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Question: In the circular table cover of 32cm, a design is formed, leaving an equilateral triangle ABC in the ...

In the circular table cover of 32cm, a design is formed, leaving an equilateral triangle ABC in the middle, as shown in the figure below. Find the area of the design (Shaded region).

[a] (1614872183)cm2\left( \dfrac{16148}{7}-218\sqrt{3} \right)c{{m}^{2}}
[b] (1922175273)cm2\left( \dfrac{19221}{7}-527\sqrt{3} \right)c{{m}^{2}}
[c] (628876723)cm2\left( \dfrac{6288}{7}-672\sqrt{3} \right)c{{m}^{2}}
[d] None of these

Explanation

Solution

Hint: Observe that the shaded region is the region of the circle excluding the triangular portion. Hence the area of the shaded region will be equal to the equilateral triangle subtracted from the area of the circle. From the centre of the circle (say O) draw a line (OD) perpendicular to BC and Join OB. Use the fact that the angle subtended by a chord at the centre is twice the angle subtended in the alternate segment to determine BOD\angle BOD. In the triangle BOD, use trigonometry to determine BD and hence determine the length of the side BC. Use the fact that the area of an equilateral triangle of side length a is 34a2\dfrac{\sqrt{3}}{4}{{a}^{2}}. Hence determine the area of the equilateral triangle and hence the area of the shaded region.

Complete step-by-step answer:

Construction: Draw OD perpendicular BC and join OB.
Now, since BAC\angle BAC is the angle in the alternate segment, we have BOC=2BAC\angle BOC=2\angle BAC
Since OD is the angle bisector of BOC\angle BOC, we have BOD=12BOC=12(2BAC)=BAC\angle BOD=\dfrac{1}{2}\angle BOC=\dfrac{1}{2}\left( 2\angle BAC \right)=\angle BAC.
Since the measure of an angle of an equilateral triangle is 6060{}^\circ , we have
BOD=BAC=60\angle BOD=\angle BAC=60{}^\circ
Now, we have in triangle OBD,
BD is the side opposite to O and OB is the hypotenuse
Hence, we have
sin(60)=BDOB\sin \left( 60{}^\circ \right)=\dfrac{BD}{OB}
Now, we have sin60=32\sin 60{}^\circ =\dfrac{\sqrt{3}}{2} and OB=r=32OB=r=32
Hence, we have
BD32=32BD=163\dfrac{BD}{32}=\dfrac{\sqrt{3}}{2}\Rightarrow BD=16\sqrt{3}
Now, we have BC = 2BD =323=32\sqrt{3}
Now, we know that the area of an equilateral triangle of side a is given by A=34a2A=\dfrac{\sqrt{3}}{4}{{a}^{2}}
Hence, we have
ar(ΔABC)=34(323)2=7683ar\left( \Delta ABC \right)=\dfrac{\sqrt{3}}{4}{{\left( 32\sqrt{3} \right)}^{2}}=768\sqrt{3}
Also the area of the circle =πr2=227×322=225287=\pi {{r}^{2}}=\dfrac{22}{7}\times {{32}^{2}}=\dfrac{22528}{7}
Observe that the shaded region is the region of the circle excluding the triangular portion. Hence the area of the shaded region will be equal to the equilateral triangle subtracted from the area of the circle.

Hence the area of the shaded region =2252877683=\dfrac{22528}{7}-768\sqrt{3}
Hence option [d] is correct.

Note: [1] In these types of questions, we need to identify which areas can be calculated from standard formulae. Then we need to think about how the required area can be found using the calculated areas. Like in the above case, the area of the shaded region cannot be calculated directly from standard formulae, but once we observe that this area is the difference of the area of the circle and the triangle, we easily calculate the area of the region.