Question

If \( \triangle ABC \) is an isosceles triangle with base \( BC \), then \( r_1 = ? \)

• A. \( R^2 \cos^2 A \)
• B. \( \frac{a^2}{2} \)
• C. \( \frac{r}{R} \)
• D. \( R^2 \sin^2 A \)

Hint:

For isosceles triangles, use symmetry to simplify trigonometric identities. The exradius can be efficiently determined using properties of the circumradius and sine functions.

Answer 1

The Correct Option is D

We need to determine \( r_1 \), the exradius corresponding to side \( BC \), for an isosceles triangle \( \triangle ABC \) with base \( BC \).

 Step 1: Recall the formula for the exradius 
The exradius \( r_1 \) corresponding to a side of a triangle is given by: \[ r_1 = \frac{\triangle}{s - a} \] where: - \( \triangle \) is the area of the triangle, - \( s \) is the semi-perimeter, - \( a \) is the length of the side opposite to the exradius. 

Step 2: Express \( \triangle \) in terms of circumradius \( R \) 
For a triangle, \[ \triangle = R^2 \sin A \sin B \sin C \] Since the triangle is isosceles, angles \( B \) and \( C \) are equal: \[ B = C \] Using the identity: \[ \sin B = \sin C = \sin A \] we get: \[ \triangle = R^2 \sin^2 A \] 

Step 3: Compute \( r_1 \) 
Since \( r_1 \) is given by: \[ r_1 = R^2 \sin^2 A \] Thus, the correct answer is: \[ \mathbf{R^2 \sin^2 A} \]