Question 137

In ΔPQR, ∠PQR = 90°, PQ = 5 cm and QR = 12. What is the radius (in cm) of the circum-circle of ΔPQR?

Solution

Given : In ΔPQR, ∠PQR = 90°, PQ = 5 cm and QR = 12 cm

To find : OR = ?

Solution : In $$\triangle$$ PQR,

=> $$(PR)^2=(PQ)^2+(QR)^2$$

=> $$(PR)^2=(5)^2+(12)^2$$

=> $$(PR)^2=25+144=169$$

=> $$PR=\sqrt{169}=13$$ cm

Also, in a right angled triangle, circumradius is half the hypotenuse of the triangle.

$$\therefore$$ OR = $$r=\frac{13}{2}=6.5$$ cm

=> Ans - (A)

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