Question

In \( \triangle ABC \), if \( a = 8 \), \( b = 10 \), \( c = 12 \), then \( \frac{r}{R} = \)

• A. $ \frac{8}{15} $
• B. $ \frac{7}{16} $
• C. $ \frac{3}{5} $
• D. $ \frac{5}{8} $

Hint:

To solve problems involving the ratio of inradius to circumradius, use the formula $ \frac{r}{R} = \frac{4K}{abc} $, where $ K $ is the area of the triangle. Alternatively, use known geometric relationships specific to the triangle.

Answer 1

The Correct Option is B

Given a triangle \( \triangle ABC \) with side lengths \( a = 8 \), \( b = 10 \), and \( c = 12 \). We need to find the ratio of the inradius (\( r \)) to the circumradius (\( R \)), i.e., \( \frac{r}{R} \). 
Step 1: Calculate the Semi-perimeter (\( s \))
The semi-perimeter \( s \) is half the perimeter of the triangle:
\[ s = \frac{a+b+c}{2} \]
Substituting the given values:
\[ s = \frac{8+10+12}{2} = \frac{30}{2} = 15 \] Step 2: Calculate the Area of the Triangle (\( \Delta \)) using Heron's Formula
Heron's formula for the area of a triangle is: \[ \Delta = \sqrt{s(s-a)(s-b)(s-c)} \] Substitute the values of \( s, a, b, c \): \[ \begin{aligned} \Delta &= \sqrt{15(15-8)(15-10)(15-12)} \\ &= \sqrt{15 \cdot 7 \cdot 5 \cdot 3} \\ &= \sqrt{(3 \cdot 5) \cdot 7 \cdot 5 \cdot 3} \\ &= \sqrt{3^2 \cdot 5^2 \cdot 7} \\ &= 3 \cdot 5 \sqrt{7} \\ &= 15\sqrt{7} \end{aligned} \] Step 3: Calculate the Inradius (\( r \))
The inradius \( r \) of a triangle can be found using the formula: \[ r = \frac{\Delta}{s} \]
Substitute the calculated values of \( \Delta \) and \( s \):
\[ r = \frac{15\sqrt{7}}{15} = \sqrt{7} \] Step 4: Calculate the Circumradius (\( R \))
The circumradius \( R \) of a triangle can be found using the formula: \[ R = \frac{abc}{4\Delta} \]
Substitute the given side lengths \( a, b, c \) and the calculated area \( \Delta \):
\[ \begin{aligned} R &= \frac{8 \cdot 10 \cdot 12}{4 \cdot 15\sqrt{7}} \\ &= \frac{960}{60\sqrt{7}} \\ &= \frac{16}{\sqrt{7}} \end{aligned} \] Step 5: Calculate the Ratio \( \frac{r}{R} \)
Finally, we compute the ratio of the inradius to the circumradius: \[ \frac{r}{R} = \frac{\sqrt{7}}{\frac{16}{\sqrt{7}}} \] To simplify, multiply the numerator by the reciprocal of the denominator: \[ \frac{r}{R} = \sqrt{7} \cdot \frac{\sqrt{7}}{16} = \frac{(\sqrt{7})^2}{16} = \frac{7}{16} \] Conclusion
The ratio \( \frac{r}{R} \) for the given triangle is \( \frac{7}{16} \).