A round table cover has six equal designs as shown in Fig. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of  0.35 per cm² . (Use $\sqrt{3}$ = 1.7)

detailed solution

We know that,  formula for the area of the sector of a circle

Area of the sector = $\frac{\theta }{{360}^0}$ $\times$ ${\mathrm{\pi r}}^2$

 Area of the segment  = Area of the sector - Area of the corresponding triangle

Now we label the diagram as follows,

From the diagram that is seen that, the designs are segments of the circle.

Thus,  Area of the design  is equal to the area of 6 segments of the circle.

Hence, The angle of each sector at the center = 360⁰/6 = 60⁰

Consider segment APB. Chord AB subtends an angle of 60° at the center.

Hence, Area of segment APB = Area of sector AOPB - Area of ΔAOB

Consider ΔAOB,

The radius of the circle is 

OB = OA

∠OAB = ∠OBA (angles opposite to equal sides in a triangle are equal)

We know that, The sum of all angles of a triangle is 180 degrees

 ∠AOB $+$ ∠OAB $+$ ∠OBA = 180⁰

2∠OAB = 180⁰ - 60⁰ (Since, ∠AOB = 60⁰)

∠OAB = 120⁰/2 = 60⁰

Hence,  ΔAOB is an Equilateral triangle

Area of ΔAOB = $\sqrt{3}$/4 (side)²

= $\sqrt{3}$/4 (28)² (Since the side of the triangle = radii of the circle = 28 cm)

= $\sqrt{3}$ $\times$ 7 $\times$28

= 196$\sqrt{3}$

= 196 $\times$ 1.7

= 333.2 cm²

Area of sector OAPB = 60⁰/360⁰ $\times$ ${\mathrm{\pi r}}^2$

= 1/6 $\times$ 22/7 $\times$ 28$\times$ 28

= (11 $\times$ 4 $\times$28)/3

= 1232/3 cm²

Now, Area of the segment APB = Area of sector OAPB - Area of ΔAOB

= (1232/3 - 333.2) cm²

Area of the designs = 6 $\times$ Area of segment

= 6 $\times$ (1232/3 - 333.2) cm²

= 2464 - 1999.2 cm²

= 464.8 cm²

Cost of making 1 cm² of designs = ₹ 0.35

Thus, The cost of making 464.8 cm² of designs

= ₹ 0.35 $\times$464.8

= ₹ 162.68