Question

In a triangle the sum of two sides is $x$ and their product is $y$ such that $(x+z)(x-z)=y$ where $z$ is the third side of the triangle.

Area of the triangle is :

• A. $\frac{\sqrt{3}}{4}[y+\frac{1}{8}{(x+\sqrt{x^2-4y})}^2]$
• B. $\frac{\sqrt{3}}{4}y$
• C. $\frac{1}{4}x$
• D. $x-z^2$

Answer 1

The correct option is B $\frac{\sqrt{3}}{4}y$

Let sides of the triangles are $a,b,c$
Given $a+b=x,ab=y,c=z$ (i)
Given $(x+z)(x-z)=y\Rightarrow x^2-z^2=y$ (ii)
Now using cosine rule, $\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{(a+b{)}^2-c^2-2ab}{2ab}$
$\Rightarrow \cos C=\frac{x^2-z^2-2y}{2y}=\frac{y-2y}{2y}=-\frac{1}{2}$
$\therefore \angle C={120}^0$
Hence area of triangle is, $\Delta =\frac{1}{2}ab\sin C=\frac{1}{2}y\sin \left({120}^0\right)=\frac{\sqrt{3}}{4}y$