question_answer

In a right angled triangle PQR, PR is the hypotenuse of length 20 cm, \[\angle PRQ=30{}^\circ ,\] then the area of the triangle is

A) \[100\sqrt{3}\,c{{m}^{2}}\]                 

B) \[100/\sqrt{3}\,c{{m}^{2}}\]

C) \[50\sqrt{3}\,c{{m}^{2}}\]       

D) \[25\sqrt{3}\,c{{m}^{2}}\]

Correct Answer: C

Solution :

\[\cos 30{}^\circ =\frac{QR}{PR}=\frac{QR}{20}\]

\[\Rightarrow \]   \[\frac{\sqrt{3}}{2}\times 20=QR\]                      (i)

\[\sin 30{}^\circ =\frac{PQ}{PR}=\frac{PQ}{20}\]\[\Rightarrow \]\[\frac{1}{2}=\frac{PQ}{20}\]

\[\therefore \]      \[PQ=10\]

Area of \[\Delta PQR=\frac{1}{2}\times 10\times 10\sqrt{3}=50\sqrt{3}\,c{{m}^{2}}\]