Question : The sides of a triangle are 9 cm, 6 cm, and 5 cm. What is the value of the circumradius of this triangle?
Option 1: $\frac{9 \sqrt{2}}{2} \mathrm{~cm}$
Option 2: $\frac{9 \sqrt{3}}{5} \mathrm{~cm}$
Option 3: $\frac{9 \sqrt{3}}{4} \mathrm{~cm}$
Option 4: $\frac{27 \sqrt{2}}{8} \mathrm{~cm}$
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Correct Answer: $\frac{27 \sqrt{2}}{8} \mathrm{~cm}$
Solution :
Given: The three sides of the triangle are 9 cm, 6 cm, and 5 cm respectively.
Let $a, b$, and $c$ be the sides of the triangle.
We know that,
Area of triangle = $\sqrt{s(s - a) (s - b) (s - c)}$
Where $s = \frac{(a + b + c)}{2} = \frac{20}{2} = 10\ \text{cm}$
Area = $\sqrt{10 × (10 - 9) × (10 - 6) × (10 - 5)}=\sqrt{200}=10\sqrt{2}$
Circumradius = $\frac{\text{Product of the sides}}{4\times \text{Area}}=\frac{9\times6\times5}{4\times \text{Area}}=\frac{270}{40\sqrt{2}}=\frac{27 \sqrt{2}}{8}$
Hence, the correct answer is $\frac{27 \sqrt{2}}{8} \mathrm{~cm}$.
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