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Ex 11.1, 1 - Construct angle 90 degree - Chapter 11 Class 9

Ex 11.1, 1 - Chapter 11 Class 9 Constructions - Part 2
Ex 11.1, 1 - Chapter 11 Class 9 Constructions - Part 3

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Ex 11.1, 1 - Chapter 11 Class 9 Constructions - Part 4 Ex 11.1, 1 - Chapter 11 Class 9 Constructions - Part 5 Ex 11.1, 1 - Chapter 11 Class 9 Constructions - Part 6

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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 Draw a ray OA. Taking O as center and any radius, draw an arc cutting OA at B. 3. Now, taking B as center and with the same radius as before, draw an arc intersecting the previously drawn arc at point C. 4. With C as center and the same radius, draw an arc cutting the arc at D 5. With C and D as centers 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° (Radius of equal arcs - By Construction) (Radius of equal arcs - By Construction) 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° (Radius of same arcs) (Arc of same radii) (Common) (We proved earlier that ∠ DOC= 60° ) Now, ∠ AOP = ∠ BOC + ∠ COP ∠ AOP = 60° + 30° ∠ AOP = 90° Hence justified

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