Question

Construct ΔPQR such that ∠P = 70°, ∠R = 50°, QR = 7.3 cm and constructs its circumcircle.

Answer 1

Rough figure:

In Δ PQR,

∠P + ∠Q +∠R = 180°      ...(Sum of the measures of the angles of a triangle is 180°)

⇒ 70° + ∠Q + 50° = 180°

⇒ 120° + ∠Q = 180° 

⇒ ∠Q = 60°

Steps of construction:

1. Construct ΔPQR of the given measurement.
2. Draw the perpendicular bisectors of side PQ and side QR of the triangle.
3. Name the point of intersection of the perpendicular bisectors as point C.
4. Join seg CP.
5. Draw circle with centre C and radius CP.