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