Question 137
In $$\triangle$$PQR, $$\angle$$RPQ = 90° PR=6CM and PQ=8CM then the radius of the circumcircle of $$\triangle$$PQR is
Solution
In $$\triangle$$PQR
=> QR = $$\sqrt{(PR)^2 + (PQ)^2}$$
=> QR = $$\sqrt{6^2 + 8^2}$$ = $$\sqrt{100}$$
=> QR = 10 cm
In a right angled triangle, circumcentre lies on the mid point of hypotenuse.
=> circumradius = $$\frac{QR}{2}$$ = $$\frac{10}{2}$$ = 5 cm
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