Question : $\triangle A B C$ inscribe in a circle with centre O. If AB = 9 cm, BC = 40 cm, and AC = 41 cm, then what is the circum-radius of the triangle?
Option 1: $20 \frac{1}{2} \mathrm{~cm}$
Option 2: $12 \frac{1}{2} \mathrm{~cm}$
Option 3: $18 \frac{1}{2} \mathrm{~cm}$
Option 4: $16 \frac{1}{2} \mathrm{~cm}$
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Correct Answer: $20 \frac{1}{2} \mathrm{~cm}$
Solution :
Given: $\triangle A B C$ inscribe in a circle with centre O.
AB = 9 cm, BC = 40 cm, and AC = 41 cm
Also, we can see $9^2+40^2=1681=41^2$
So, it is a right triangle.
Hypotenuse = AC = 41 cm
For any right-angled triangle,
Circumradius $= \frac{1}{2} \times \text{Hypotenuse}= \frac{1}{2} \times 41= \frac{41}{2} = 20\frac{1}{2}\ \text{cm}$
Hence, the correct answer is $20\frac{1}{2}\ \text{cm}$.
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