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Q.
If the sides of a traingle are three consecutive natural numbers and its largest angle is twice the smallest one, then the area (in sq. units) of that triangle is
TS EAMCET 2020
Solution:
Let three sides are $x-1, x, x+1$
Given, largest angle is twice the smallest angle
$\therefore \cos \theta=\frac{(x+1)^{2}+(x)^{2}-(x-1)^{2}}{2 x(x+1)}$
$=\frac{x^{2}+2 x+1+x^{2}-x^{2}+2 x-1}{2 x(x+1)}$
$=\frac{x^{2}+4 x}{2 x(x+1)}=\frac{x+4}{2(x+1)}$
By Sine rule,
$\frac{x+1}{\sin 2 \theta}=\frac{x-1}{\sin \theta}$
$\Rightarrow \frac{x+1}{2 \sin \theta \cos \theta}=\frac{x-1}{\sin \theta}$
$\Rightarrow \cos \theta=\frac{x+1}{2(x-1)}$
From Eqs. (i) and (ii),
$\frac{x+1}{2(x-1)}=\frac{x+4}{2(x+1)}$
$\Rightarrow x=5$
$\therefore $ Sides are $4,5,6$
$\Delta =\frac{1}{2} \sqrt{\frac{15}{2}\left(\frac{15}{2}-4\right)\left(\frac{15}{2}-5\right)\left(\frac{15}{2}-6\right)}$
$=\frac{1}{4} \sqrt{15 \times 7 \times 5 \times 3}=\frac{15}{4} \sqrt{7}$
