of all sectors
Area of all 3 sectors area equal as all the three circles are having same radius which is equal to the half of the side of the equilateral triangle.
Let the side of equilateral triangle be = a cm
⇒ a = 7×2
⇒ a = 14 cm
So radius of the circles = 7 cm
∴ Area of all 3 sectors = 77 cm2→ eqn1
Area of triangle not included = (49×1.73) - 77
⇒ Area of triangle not included = 84.77 – 77
∴ Area of triangle not included = 7.77 cm2
Area of triangle not included in circles is 7.77 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.]