Quantitative Aptitude Ques 226
Question: 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
Options:
A) $ 100\sqrt{3},cm^{2} $
B) $ 100/\sqrt{3},cm^{2} $
C) $ 50\sqrt{3},cm^{2} $
D) $ 25\sqrt{3},cm^{2} $
Show Answer
Answer:
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},cm^{2} $
BETA
