Surface Areas and Volumes

Question

Construct the following angles and verify by measuring them by a protractor:

135°

Solution

135°

Given: A ray OA.
Required: To construct an angle of 135° at O. Steps of Construction:
1. Produce AO to A' to form ray OA'.
2. Taking O as centre and some radius, draw an arc of a circle, which intersects OA at a point B and OA' at a point B'.

3. Taking B as centre and with the same radius as before, draw an arc intersecting the previously 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 1, 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 the radius more than $1 half$ CD, draw arcs to intersect each other, say at G.
8. Draw the ray OG intersecting the arc drawn in step 1 at H. This ray OG is the bisector of the angle FOE,
$straight i. straight e. comma space space space angle FOG equals angle EOG equals 1 half angle FOE equals 1 half left parenthesis 60 degree right parenthesis equals 30 degree. Thus comma space angle GOA equals angle GOE plus angle EOA equals 30 degree plus 60 degree equals 90 degree therefore space angle straight B apostrophe OH equals 90 degree$

9. Next, taking B' and H as centres and with the radius more than the radius more than $1 half$ B'H, draw arcs to intersect each other, say at I.

10. Draw the ray OI. This ray 01 is the bisector of the angle B'OG, i.e., ∠B'OI = ∠GOI
$equals 1 half angle straight B apostrophe OG equals 1 half left parenthesis 90 degree right parenthesis equals 45 degree Thus comma space angle IOA equals angle IOG plus angle GOA space space space space space space space space space space space space space space space space equals space 45 degree plus 90 degree equals 135 degree$

On measuring the ∠IOA by protractor, we find that ∠IOA = 135°.
Thus, the construction is verified.

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