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Ex 11.1, 1

Construct an angle of 90° at the initial point of a given ray and justify the construction .

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°

Justification

We need to prove ∠ AOP = 90°

Join OC and BC

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°

∠ AOP = 90°

Hence justified

 

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  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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