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 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°
Now we draw bisector of ∠ AOP
7. Let OP intersect the original arc at point Q
8. Now, taking B and Q as centers, and radius greater than 1/2 BQ,
draw two arcs intersecting at R.
9. Join OR.
Thus, ∠ AOR = 45°
Justification
We need to prove ∠ AOR = 45°
Join OC & OB
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°
(Radius of same arcs)
(Arc of same radii)
(Common)
(SSS Congruency)
(CPCT)
(We proved earlier that ∠ DOC= 60° )
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
(Radius of same arcs)
(Arc of same radii)
(Common)
(SSS Congruency)
∠ QOR = ∠ BOR = 1/2 ∠ AOP
∠ DOP = ∠ COP = 1/2 × 90°
= 45°
Thus, ∠ AOR = 45°
Hence justified

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.