 ## Book: RS Aggarwal - Mathematics

### Chapter: 18. Area of Circle, Sector and Segment

#### Subject: Maths - Class 10th

##### Q. No. 55 of Exercise 18B

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55
##### 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.

55

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