Question

The sides of a triangle are three consecutive integers.The largest angle is twice the smallest one.Its area is

• A. $\frac{5\sqrt{7}}{4}$
• B. $\frac{15\sqrt{7}}{2}$
• C. $\frac{15\sqrt{7}}{4}$
• D. $5\sqrt{7}$

Answer 1

Answer: (C)

$\frac{15\sqrt{7}}{4}$

Let $a=n-1,b=n,c=n+1$
$C=2A\Rightarrow \frac{a}{\sin A}=\frac{c}{\sin C}$
$\Rightarrow \frac{n-1}{\sin A}=\frac{n+1}{\sin 2A}=\frac{n+1}{2\sin A\cos A}$
$\Rightarrow 2(n-1)\cos A=n+1$
$\Rightarrow 2(n-1)\left(\frac{b^2+c^2-a^2}{2bc}\right)=n+1$
$\Rightarrow (n-1)\left(\frac{b^2+c^2-a^2}{bc}\right)=n+1$
Substituting for $a=n-1,b=n,c=n+1$ we get
$\Rightarrow (n-1)\left(\frac{n^2+{(n+1)}^2-{(n-1)}^2}{n(n+1)}\right)=n+1$
$\Rightarrow (n-1)[n^2+{(n+1)}^2-{(n-1)}^2]=(n+1).n.(n-1)$
$\Rightarrow (n-1).n.(n+4)=n{(n+1)}^2$
$\Rightarrow n^2+3n-4=n^2+2n+1$
$\Rightarrow n=5$ on simplification.
Thus the sides are $a=n-1=5-1=4$
$b=n=5$ and $c=n+1=5+1=6$
and $s=\frac{15}{2}$
$\therefore △=\sqrt{s(s-a)(s-b)(s-c)}.$
$\therefore △=\sqrt{\frac{15}{2}(\frac{15}{2}-4)(\frac{15}{2}-5)(\frac{15}{2}-6)}$
$=\sqrt{\frac{15}{2}.\frac{7}{2}.\frac{5}{2}.\frac{3}{2}}=\frac{15\sqrt{7}}{4}$