Question 3 - Chapter 11 Class 9 Constructions (Important Question)

Last updated at Dec. 16, 2024 by

Ex 11.2, 3 - Construct triangle PQR - QR = 6cm, Q = 60 and PR - PQ = 2

Ex 11.2, 3 - Chapter 11 Class 9 Constructions - Part 2
Ex 11.2, 3 - Chapter 11 Class 9 Constructions - Part 3

Go Ad-free

Transcript

Question 3 Construct a triangle PQR in which QR = 6 cm, ∠Q = 60° and PR − PQ = 2 cm. Steps of Construction: Draw base QR of length 6 cm 2. Now, let’s draw ∠ Q = 60° Let the ray be QX Open the compass to length PR – PQ = 2 cm. From point Q as center, cut an arc on ray QX. (opposite side of QR). Let the arc intersect QX at D 4. Join RD Note: Since PR – PQ = 2 cm, (PQ – PR) is negative So, QD will be below line QR Now, we will draw perpendicular bisector of RD 6. Mark point P where perpendicular bisector intersects RD Join PR ∴ Δ PQR is the required triangle

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo