Question

EXERCISE 11.1
1. Construct an angle of 90° at the initial point of a given ray and justify the construction.​

Answer 1

Answer:

1. First draw a ray OA with initial point O.

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

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

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

5.Draw the ray OE passing through C. Then ∠EOA = 60∘ & the ray OF passing through D. Then ∠FOE =60∘.

6.Next taking C and D as centres and with the radius more than 1/2 CD, draw arcs to intersect each other, at a point G.

7.Draw the ray 0G, which is the angle bisector of the ∠FOE,

i.e., ∠FOG = ∠EOG = 1/2 ∠FOE =1/2 (60∘) = 30∘.

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

Justification:

(i) Join BC.

Then, OC = OB = BC (By construction)

∴ ΔCOB is an equilateral triangle.

∴ ∠COB =60∘.

∴ ∠ EOA = 60∘.

ii) Join CD.

Then, OD = OC = CD (By construction)

So, ΔDOC 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]

∴ ΔODG≅ Δ OCG [SSS Rule]

∴ ∠DOG =∠COG [CPCT]

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

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

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