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(B) Let the sides of the triangle be $x-1, x, x+1$ where $x > 2$ (to ensure all sides are positive). The smallest side is $x-1$ and the largest side i

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

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(B) Let the sides of the triangle be $x-1, x, x+1$ where $x > 2$ (to ensure all sides are positive). The smallest side is $x-1$ and the largest side is $x+1$. Let the smallest angle be $	heta$ (opposite to side $x-1$) and the largest angle be $2	heta$ (opposite to side $x+1$).Using the Law of Cosines for the angle $2	heta$ (opposite to side $x+1$):$\cos(2	heta) = \frac{x^2 + (x-1)^2 - (x+1)^2}{2x(x-1)} = \frac{x^2 + x^2 - 2x + 1 - (x^2 + 2x + 1)}{2x(x-1)} = \frac{x^2 - 4x}{2x(x-1)} = \frac{x-4}{2(x-1)}$Using the Law of Sines:$\frac{x+1}{\sin(2	heta)} = \frac{x-1}{\sin(	heta)} \Rightarrow \frac{x+1}{2\sin(	heta)\cos(	heta)} = \frac{x-1}{\sin(	heta)} \Rightarrow \cos(	heta) = \frac{x+1}{2(x-1)}$Since $\cos(2	heta) = 2\cos^2(	heta) - 1$,we substitute the expressions:$\frac{x-4}{2(x-1)} = 2\left(\frac{x+1}{2(x-1)}ight)^2 - 1 = \frac{(x+1)^2}{2(x-1)^2} - 1 = \frac{x^2+2x+1 - 2(x^2-2x+1)}{2(x-1)^2} = \frac{-x^2+6x-1}{2(x-1)^2}$Solving $\frac{x-4}{x-1} = \frac{-x^2+6x-1}{(x-1)^2} \Rightarrow (x-4)(x-1) = -x^2+6x-1 \Rightarrow x^2-5x+4 = -x^2+6x-1 \Rightarrow 2x^2-11x+5 = 0 \Rightarrow (2x-1)(x-5) = 0$. Since $x$ is a natural number and $x>2$,we get $x=5$.The sides are $4, 5, 6$. The semi-perimeter $s = \frac{4+5+6}{2} = \frac{15}{2}$.Area = $\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{15}{2} 	imes (\frac{15}{2}-4) 	imes (\frac{15}{2}-5) 	imes (\frac{15}{2}-6)} = \sqrt{\frac{15}{2} 	imes \frac{7}{2} 	imes \frac{5}{2} 	imes \frac{3}{2}} = \sqrt{\frac{1575}{16}} = \frac{15\sqrt{7}}{4}$.

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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