ΔPQR inscribes in a circle with centre O. If PQ = 12 cm, QR = 16 cm and PR = 20 cm, find the circumradius of a triangle.

Given:

In ΔPQR,

PQ = 12 cm, QR = 16 cm and PR = 20 cm

Concept used:

a, b and c denote the sides of the triangle and A denotes the area of the triangle,

Then the measure of the circumradius(r) is.

r = [abc/4A]

If the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the biggest side is a right angle.

Hypotenuse² = Perpendicular² + Base² 

Calculation:

Here, we can see that 

(20)2 = (16)2 + (12)2  = 400 

⇒ PR² = QR² + PQ²

So, ΔPQR is a right angle triangle.

Area of ΔPQR = (½ ) × base × Perpendicular

⇒ Area of ΔPQR = (½ ) × 16 × 12

⇒ Area of ΔPQR = 96 cm²

Circumradius (r) = [abc/4 × Area] 

Circumradius (r) = [(12 × 16 × 20)/4 × 96] = 10 cm

∴ The circumradius of a triangle is 10 cm.

For a right-angled triangle, the circumcenter lies at the midpoint of the hypotenuse. All the vertices of a triangle are at an equal distance from the circumcenter.

PO = QO = OR = r

⇒ PO = PR/2

⇒ PO = 20/2 = 10 cm

∴ The circumradius of a triangle is 10 cm.