A round table cover has six equal designs as shown in Fig. 12.14. If the radius of the cover is `28` cm, find the cost of making the designs at the rate of `R s\ 0. 35\ p e r\ c m^2`.

Explanation

We know that, in a circle with radius r and the angle at the center with degree measure `theta`,
(i) Area of the sector = `theta/360 × pir^2`
(ii) Area of the segment = Area of the sector - Area of the corresponding triangle
Area of the design = Area of 6 segments of the circle.
Since the table cover has 6 equal design, therefore angle of each sector at the center = `(360^@)/(6) = 60^@`

Consider segment `APB.` Chord `AB` subtends an angle of `60^@` at the center.

Area of segment `APB` = Area of sector `AOPB` - Area of `triangleAOB`
Consider `triangleAOB`
`OB = OA` (radii of the circle)

`angleOAB = angleOBA` (angles opposite to equal sides in a triangle are equal)
`angleAOB + angleOAB + angleOBA = 180^@` (angle sum property of a triangle)
`2angleOAB = 180^@ - 60^@ (Since, angleAOB = 60^@)`
`angleOAB = (120^@)/(2) = 60^@`

`triangleAOB` is an equilateral triangle
Area of `triangleAOB = (sqrt3)/4 (side)^2`
`= (sqrt3)/4 (28)^2` (Since the side of the triangle = radii of the circle = 28 cm)
= `sqrt(3) × 7 × 28`
= `196sqrt3`
= `196 xx 1.7`
= `333.2 cm^2`

Area of sector `OAPB = (60^@)/(360^@) × pir^2`

= `1/6 xx22/7 xx 28 xx 28`
= `(11 xx 4 xx 28)/3`
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