Maharashtra State BoardSSC (English Medium) 9th Standard
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Construct δ Pqr, Such that Qr = 7.4 Cm, ∠ Pqr= 60 ∘ and Pq − Pr = 2.5 Cm. - Geometry

Diagram

Construct Δ PQR, such that  QR = 7.4 cm, ∠PQR= 60 ° and  PQ - PR = 2.5 cm.

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Solution

Steps of construction: 
(1) Draw seg QR of length 6.5 cm.
(2) Draw ray QA such that ∠RQA = 60°
(3) Take point B on ray QA such that QB = 2.5 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
Practice Set 4.2 | Q 2 | Page 54
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