Question

In \(\triangle ABC\), the sum of the lengths of two sides is \(x\) and the product of those lengths is \(y\). If \(c\) is the length of its third side and \(x^2 - c^2 = y\), then the circumradius of that triangle is:

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

Hint:

Use cosine law to relate sides and apply formula for circumradius \(R = \frac{c}{2 \sin C}\).

Answer 1

The Correct Option is A

Step 1: Use notation for sides
Let the two sides be \(a\) and \(b\) such that: \[ a + b = x, \quad ab = y \] Step 2: Given
\[ x^2 - c^2 = y \] Step 3: Express \(c\) in terms of \(a\) and \(b\)
Using \(a + b = x\), \[ x^2 = (a + b)^2 = a^2 + 2ab + b^2 = a^2 + b^2 + 2y \] So, \[ x^2 - y = a^2 + b^2 + y \] Given, \[ x^2 - c^2 = y \implies c^2 = x^2 - y \] Step 4: Use cosine law
\[ c^2 = a^2 + b^2 - 2ab \cos C = (a^2 + b^2) - 2 y \cos C \] Substitute \(c^2 = x^2 - y\): \[ x^2 - y = (a^2 + b^2) - 2 y \cos C \] Using previous relations: \[ a^2 + b^2 = x^2 - 2 y \] Therefore: \[ x^2 - y = (x^2 - 2 y) - 2 y \cos C \implies -y = -2 y - 2 y \cos C \implies 2 y \cos C = - y \implies \cos C = -\frac{1}{2} \] Step 5: Angle \(C\)
\[ \cos C = -\frac{1}{2} \implies C = 120^\circ = \frac{2\pi}{3} \] Step 6: Use formula for circumradius \(R\)
\[ R = \frac{c}{2 \sin C} = \frac{c}{2 \sin 120^\circ} = \frac{c}{2 \times \frac{\sqrt{3}}{2}} = \frac{c}{\sqrt{3}} \]