Question

In a triangle the sum of length of two sides is x and the product of the lengths of the same two sides is y. If $x^2 - c^2 = y$, where c is the length of the third side of the triangle. Then the circumradius of the triangle is

• A. $\frac{c}{3}$
• B. $\frac{c}{\sqrt{3}}$
• C. $\frac{3}{2}y$
• D. $\frac{y}{\sqrt{3}}$

Answer 1

Answer: (B)

$\frac{c}{\sqrt{3}}$

Answer 2

Let the two sides be $a$ and $b$ with

$$a+b = x \quad \text{and} \quad ab = y.$$

The given relation is:

$$x^2 - c^2 = y.$$

Notice that:

$$x^2 = (a+b)^2 = a^2 + 2ab + b^2.$$

Thus,

$$(a+b)^2 - c^2 = a^2+2ab+b^2 - c^2.$$

Using the Law of Cosines in the triangle for side $c$, we have:

$$c^2 = a^2 + b^2 - 2ab\cos C,$$

where $C$ is the angle opposite side $c$. Substituting this in:

$$(a+b)^2 - c^2 = a^2+2ab+b^2 - (a^2+b^2-2ab\cos C) = 2ab(1+\cos C).$$

Given $(a+b)^2 - c^2 = ab$, we equate:

$$2ab(1+\cos C) = ab.$$

Since $ab \neq 0$, dividing by $ab$ gives:

$$2(1+\cos C)=1 \quad \Rightarrow \quad 1+\cos C=\frac{1}{2} \quad \Rightarrow \quad \cos C=-\frac{1}{2}.$$

Thus, $C = 120^\circ$.

The circumradius $R$ of a triangle is given by:

$$R = \frac{c}{2\sin C}.$$

Since $\sin 120^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}$, we obtain:

$$R = \frac{c}{2\left(\frac{\sqrt{3}}{2}\right)} = \frac{c}{\sqrt{3}}.$$