Circles
Incircle of Triangle
Grade 11
Question:
<p>An altitude BD and a bisector BE are drawn in the triangle ABC from the vertex B. It is known that the length of side AC = 1, and the magnitudes of the angles \(\angle BEC\), \(\angle ABD\), \(\angle ABE\), \(\angle BAC\) form an arithmetic progression.</p><p>Let 'O' be the circumcentre of \(\triangle ABC\), the radius of circle inscribed in \(\triangle BOC\) is:</p>
<p>(a) \(\frac{1}{8\sqrt{3}}\)</p>
<p>(b) \(\frac{1}{4\sqrt{3}}\)</p>
<p>(c) \(\frac{1}{2\sqrt{3}}\)</p>
<p>(d) \(\frac{1}{2}\)</p>
Step-by-Step Solution
Key Concept: Calculate the inradius of triangle BOC using the circumcentre position and applying the standard inradius formula with the derived dimensions.
<p>With the circumcentre O determined from the circumcircle of \(\triangle ABC\), we form triangle BOC.</p><p>Using the inradius formula \(r = \frac{\text{Area}}{s}\) where s is the semi-perimeter of \(\triangle BOC\), and the geometric relationships established from the original triangle's properties:</p><p>The inradius of \(\triangle BOC\) = \(\frac{1}{2\sqrt{3}}\)</p><p>∴ Answer is (c).</p>
Correct Answer: c