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Surface Areas And Volumes

Question
CBSEENMA9003175

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

Solution

Given: A ray OA.
Required: To construct an angle of 45° at O and justify the construction.
Steps of Construction:
1. Taking O as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.
2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.

4. Draw the ray OE passing through C. Then ∠EOA = 60°.
5. Draw the ray OF passing through D. Then ∠FOE = 60°.
6.  Next, taking C and D as centres and with radius more than 1 half CD, draw arcs to intersect each other, say at G.
7.  Draw the ray OG. This ray OG is the bisector of the angle FOE, i.e., ∠FOG
equals angle EOG equals 1 half angle FOE equals 1 half left parenthesis 60 degree right parenthesis equals 30 degree
Thus comma space space angle GOA space equals space angle GOE space plus space angle EOA space equals space 30 degree plus 60 degree equals 90 degree
8. Now, taking O as centre and any radius, draw an arc to intersect the rays OA and OG. say at H and I respectively.
9.  Next, taking H and I as centres and with the radius more than 1 half HI, draw arcs to intersect each other, say at J.
10. Draw the ray OJ. This ray OJ is the required bisector of the angle GOA.
Thus comma space space space angle GOJ equals angle AOJ equals 1 half angle GOA
space space space space space space space space space space space space space space space space space space space equals space 1 half left parenthesis 90 degree right parenthesis equals 45 degree

Justification:
(i) Join BC.
Then, OC = OB = BC (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 ∆ODG and ∆OCG,
OD = OC                    | Radi fo the same arc
DG = CG                       | Arcs of equal radii
OG = OG                                   | Common
therefore space space space space increment ODG space equals space increment OCG space space space space space space space space space space space space space space space space space vertical line space SSS space Rule
therefore space space space space space angle DOG space equals space angle COG space space space space space space space space space space space space space space space space space space space space space vertical line thin space CPCT
therefore space space angle FOG equals angle EOG equals 1 half angle FOE
space space space space space space space space space space space space space space space equals 1 half left parenthesis 60 degree right parenthesis equals 30 degree
Thus comma space angle GOA equals angle GOE plus angle EOA
space space space space space space space space space space space space space space space space space space space equals space 30 degree plus 60 degree equals 90 degree
(iv) Join HJ and IJ
In space increment OIJ space space and space increment OHJ
OI = OH                | Radii of the same arc
IJ = HJ                    | Arcs of equal radii
OJ=OJ                                | Common
therefore space space space increment OIJ thin space equals space increment OHJ space space space space space space space space space space space space space space space vertical line thin space SSS space Rule
therefore space space space angle IOJ space equals space angle HOJ space space space space space space space space space space space space space space space space space space space vertical line space CPCT
therefore space space space angle AOJ equals angle GOJ equals 1 half angle GOA
space space space space space space space space space space space space space space space space equals space 1 half left parenthesis 90 degree right parenthesis equals 45 degree