Question

If the side of the triangle $ABC$ are $6,8,10$ units, then find the radius of circumcircle is

• A. $4$
• B. $3$
• C. $6$
• D. $5$

Answer 1

Answer: (D)

$5$

REF.Image.

given : sides of $\Delta$ ABC : 6,8,10

$[6^2+8^2={10}^2]$

$(AB{)}^2+(BC{)}^2=(AC{)}^2$

$\therefore \Delta$ABC is right triangle.

radius of circum circle of a right triangle=$\left[\frac{hypotenuse}{2}\right]$

radius of circum circle of $△ABC=\frac{AC}{2}=5$ units.

$\therefore$ The radius of circum circle of $\Delta$ABC is 5 units.