Find the radius of the circle passing through the vertices of a right angled triangle when length of perpendicular sides are 6 and 8.

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Solution

We know that an angle in a semicircle is a right angle.
So, we can say that the hypotenuse (AB) of triangle ABC is the diameter of the circle.
Now, using Pythagoras theorem in triangle ABC, we have:

$AB = \sqrt{{AC}^{2} + {BC}^{2}} = \sqrt{8^{2} + 6^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10 units$

∴ Radius of the circle = 5 units (AB is the diameter)

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Find the radius of the circle passing through the vertices of a right angled triangle when length of perpendicular sides are 6 and 8.

We know that an angle in a semicircle is a right angle. So, we can say that the hypotenuse (AB) of triangle ABC is the diameter of the circle. Now, using Pythagoras theorem in triangle ABC, we have: AB=AC2+BC2 =82+62 =64+36 =100 = 10 units ∴ Radius of the circle = 5 units (AB is the diameter)