Let `(5,(a)/(4))` be the circumcenter of a triangle with vertices `A(a,-2),B(a,6)` and `C((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

Options

A. 62
B. 30
C. 60
D. 53

Answer

To solve the problem, we need to find the values of the circumradius (α), area (β), and perimeter (γ) of the triangle with vertices \( A(a, -2) \), \( B(a, 6) \), and \( C\left(\frac{a}{4}, -2\right) \). The circumcenter is given as \( O(5, \frac{a}{4}) \).

### Step 1: Find the value of \( a \)

Since \( O \) is the circumcenter, the distances from \( O \) to each vertex \( A \) and \( B \) must be equal. Therefore, we can set up the equation:

\[ AO^2 = BO^2 \]

Calculating \( AO^2 \):

\[ AO^2 = (5 - a)^2 + \left(\frac{a}{4} + 2\right)^2 \]

Calculating \( BO^2 \):

\[ BO^2 = (5 - a)^2 + \left(\frac{a}{4} - 6\right)^2 \]

Setting \( AO^2 = BO^2 \):

\[ (5 - a)^2 + \left(\frac{a}{4} + 2\right)^2 = (5 - a)^2 + \left(\frac{a}{4} - 6\right)^2 \]

Cancelling \( (5 - a)^2 \) from both sides:

\[ \left(\frac{a}{4} + 2\right)^2 = \left(\frac{a}{4} - 6\right)^2 \]

Expanding both sides:

\[ \left(\frac{a}{4}\right)^2 + 2 \cdot 2 \cdot \frac{a}{4} + 4 = \left(\frac{a}{4}\right)^2 - 2 \cdot 6 \cdot \frac{a}{4} + 36 \]

Simplifying:

\[ \frac{a^2}{16} + \frac{a}{4} + 4 = \frac{a^2}{16} - \frac{3a}{2} + 36 \]

Eliminating \( \frac{a^2}{16} \) from both sides:

\[ \frac{a}{4} + 4 = -\frac{3a}{2} + 36 \]

Multiplying through by 4 to eliminate the fraction:

\[ a + 16 = -6a + 144 \]

Rearranging gives:

\[ 7a = 128 \implies a = \frac{128}{7} \]

### Step 2: Calculate the circumradius \( \alpha \)

The circumradius \( \alpha \) for a right triangle can be calculated using the formula:

\[ \alpha = \frac{c}{2} \]

where \( c \) is the length of the hypotenuse. The vertices \( A \) and \( B \) form a vertical line, and \( C \) is on the same horizontal line as \( A \). The lengths of the sides are:

- \( AB = 8 \)
- \( AC = 6 \)
- \( BC = 10 \) (hypotenuse)

Thus,

\[ \alpha = \frac{10}{2} = 5 \]

### Step 3: Calculate the area \( \beta \)

The area \( \beta \) of the triangle can be calculated using the formula:

\[ \beta = \frac{1}{2} \times \text{base} \times \text{height} \]

Using \( AB \) as the base and the vertical distance between \( A \) and \( B \):

\[ \beta = \frac{1}{2} \times 8 \times 6 = 24 \]

### Step 4: Calculate the perimeter \( \gamma \)

The perimeter \( \gamma \) is the sum of the lengths of all sides:

\[ \gamma = AB + AC + BC = 8 + 6 + 10 = 24 \]

### Step 5: Calculate \( \alpha + \beta + \gamma \)

Now we can sum up the values:

\[ \alpha + \beta + \gamma = 5 + 24 + 24 = 53 \]

Thus, the final answer is:

\[ \boxed{53} \]