"Question 1 Construct an angle of 90° at the initial point of a given ray and justify the construction.
Class 9 - Math - Constructions Page 191"
Answers
First,Steps of construction .
1}}Draw a ray OA
2}}With it's initial point O as centre and any radius, draw an arc CDE, cutting OA at C.
3}} With center C and same radius ,draw an arc, cutting the arc CDE at D.
4}}With D as centre and the same radius ,draw am arc cutting the arc CDE and D.
5}}}With D and E as centres,and any convenient radius (more than 1/2DE),draw two arcs intersecting at P.
6}}}Join OP .Then /_Aop =90°
Hope it is helpful...
Steps of Construction:
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∘.
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