In $ \triangle ABC $, if $ r = 1 $, $ R = 4 $, and $ \Delta = 8 $, then
\[
\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} =
\]
Show Hint
- \( {8} \)
- \( \frac{1}{4} \)
- \( \frac{1}{8} \)
- \( \frac{1}{16} \)
The Correct Option is C
Solution and Explanation
We are given the following values:
Inradius \( r = 1 \)
Circumradius \( R = 4 \)
Area \( \Delta = 8 \)
We need to find the value of \( \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} \), where \( a, b, c \) are the sides of the triangle.
Step 1: Use the identity involving the area \( \Delta \) of a triangle: \[ \Delta = \frac{1}{2} ab \sin C = \frac{1}{2} bc \sin A = \frac{1}{2} ca \sin B \] This gives us relationships between the area and the sides of the triangle.
Step 2: We can use the formula for the sum of the reciprocals of the sides: \[ \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{1}{\Delta} \left( \frac{1}{r} \right) \] This formula uses the inradius \( r \) and the area \( \Delta \) to express the desired sum of reciprocals.
Step 3: Substitute the given values \( r = 1 \) and \( \Delta = 8 \) into the formula: \[ \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{1}{8} \times 1 = \frac{1}{8} \] Thus, the value of \( \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} \) is \( \frac{1}{8} \).
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