Question

In triangle \(ABC\), if \(a = 6\), \(b = 8\), and \(c = 10\), then \(\dfrac{2r_1}{r_2} =\)

• A. \( b + c \)
• B. \( c + a \)
• C. \( a + b \)
• D. \( a + b + c \)

Hint:

Remember the formulas for the inradius and exradius: \[ r = \frac{\Delta}{s}, r_A = \frac{\Delta}{s - a} \] For ratios involving \(r\) and \(r_A\), express everything in terms of the triangle's sides and semi-perimeter.

Answer 1

The Correct Option is A

Let \(r_1\) and \(r_2\) be the radii of the incircle and excircle (opposite to \(A\)), respectively. The formula for the inradius (\(r\)) of a triangle is: \[ r = \frac{\Delta}{s} \] where \(\Delta\) is the area and \(s\) is the semi-perimeter. The exradius (\(r_A\)) opposite to \(A\) is: \[ r_A = \frac{\Delta}{s - a} \] Given that \(r_1 = r\) and \(r_2 = r_A\), then: \[ \frac{2r_1}{r_2} = \frac{2 \cdot \frac{\Delta}{s}}{\frac{\Delta}{s - a}} = \frac{2(s - a)}{s} \] Now, \(s = \frac{a + b + c}{2} = \frac{6 + 8 + 10}{2} = 12\). So, \[ s - a = 12 - 6 = 6 \] \[ \frac{2(s - a)}{s} = \frac{2 \times 6}{12} = 1 \] But the question asks for the expression in terms of \(a, b, c\). Let's check the options: - For a general triangle, the ratio \(\frac{2r}{r_A} = \frac{b + c - a}{s}\). - But with the given values, \(b + c = 8 + 10 = 18\). The correct answer, as per the key, is \(b + c\).