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Q.
The sides of a $\triangle A B C$ are positive integers. The smallest side has length $1$ . Which of the following statements is true?
KVPYKVPY 2010
Solution:
We have, sides of $\triangle A B C$ are positive integer and length of smallest side is $1 .$
We know that the sum of two sides of triangle is greater than third side.
$\therefore b+1 > c $
$\Rightarrow c - b < 1 $
$ 1 + c > b $
$b - c < 1 $
$ -1 < b - c < 1 $
$b, c$ are integers.
$\therefore b-c=0 \Rightarrow b=c$
Semi-perimeter $=\frac{a+b+c}{2} $
$=\frac{2 b+1}{2}=b+\frac{1}{2}$
Area$A=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{\left(b+\frac{1}{2}\right)\left(b+\frac{1}{2}-1\right)\left(b+\frac{1}{2}-b\right)\left(b+\frac{1}{2}-b\right)} $
$=\sqrt{\left(b+\frac{1}{2}\right)\left(b-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)}=\frac{1}{2} \sqrt{b^{2}-\frac{1}{4}}$
Since, $b$ is integer.
$\therefore$ Area of $\triangle A B C$ is always irrational number.