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  1. Chapter 11 Class 9 Constructions
  2. Serial order wise
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Ex 11.1, 4 Construct the following angles and verify by measuring them by a Protractor : (i) 75° 75° = 60°  + 15° 75° = 60°  + (30°)/2 So, to we make 75° , we make 60° and then bisector of 30° 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. Now, with C as center, and same radius, draw another arc intersecting the previously drawn arc at point D. Thus, ∠ AOE = 60° and ∠ EOF = 60° Taking C and D as Centre , with radius more than 1/2CD, draw arcs intersecting at P. Join OP Thus, ∠ EOP = 30° Taking Q and C as Centre , with radius more than 1/2QC, draw arcs intersecting at R. Join OR Thus, ∠ AOR = 75° On measuring the ∠ AOR by protractor, we find that ∠ AOR = 75° Thus, the construction is verified     (ii) 105° 105° = 90°  + 15° 105° = 90°  + (30°)/2 So, to make 105° , we make 90° and then bisector of 30° 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° Join OD ∠ POD = 30° as OP is bisector of ∠ COD. With Q and D as centres and radius more than 1/2 DQ draw two arcs intersecting at R.   Join OR Thus, ∠ POR = 15° ∴ ∠ AOR = ∠ AOP + ∠ POR = 90° + 15° ∴ ∠ AOR = 105°   (iii) 135° 135° = 90°  + 45° So, to make 135° , we make 90° and then 45° Steps of construction Draw a line OAA’. 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° Also, ∠ A’OP = 90° So, to we bisect ∠ A’OP 7.With B’ and Q as centres and radius more than 1/2 B’Q draw two arcs intersecting at R. 8.Join OR. ∴ ∠ A’OR = 45° ⇒ ∠ POR = 45° Thus, ∠ AOR = ∠ AOP + ∠ POR = 90° + 35° ∴ ∠ AOR = 135°

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