In a triangle PQR, ∠Q = 90°. If PQ = 12 cm and QR = 5 cm, then what is the radius (in cm) of the circumcircle of the triangle?
Given : ∠Q = 90°, PQ = 12 cm and QR = 5 cm
To find : CIrcumradius = ?
Solution : In right angled $$\triangle$$ PQR
=> $$(PR)^2=(PQ)^2+(QR)^2$$
=> $$(PR)^2=(12)^2+(5)^2$$
=> $$(PR)^2=144+25=169$$
=> $$PR=\sqrt{169}=13$$ cm
In a right angled triangle, the circumcentre lies at the mid point of hypoenuse.
=> Circumradius = $$\frac{PR}{2}=\frac{13}{2}=6.5$$ cm
=> Ans - (C)
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