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

  1. Chapter 11 Class 9 Constructions (Term 2)
  2. Serial order wise

Transcript

Ex 11.2, 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

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.