A round table cover has six equal designs as shown in the given figure. If the radius of the cover is 35 cm then find the total area of the design. [Use √3 = 1.732 and π = 3.14.]
Total area of design = Area of all the minor segments
Here we will find out the area of one segment and then multiply it with 6 to get the total area of design. And as the figure inscribed in the circle is a regular hexagon this implies that it will be having all edges of same length. Therefore we can say that the angle subtended by each chord which are actually the edges of regular hexagon are equal(theorem).
Let angle subtended by chord AB on centre O be θ
∴ Angle subtended = θ = 60°
Radius of circle = 35 cm
Area of minor segment OAB = Area of sector – Area of ∆OAB
Area of minor segment OAB = 641.0833333 – 530.425
∴ Area of minor segment OAB = 110.6583333 cm2→ eqn2
Total area of design = 6 × Area of minor segment OAB
⇒ Total area of design = 6×110.6583333 (from eqn2)
∴ Total area of design = 663.95 cm2
Total area of design is 663.95 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.]