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`. (Use√ 3 = 1.7)

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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