In `DeltaABC, angle BCA = 90^@, AC = 24 cm and BC = 10 cm`. What is the radius (in cm) of the circum-circle of `DeltaABC`?

Options

A. `12.5 `
B. `13 `
C. `25`
D. `26`

Answer

To find the radius of the circumcircle of triangle ABC, where angle BCA is 90 degrees, AC = 24 cm, and BC = 10 cm, we can follow these steps:

### Step 1: Identify the triangle and its sides
We have a right triangle ABC with:
- Angle BCA = 90 degrees
- AC (the side opposite to angle B) = 24 cm
- BC (the side opposite to angle A) = 10 cm

### Step 2: Use the Pythagorean theorem to find the hypotenuse (AB)
In a right triangle, the hypotenuse can be calculated using the Pythagorean theorem:
\[ AB^2 = AC^2 + BC^2 \]

Substituting the known values:
\[ AB^2 = 24^2 + 10^2 \]
\[ AB^2 = 576 + 100 \]
\[ AB^2 = 676 \]

Now, take the square root to find AB:
\[ AB = \sqrt{676} = 26 \text{ cm} \]

### Step 3: Find the radius of the circumcircle
For a right triangle, the radius \( R \) of the circumcircle is half the length of the hypotenuse:
\[ R = \frac{AB}{2} \]
Substituting the value of AB:
\[ R = \frac{26}{2} = 13 \text{ cm} \]

### Conclusion
The radius of the circumcircle of triangle ABC is \( 13 \) cm.

---