web analytics

Slide6.JPG

Slide7.JPG
Slide8.JPG Slide9.JPG Slide10.JPG Slide11.JPG Slide12.JPG

-v-

Ex 11.1, 2

Construct an angle of 45° 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°
  7. Now, take B and Q as centers, and radius greater than 1/2 BQ, draw two arcs intersecting at R.
  8. 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

-ev-

  1. Class 9
  2. Important Questions for Exam - Class 9
Ask Download

Transcript

Ex 11.1, 2 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

About the Author

CA Maninder Singh's photo - Expert in Practical Accounts, Taxation and Efiling
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 .
Jail