| Assertion (A): In the given figure, if arcs are drawn by taking vertices A, B and C of an equilateral triangle of side 8 cm, to intersect the sides BC, CA and AB at their respective mid-points D, E and F, then are of the shaded region is\[25.12\,c{{m}^{2}}\]. |
| Reason (R): Area of shaded region\[=\text{Area of }\Delta \text{ABC}-\text{Area of 3 equal sectors}.\] |
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A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A)
C) Assertion (A) is true but Reason (R) is false
D) Assertion (A) is false but Reason (R) is true
Correct Answer: C
Solution :
| [c] Given, \[\Delta ABC\]is an equilateral triangle. |
| \[\therefore \]\[\angle A=\angle B=\angle C=60{}^\circ \]and radius, |
| \[r=\frac{8}{2}cm=4cm\] |
| Area of sector |
| \[AFEA=\frac{\theta }{360{}^\circ }\times \pi {{r}^{2}}=\frac{60{}^\circ }{360{}^\circ }\times \pi {{(4)}^{2}}\] |
| \[=\frac{8}{3}\pi c{{m}^{2}}\] ...(1) |
| Since, area of all these sectors are equal. |
| \[\therefore \] Total area of shaded region |
| \[=3\times \text{Area of sector AFEA = 3}\left( \frac{8}{3}\pi \right)\] |
| \[=3\times \frac{8}{3}\times 3.14=25.12\,c{{m}^{2}}\] |
| Hence, the area of shaded region is \[25.12c{{m}^{2}}\]. |
| Assertion: True: Reason: False. |
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