The sides of a triangle ABC are positive inte | vedclass

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(B) Let the sides of the triangle be $a, b, c$ where $a=1$ is the smallest side. Since the sides are positive integers,$b \ge 1$ and $c \ge 1$.By the

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The area of $	riangle ABC$ is always an irrational number

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(B) Let the sides of the triangle be $a, b, c$ where $a=1$ is the smallest side. Since the sides are positive integers,$b \ge 1$ and $c \ge 1$.By the triangle inequality,the sum of any two sides must be greater than the third side:$b+c > a \Rightarrow b+c > 1$$a+b > c$ $\Rightarrow 1+b > c$ $\Rightarrow b-c > -1$$a+c > b$ $\Rightarrow 1+c > b$ $\Rightarrow c-b > -1$Combining these,we get $-1 Thus,the sides are $1, b, b$ where $b$ is a positive integer. For a triangle to exist,$b+b > 1 \Rightarrow 2b > 1$,which is true for all $b \ge 1$. However,if $b=1$,the sides are $1, 1, 1$ (equilateral triangle),and the area is $\frac{\sqrt{3}}{4}$,which is irrational.For $b > 1$,the semi-perimeter $s = \frac{1+b+b}{2} = b + \frac{1}{2}$.The area $A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(b+\frac{1}{2})(b+\frac{1}{2}-1)(b+\frac{1}{2}-b)(b+\frac{1}{2}-b)}$$A = \sqrt{(b+\frac{1}{2})(b-\frac{1}{2})(\frac{1}{2})(\frac{1}{2})} = \sqrt{(b^2-\frac{1}{4}) \cdot \frac{1}{4}} = \frac{1}{2} \sqrt{b^2 - \frac{1}{4}} = \frac{1}{4} \sqrt{4b^2 - 1}$.Since $4b^2-1$ is not a perfect square for any integer $b \ge 1$ (as $(2b-1)^2 < 4b^2-1 < (2b)^2$),the area is always irrational.

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$

The sides of a $\triangle ABC$ are positive integers. The smallest side has length $1$. Which of the following statements is true?

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