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Construct a ∆PQR in which Q =60°=45° and QR= 8 cm

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Construct a ∆ PQR such that : PQ = 6 cm, ∠Q = 60° and ∠P = 45°.

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asked Jun 11, 2019 in Class VII Maths by navnit40 (-4,942 points)

Construct a ∆ PQR such that :

(i) PQ = 6 cm, ∠Q = 60° and ∠P = 45°. Measure ∠R.

(ii) QR = 4.4 cm, ∠R = 30° and ∠Q = 75°. Measure PQ and PR.

(iii) PR = 5.8 cm, ∠P = 60° and ∠R = 45°.

Measure ∠Q and verify it by calculations

triangles

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answered Jun 11, 2019 by priya12 (-12,635 points)

(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°