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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

  1. Draw a ray OA.
  2. Taking O as centre and any radius, draw an arc cutting OA at B.
  3. Now, taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point C.
  4. 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°
  5. Taking C and D as Centre , with radius more than 1/2CD, draw arcs intersecting at P.
  6. Join OP
    Thus, ∠ EOP = 30°
  7. Taking Q and C as Centre , with radius more than 1/2QC, draw arcs intersecting at R.
  8. 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

  1. Draw a ray OA.
  2. Taking O as centre and any radius, draw an arc cutting OA at B.
  3. Now, taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point C.
  4. With C as centre and the same radius, draw an arc cutting the arc at D.
  5. With C and D as centres and radius more than 1/2 CD draw two arcs intersecting at P.
  6. Join OP.  Thus, ∠ AOP = 90°
  7. Join OD ∠ POD = 30° as OP is bisector of ∠ COD.
  8. 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

  1. Draw a line OAA’ .
  2. Taking O as centre and any radius, draw an arc cutting OA at B.
  3. Now, taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point C.
  4. With C as centre and the same radius, draw an arc cutting the arc at D.
  5. With C and D as centres and radius more than 1/2 CD draw two arcs intersecting at P.
  6. 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°  

-ev-

  1. Chapter 11 Class 9 Constructions
  2. Concept wise
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CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 7 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
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