Question

"Question 2 Construct an angle of 45° at the initial point of a given ray and justify the construction.

Class 9 - Math - Constructions Page 191"

Answer 1

Hi friend.

STEPS OF CONSTRUCTION:-

1)Draw a ray OA .


2)With O as centre and any suitable radius draw an arc cutting OA at B .

3)With B as centre and same radius cut the previous drawn. arc at C and then with C as centre and radius cut the arc at D.

4(With C as centre and radius more than half Cd draw an arc .

5)With D as centre and same radius draw another arc to cut the previous arc at E.

6=)Join OE ,then angle AOE ==90°.

7))Draw the bisector OF of Angle AOE ,then Angle AOF =45°.

Justify......

By construction.
Angle AOE ==90° and OF is the bisector of Angle AoE

So, Angle =1/2/_AOE =1/2×90°==45°


Hope helped ...

Answer 2

 [Fig. is in the attachment]

 Steps of Construction:

 

1.First , draw a ray OA with intial point O.

 

2.Taking O as a centre and some radius, draw an arc of a circle, which intersects OA, at a point B.

3.Taking B as centre and with the same radius as before, draw an arc intersecting the previous drawn arc, at a point C.

 

4.Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 2, say at D.

5.Draw the ray OE passing through C. Then ∠EOA = 60∘ .

6.Draw the ray OF passing through D. Then ∠ FOE = 60∘ .

7. Next, taking C and D as centres and with radius more than ½ CD, draw arcs to intersect each other, at G.

8. Draw the ray OG, which is the bisector of the angle FOE, i.e., ∠FOG = ∠EOG = 1/2 ∠FOE = 1/2 (60∘ ) = 30∘ .

Thus, ∠GOA = ∠GOE + ∠ EOA = 30∘ + 60∘ = 90∘.

9.Now taking O as centre and any radius more than OB, draw an arc to intersect the rays OA and OG, at H and I.

10. Next, taking H and I as centres and with the radius more than 1/2 HI, draw arcs to intersect each other, at J.

11.  Draw the ray OJ. This ray OJ is the required bisector of the ∠ GOA.

 Thus, ∠GOJ = ∠AOJ = 1/2  ∠GOA =  1/2(90∘) = 45∘.

 

 Justification:

(i)Join BC.

Then, OC = OB = BC triangle. (By construction)

∴   ∠COB is an equilateral triangle.

∴   ∠COB = 60∘.

∴   ∠EOA = 60∘.

 

(ii)Join CD.

Then, OD = OC = CD (By construction)

∆DOC is an equilateral triangle.

∴ ∠DOC = 60∘.

∴ ∠ FOE = 60∘.

 

(iii)Join CG and DG.

In ΔODG and ΔOCG,

OD = OC                           

 [ Radii of the same arc]

DG=CG                             [Arcs of equal radii]

OG=OG                            [Common]

∴ Δ ODG = ΔOCG          [SSS Rule]

 ∴ ∠ DOG=  ∠COG           [CPCT]

∴ ∠FOG = ∠ EOG = 1/2 ∠FOE = 1/2 (60∘) = 30∘

 

Thus, ∠GOA =  ∠GOE +  ∠EOA = 30∘ + 60∘ = 90∘.

 

Iv)  Join HJ and IJ.

In ΔOIJ and ΔOHJ,

OI = OH          

[Radii of the same arc]

IJ = HJ              

[Arcs of equal radii]

OJ = OJ             

 [Common]

 ∴ ΔOIJ ≅ ΔOHJ

[SSS Rule]

 ∴  ∠IOJ = ∠ HOJ

[CPCT]

 

∴ ∠AOJ= ∠GOJ= 1/2  ∠GOA = ½(90°)=45

=========================================================

Hope this will help you.....