Question : The sides of $\triangle$ABC are 10 cm, 10.5 cm and 14.5 cm. What is the radius of its circumcircle?
Option 1: 5 cm
Option 2: 7.5 cm
Option 3: 5.25 cm
Option 4: 7.25 cm
Correct Answer: 7.25 cm
Solution :
The sides of $\triangle$ABC are 10 cm, 10.5 cm and 14.5 cm.
Now, $10.5^2 + 10^2$
$= 110.25 + 100$
$= 210.25 = 14.5^2$
So, the triangle is a right-angled triangle.
Now, in a right-angled triangle,
Circum radius of the triangle = $\frac{\text{Hypotenuse}}{2}$ = $\frac{14.5}{2}$ = 7.25 cm
Hence, the correct answer is 7.25 cm.
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