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If the sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one, then the area (in sq. units) of that triangle is
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Verified Answer
The correct answer is:
$\frac{15}{4} \sqrt{7}$
Let three sides are $x-1, x, x+1$.
Given, largest angle is twice the smallest angle
$\begin{aligned}
\therefore \quad \cos \theta & =\frac{(x+1)^2+(x)^2-(x-1)^2}{2 x(x+1)} \\
& =\frac{x^2+2 x+1+x^2-x^2+2 x-1}{2 x(x+1)}
\end{aligned}$
By Sine rule,
$\frac{x+1}{\sin 2 \theta}=\frac{x-1}{\sin \theta} \Rightarrow \frac{x+1}{2 \sin \theta \cos \theta}=\frac{x-1}{\sin \theta}$
From Eqs. (i) and (ii),
$\frac{x+1}{2(x-1)}=\frac{x+4}{2(x+1)} \Rightarrow x=5$
$\therefore$ Sides are $4,5,6$
$\begin{aligned}
\Delta & =\frac{1}{2} \sqrt{\frac{15}{2}\left(\frac{15}{2}-4\right)\left(\frac{15}{2}-5\right)\left(\frac{15}{2}-6\right)} \\
& =\frac{1}{1} \sqrt{15 \times 7 \times 5 \times 3}=\frac{15}{4} \sqrt{7}
\end{aligned}$
Given, largest angle is twice the smallest angle
$\begin{aligned}
\therefore \quad \cos \theta & =\frac{(x+1)^2+(x)^2-(x-1)^2}{2 x(x+1)} \\
& =\frac{x^2+2 x+1+x^2-x^2+2 x-1}{2 x(x+1)}
\end{aligned}$
By Sine rule,
$\frac{x+1}{\sin 2 \theta}=\frac{x-1}{\sin \theta} \Rightarrow \frac{x+1}{2 \sin \theta \cos \theta}=\frac{x-1}{\sin \theta}$
From Eqs. (i) and (ii),
$\frac{x+1}{2(x-1)}=\frac{x+4}{2(x+1)} \Rightarrow x=5$
$\therefore$ Sides are $4,5,6$
$\begin{aligned}
\Delta & =\frac{1}{2} \sqrt{\frac{15}{2}\left(\frac{15}{2}-4\right)\left(\frac{15}{2}-5\right)\left(\frac{15}{2}-6\right)} \\
& =\frac{1}{1} \sqrt{15 \times 7 \times 5 \times 3}=\frac{15}{4} \sqrt{7}
\end{aligned}$
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