In the given figure, PQR is a triangle in which PQ = PR and ∠PQR = 75°, then show that QR is equal to the radius of circumcircle of ∆PQR, whose centre is O.
Answers
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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Answer:
PQ=PR and angle PQR=75^ ,
=> angle PRQ=75^
angle PQR+ angle PRQ+ angle QPR=180^
Rightarrow75^ +75^ + angle QPR=180^
Rightarrow angle QPR=30^
angle QOR=2^ * angle QPR (angle by same chord at
circle And center)
Rightarrow angle QOR=2^ * 30^ =60^
in A QOR
QO = OR = Radius
angle OQR= angle ORQ
angle OQR= angle ORQ+ angle QOR=180^
angle OQR= angle ORQ+60^ =180^
angle OQR= angle ORQ=120^
angle OQR= angle ORQ=60^
angle OQR= angle ORQ= angle QOR=60^
=> QOR is Equilateral Triangle
QR=OQ=OR= Radius of circum circle
QED
Hence shown that QR is equal to the radius of circumcircle of Delta*F , whose centre is O.
Step-by-step explanation:
I hope it helps