Construct ∆PQR such that PQ=6cm,Q=35°,QR=5.5.Draw incircle of ∆PQR.Draw the diagram neatly.
Answers
(i) Steps of Construction:
(i) Draw a line segment PQ = 6 cm.
(ii) At P, draw a ray making an angle of 45°
(iii) At Q, draw another ray making an angle of 60° which intersects the first ray at R.
∆ PQR is the required triangle.
On measuring ∠R, it is 75°.
(ii) Steps of Construction :
(i) Draw a line segment QR = 44 cm.
(ii) At Q, draw a ray making an angle of 75°
(iii) At R, draw another arc making an angle of 30° ; which intersects the first ray at R
∆ PQR is the required triangle.
On measuring the lengths of PQ and PR, PQ = 2.1 cm and PR = 4. 4 cm.
(iii) Steps of Construction :
(i) Draw a line segment PR = 5.8 cm
(ii) At P, construct an angle of 60°
(iii) At R, draw another angle of 45° meeting each other at Q.
∆ PQR is the required triangle. On measuring ∠Q, it is 75°
Verification : We know that sum of angles of a triangle is 180°
∴∠P + ∠Q + ∠R = 180°
⇒ 60° + ∠Q + 45° = 180°
⇒ ∠Q + 105° = 180°
⇒ ∠Q = 180° – 105° = 75°