Question
Construct ∆PQR such that ∠P = 70°, ∠R = 50°, QR = 7.3 cm and construct its circumcircle.
✓
Answer
In ∆PQR,
m∠P + m∠Q + m∠R = 180° … [Sum of the measures of the angles of a triangle is 180°]
∴ 70° + m∠Q + 50° = 180°
∴ m∠Q = 180° – 70° + m∠Q + 50° = 180°
∴ m∠Q = 180° – 70° – 50°
∴ m∠Q = 60°
m∠P + m∠Q + m∠R = 180° … [Sum of the measures of the angles of a triangle is 180°]
∴ 70° + m∠Q + 50° = 180°
∴ m∠Q = 180° – 70° + m∠Q + 50° = 180°
∴ m∠Q = 180° – 70° – 50°
∴ m∠Q = 60°
Steps of construction:
i. Construct A PQR of the given measurement.
ii. Draw the perpendicular bisectors of side PQ and side QR of the triangle.
iii. Name the point of intersection of the perpendicular bisectors as point C.
iv. Join seg CP
v. With C as centre and CP as radius, draw a circle which passes through the three vertices of the triangle.
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