1. Construct an
angle of 90° at the initial point of a given ray and justify the construction.
Construct an angle of 45° at the initial point of a given ray and justify the construction.
3. Construct the angles of the following measurements:
(i) 22.10
2
(i) 15°
(1) 30°
4. Construct the following angles and verify by measuring them by a protractor:
() 75°
Construct an equilateral triangle, given its side and justify the construction.
,
5/
Answers
Step-by-step explanation:
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∘.
