Question

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

• A. $6$
• B. $\frac{15}{4}\sqrt{7}$
• C. $\frac{18}{5}\sqrt{7}$
• D. $\frac{14}{3}\sqrt{5}$

Answer 1

Answer: (B)

Let three sides are $x-1,x,x+1$

Given, largest angle is twice the smallest angle
$\therefore \cos \theta =\frac{(x+1{)}^2+(x{)}^2-(x-1{)}^2}{2x(x+1)}$
$=\frac{x^2+2x+1+x^2-x^2+2x-1}{2x(x+1)}$
$=\frac{x^2+4x}{2x(x+1)}=\frac{x+4}{2(x+1)}$
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 $4,5,6$
$\Delta =\frac{1}{2}\sqrt{\frac{15}{2}\left(\frac{15}{2}-4\right)\left(\frac{15}{2}-5\right)\left(\frac{15}{2}-6\right)}$
$=\frac{1}{4}\sqrt{15\times 7\times 5\times 3}=\frac{15}{4}\sqrt{7}$