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  1. Chapter 11 Class 9 Constructions
  2. Serial order wise
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Construct an angle of 45° at the initial point of a given ray and justify the construction . Steps of construction Draw a ray OA. Taking O as centre and any radius, draw an arc cutting OA at B. Now, taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point C. With C as centre and the same radius, draw an arc cutting the arc at D. With C and D as centres and radius more than 1/2 CD draw two arcs intersecting at P. Join OP.  Thus, ∠ AOP = 90° Now, take B and Q as centers, and radius greater than 1/2 BQ, draw two arcs intersecting at R. Join OR.  Thus, ∠ AOR = 45° Justification We need to prove ∠ AOR = 45° Join OC Thus, OB = BC = OC ∴ Δ OCB is an equilateral triangle ∴ ∠ BOC = 60° Join OD, OC and CD Thus, OD = OC = DC ∴ Δ DOC is an equilateral triangle ∴ ∠ DOC = 60° Join PD and PC Now, In Δ ODP and Δ OCP   OD = OC   DP  = CP   OP =  OP ∴ Δ ODP ≅ Δ OCP ∴ ∠ DOP = ∠ COP So, we can say that ∠ DOP = ∠ COP = 1/2 ∠ DOC ∠ DOP = ∠ COP = 1/2  × 60° = 30° Now, ∠ AOP = ∠ BOC + ∠ COP ∠ AOP = 60° + 30° = 90° Now, Join QR and BR In Δ OQR and Δ OBR   OQ = OB   QR  = BR   OR =  OR ∴ Δ OQR ≅ Δ OBR ∴ ∠ QOR = ∠ BOR ∠ QOR = ∠ BOR = 1/2 ∠ AOP ∠ DOP = ∠ COP = 1/2  × 90° = 45° Thus, ∠ AOR = 45° Hence justified

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CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
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