The sides of `DeltaABC` are 10 cm, 10.5 cm and 14.5 cm. What is the radius of its circumcircle ?

Options

A. 5 cm
B. 7.5 cm
C. 5.25 cm
D. 7.25 cm

Answer

To find the radius of the circumcircle of triangle ABC with sides 10 cm, 10.5 cm, and 14.5 cm, we can follow these steps:

### Step 1: Identify the sides of the triangle
Let the sides of triangle ABC be:
- \( a = 10 \) cm
- \( b = 10.5 \) cm
- \( c = 14.5 \) cm

### Step 2: Check if the triangle is a right triangle
To determine if triangle ABC is a right triangle, we can use the Pythagorean theorem. For a triangle to be right-angled, the square of the longest side must equal the sum of the squares of the other two sides.

Here, the longest side is \( c = 14.5 \) cm.

We calculate:
\[ a^2 + b^2 = 10^2 + 10.5^2 = 100 + 110.25 = 210.25 \]
\[ c^2 = 14.5^2 = 210.25 \]

Since \( a^2 + b^2 = c^2 \), triangle ABC is a right triangle.

### Step 3: Use the formula for the circumradius of a right triangle
For a right triangle, the radius \( R \) of the circumcircle is given by the formula:
\[ R = \frac{c}{2} \]
where \( c \) is the length of the hypotenuse.

### Step 4: Calculate the circumradius
Substituting the value of \( c \):
\[ R = \frac{14.5}{2} = 7.25 \text{ cm} \]

### Conclusion
The radius of the circumcircle of triangle ABC is \( 7.25 \) cm.

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