A) \[\frac{P}{2}\]
B) \[\frac{P}{3}\]
C) P
D) None of these
Correct Answer: C
Solution :
Given a =b=c |
Area of triangle\[ABC=\frac{\sqrt{3}}{4}{{a}^{2}}=\] area of triangle ABQ +area of triangle BQC + area to triangle CQA |
\[\frac{\sqrt{3}}{4}{{a}^{2}}=\frac{1}{2}\,\,ax+\frac{1}{2}\,\,by+\frac{1}{2}\,\,cz\] |
\[\frac{\sqrt{3}}{4}a=(x+y+z)=P\] |
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