RRB Non Technical Model Papers 6 with solutions

Question : 50 of 100

Marks: +1, -0

Triangle PQR circumscribes a circle with centre O and radius r cm such that ∠PQR = 90°. If PQ = 3 cm, QR = 4 cm, then the value of r is

Solution:

(4) Given, PQ = 3 cm
QR = 4 cm ∠PQR = 90°

PR^2 = PQ^2 + QR^2 = 3^2 + 4^2

= 9 + 16 = 25 cm
PR = 5 cm

= {Area of triangle}/{Semi-Perimeter of triangle}

r={1/2× 3 ×4 }/{3 +4 +5}/2=12/12= 1 cm

Go to Question:

Prev Question

Next Question