Trigonometric Equations Question 239
Question: If the sides of the triangle are $ 5K,\ 6K,\ 5K $ and radius of incircle is 6 then value of K is equal to
[Pb. CET 2004]
Options:
A) 4
B) 5
C) 6
D) 7
Show Answer
Answer:
Correct Answer: A
Solution:
- $ a=5k,,b=6k $ and $ c=5k $ $ s=\frac{a+b+c}{2} $ $ =\frac{5k+6k+5k}{2}=8k $ $ r=\frac{\Delta }{s}=\sqrt{\frac{s(s-a),(s-b),(s-c)}{s}} $ $ r=\sqrt{\frac{8k,(8k-5k),(8k-6k),(8k-5k)}{8k}} $ $ r=\frac{3k}{2}\Rightarrow k=\frac{2r}{3}=\frac{2\times 6}{3}=4 $ .
BETA
