Question

B
А
APQR - APMN. In APQR, PQ = 4 cm, QR = 5 cm and PR = 6 cm. Construct APQR
PR 3
and APMN such that
PN 5​

Answer 1

Answer:

Selina solutions

Grade 7

Biology

Mathematics

Physics

Chapters in CONCISE Mathematics - Middle School - 7

Exercises in Triangles

Question 3

Q3) 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

Solution:

(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 the sum of angles of a triangle is 180°

∴∠P + ∠Q + ∠R = 180°

⇒ 60° + ∠Q + 45° = 180°

⇒ ∠Q + 105° = 180°

⇒ ∠Q = 180° – 105° = 75°.

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