Question

Let sides of the triangles are $a,b,c$

Now $a+b=x,ab=y,c=z$ (i)

and $(x+z)(x-z)=y$ (ii)

The circum radius of the triangle is :

• A. $x$
• B. $y$
• C. $z$
• D. none of these

Answer 1

The correct option is D none of these

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, $\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{(a+b{)}^2-c^2-2ab}{2ab}$
$\Rightarrow \cos C=\frac{x^2-z^2-2y}{2y}=\frac{y-2y}{2y}=-\frac{1}{2}$
$\therefore \angle C={120}^0$
Now using sine rule $\frac{c}{\sin C}=2R\Rightarrow R=\frac{z}{2\sin \left({120}^0\right)}=\frac{z}{\sqrt{3}}$