Trigonometric Equations Question 118
Question: In a $ \Delta ABC, $ let $ \angle C=\frac{\pi }{2}. $ If $ r $ and $ R $ are in radius and the circum-radius respectively of the triangle, then $ 2(r+R) $ is equal to
[IIT Screening 2000; AIEEE 2005]
Options:
A) $ a+b $
B) $ b+c $
C) $ c+a $
D) $ a+b+c $
Show Answer
Answer:
Correct Answer: A
Solution:
- Here, $ R=OA=OB=OC=\frac{1}{2}AB=\frac{c}{2} $ $ r=\frac{\Delta }{s}=\frac{\frac{1}{2}ab}{\frac{1}{2}(a+b+c)}=\frac{ab}{a+b+c} $
$ \therefore r+R=\frac{ab}{a+b+c}+\frac{c}{2}=\frac{2ab+c(a+b+c)}{2(a+b+c)} $ $ =\frac{2ab+ca+bc+a^{2}+b^{2}}{2(a+b+c)},(\because ,c^{2}=a^{2}+b^{2}) $ $ =\frac{{{(a+b)}^{2}}+c(a+b)}{2(a+b+c)},=\frac{(a+b)(a+b+c)}{2(a+b+c)} $
$ \therefore 2(r+R)=a+b. $
BETA
