Question

In the given figure, PQR is a triangle in whichPQ = PR and ∠PQR = 75°, then show that QR is equal to the radius of circumcircle of ∆PQR, whose centre is O​

Answer 1

Step-by-step explanation:

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Answer 2

Given :   PQR is a triangle in which PQ = PR and ∠PQR = 75°,

To Find : show that QR is equal to the radius of circumcircle of ∆PQR, whose centre is O.

Solution :

PQ = PR and ∠PQR = 75°,

=> ∠PRQ = 75°  

∠PQR  + ∠PRQ + ∠QPR = 180°

=> 75° + 75° + ∠QPR = 180°

=> ∠QPR = 30°

∠QOR = 2 * ∠QPR   ( angle by same chord at circle And center )

=> ∠QOR = 2 *  30° = 60°

in Δ QOR

QO = OR  = Radius

=> ∠OQR = ∠ORQ

∠OQR = ∠ORQ + ∠QOR = 180°

=>  ∠OQR = ∠ORQ + 60° = 180°

=>  ∠OQR = ∠ORQ  = 120°

=> ∠OQR = ∠ORQ = 60°

∠OQR = ∠ORQ =  ∠QOR = 60°

=> Δ QOR is Equilateral Triangle

=> QR = OQ = OR  = Radius of circum circle

QED

Hence shown that  QR is equal to the radius of circumcircle of ∆PQR, whose centre is O.

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