Question:
An isosceles triangle PQR is inscribed inside a circle. If PQ = PR = \(8\sqrt5\) cm and QR = 16 cm, then find the radius of the circle.
An isosceles triangle PQR is inscribed inside a circle. If PQ = PR = \(8\sqrt5\) cm and QR = 16 cm, then find the radius of the circle.
Updated On: Apr 28, 2025
- 8 cm
- 12 cm
- 15 cm
- 10 cm
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The Correct Option is D
Solution and Explanation
We know circum-radius of a triangle = {product of sides/(4 × area of the triangle)}
For triangle PQR, let PM be the perpendicular bisector of QR
Therefore, PR = \(8\sqrt5\) cm, MR = \(\frac{16}{2}\) = 8 cm
In triangle PMR, using Pythagoras theorem
PM2 = PR2 - MR2
Or, PM2 = (\(8\sqrt5\))2 - 82 = 320 - 64
Or, PM2 = 256
Or, PM = 16 (Since, length cannot be negative)
Therefore, area of the triangle = \((\frac{1}{2})\) × 16 × 16 = 128 cm2
Required circum-radius = \(\frac{(8\sqrt5 × 8\sqrt5 × 16)}{(4 × 128)}\) = 10 cm
So, the correct option is (D) : 10 cm.
For triangle PQR, let PM be the perpendicular bisector of QR
Therefore, PR = \(8\sqrt5\) cm, MR = \(\frac{16}{2}\) = 8 cm
In triangle PMR, using Pythagoras theorem
PM2 = PR2 - MR2
Or, PM2 = (\(8\sqrt5\))2 - 82 = 320 - 64
Or, PM2 = 256
Or, PM = 16 (Since, length cannot be negative)
Therefore, area of the triangle = \((\frac{1}{2})\) × 16 × 16 = 128 cm2
Required circum-radius = \(\frac{(8\sqrt5 × 8\sqrt5 × 16)}{(4 × 128)}\) = 10 cm
So, the correct option is (D) : 10 cm.
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