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(A) The designs are segments of the circle.Consider segment $APB$. Chord $AB$ is a side of the regular hexagon inscribed in the circle.Each chord subt

Vedclass · Accepted Answer

$162.68$

Vedclass · Answer

(A) The designs are segments of the circle.Consider segment $APB$. Chord $AB$ is a side of the regular hexagon inscribed in the circle.Each chord subtends an angle of $\frac{360^{\circ}}{6} = 60^{\circ}$ at the center $O$.In $	riangle OAB$,$OA = OB = 28\, cm$ (radii).Since $\angle AOB = 60^{\circ}$ and $OA = OB$,$	riangle OAB$ is an equilateral triangle.Area of $	riangle OAB = \frac{\sqrt{3}}{4} 	imes (side)^2 = \frac{1.7}{4} 	imes 28 	imes 28 = 1.7 	imes 7 	imes 28 = 333.2\, cm^2$.Area of sector $OAPB = \frac{60^{\circ}}{360^{\circ}} 	imes \pi r^2 = \frac{1}{6} 	imes \frac{22}{7} 	imes 28 	imes 28 = \frac{1232}{3} \approx 410.67\, cm^2$.Area of one segment = Area of sector $OAPB - 	ext{Area of } 	riangle OAB = \frac{1232}{3} - 333.2 = 410.67 - 333.2 = 77.47\, cm^2$.Total area of $6$ designs = $6 	imes 77.47 = 464.82\, cm^2$.Cost of making designs = $464.82 	imes 0.35 = Rs.\, 162.687 \approx Rs.\, 162.68$.

$A$ round table cover has six equal designs as shown in the figure. If the radius of the cover is $28\, cm$,find the cost of making the designs at the rate of $Rs.\, 0.35$ per $cm^2$. (in $Rs.$) (Use $\sqrt{3}=1.7$)

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