ΔABC is inscribed in a circle with centre O. If the sides of the triangle are $AB = 21 ext{ cm}$, $BC = 20 ext{ cm}$, and $AC = 29 ext{ cm}$, what is the length of the circumradius of the triangle?

Question

ΔABC is inscribed in a circle with centre O. If the sides of the triangle are $AB = 21 	ext{ cm}$, $BC = 20 	ext{ cm}$, and $AC = 29 	ext{ cm}$, what is the length of the circumradius of the triangle?

Accepted Answer

**Solution:**

1. Given sides of triangle $	riangle ABC$ are:
   - $AB = 21 	ext{ cm}$
   - $BC = 20 	ext{ cm}$
   - $AC = 29 	ext{ cm}$

2. The formula for the circumradius $R$ of a triangle with sides $a$, $b$, and $c$ is:
   $$
   R = \frac{abc}{4\Delta}
   $$
   where $\Delta$ is the area of the triangle.

3. Assign sides:
   - $a = 21$
   - $b = 20$
   - $c = 29$

4. Calculate semi-perimeter $s$:
   $$
   s = \frac{a + b + c}{2} = \frac{21 + 20 + 29}{2} = \frac{70}{2} = 35
   $$

5. Calculate area $\Delta$ using Heron's formula:
   $$
   \Delta = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{35(35 - 21)(35 - 20)(35 - 29)}
   $$
   $$
   = \sqrt{35 	imes 14 	imes 15 	imes 6}
   $$

6. Calculate the product inside the square root:
   $$
   35 	imes 14 = 490
   $$
   $$
   15 	imes 6 = 90
   $$
   $$
   490 	imes 90 = 44100
   $$

7. So,
   $$
   \Delta = \sqrt{44100} = 210
   $$

8. Now calculate the circumradius $R$:
   $$
   R = \frac{abc}{4\Delta} = \frac{21 	imes 20 	imes 29}{4 	imes 210}
   $$
   $$
   = \frac{12180}{840} = 14.5 	ext{ cm}
   $$

**Answer:** $14.5 	ext{ cm}$

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