A) \[\frac{25}{4}\]sq.cm
B) \[\frac{25}{\sqrt{3}}\]sq.cm
C) \[\frac{9\sqrt{3}}{4}\]sq.cm
D) \[25\sqrt{3}\]sq.cm
Correct Answer: C
Solution :
(c): \[PQ\parallel BC\] Also, \[\angle APQ=\angle ABC={{60}^{{}^\circ }}\] \[\angle AQP=\angle ACB={{60}^{{}^\circ }}\] \[\therefore \]Area of \[\Delta APQ=\frac{\sqrt{3}}{4}\times {{\left( PQ \right)}^{2}}=\frac{\sqrt{3}}{4}\times {{\left( 3 \right)}^{2}}=\frac{9\sqrt{3}}{4}\]sq.cmYou need to login to perform this action.
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