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A round table cover has six equal designs as shown in figure. If the radius of the cover is $28 cm$ , find the cost of making the design at the rate of $Rs . 3.50 per c m^{2}$ . (Use $\sqrt{3} = 1.7$ ).

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Rs . 1145.3

Rs . 1348

Rs . 1626.8

Rs . 1723.5

answer is C.

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Detailed Solution

Value of each θ is,

θ = \frac{360 °}{6}

(∵ all θ are equal)

θ = 60 °

Area of sector

\Rightarrow \frac{θ}{360 °} \times π r^{2}

\Rightarrow \frac{60 °}{360 °} \times π {(28)}^{2}

(∵ given radius

= 28 cm

)

\Rightarrow \frac{π {(28)}^{2}}{6}

\Rightarrow 410.5014 c m^{2}

(∵ π = 3.14)

Draw a perpendicular

OM

PQ

ΔOPQ

.
Now, we will use concept of congruent

Δ

.
In

ΔOPM

and

ΔPQM

∠ M = ∠ N (90 ° eac h)

OP = OQ (radius)

OM = OM (Common)

ΔOPM ⩭ ΔOQM

∴ PM = QM

(Corresponding part of congruent triangle)
As we see,

ΔOPQ

is equilateral

Δ

,
So, area of

ΔOPQ

is -

\frac{\sqrt{3}}{4} \times 28 \times 28

(∵ Radius

= 28 cm

)

= 196 \sqrt{3}

= 339.34 c m^{2}

Area of design in hexagon

= 6 (area of sector - area of tr i angle)

= 6 (410.50 - 339.34)

= 6 \times 77.47 c m^{2}

= 464.82 c m^{2}

As given,
Cost of per

c m^{2} = Rs . 3.50

∴ Total cost of the design

= 3.50 \times 464.82

= Rs . 1626.87

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