Question

Let\((5,\frac{a}{4})\),be the circumcenter of a triangle with vertices A (a, -2),B (a, 6)and C \((\frac{a}{4},-2)\).Let \(\alpha\) denote the circumradius, \(\beta\) denote the area and \(\gamma\) denote the perimeter of the triangle. Then \(\alpha+\beta+\gamma\) is

• A. 60
• B. 53
• C. 62
• D. 30

Answer 1

The Correct Option is B

Given points are \( A(1, -2) \), \( B(a, 6) \), and \( C\left(\frac{3}{2}, -2\right) \).  
- The circumcenter \( O \) is \( \left(\frac{5}{3}, 4\right) \).

Calculate \( AO \) and \( BO \) (Using Distance Formula): - \( AO = BO \):

\((a - 5)^2 + \left(\frac{a}{4} + 2\right)^2 = (a - 5)^2 + \left(\frac{a}{4} - 6\right)^2\)
Solving this gives \( a = 8 \).

Determine Side Lengths of the Triangle: - With \( a = 8 \): \( AB = 8 \), \( AC = 6 \), \( BC = 10 \).

Calculate Circumradius (\( \alpha \)), Area (\( \beta \)), and Perimeter (\( \gamma \)): - Circumradius \( \alpha = 5 \), Area \( \beta = 24 \), Perimeter \( \gamma = 24 \)

Compute \( \alpha + \beta + \gamma \):

\(\alpha + \beta + \gamma = 5 + 24 + 24 = 53\)

So, the correct option is: \( \mathbf{53} \)