Maharashtra State BoardSSC (English Medium) 9th Standard
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Construct δ Pqr, in Which Pq − Pr = 2.4 Cm, Qr = 6.4 Cm and ∠ Pqr = 55 ∘ . - Geometry

Diagram

Construct Δ PQR, in which PQ - PR = 2.4 cm, QR = 6.4 cm and ∠PQR = 55° . 

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Solution

​Steps of construction: 
(1) Draw seg QR of length 6.4 cm.
(2) Draw ray QA such that ∠RQA = 55°
(3) Take point B on ray QA such that QB = 2.4 cm.
(4) Construct the perpendicular bisector of seg BR.
(5) Name the point of intersection of ray QA and the perpendicular bisector of seg BR as P.
 (6) Draw seg PR.
Therefore, △PQR is required triangle.

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APPEARS IN

Balbharati Mathematics 2 Geometry 9th Standard Maharashtra State Board
Chapter 4 Constructions of Triangles
Problem Set 4 | Q 4 | Page 56
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