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Gaurav Seth 5 years, 5 months ago
Given: A ray OA.
Required: To construct an angle of 90° 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 are intersecting the previously drawn are, 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 the radius more than 
ID, 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
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.
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Sia ? 5 years, 5 months ago
Let {tex}\triangle{/tex}ABC be a right angled triangle in which {tex}\therefore{/tex}B = 90o
Then {tex}\angle{/tex}A + {tex}\angle{/tex}C = 90o . . . .[sum of three angles of a triangle]
{tex}\therefore{/tex} {tex}\angle{/tex}B = {tex}\angle{/tex}A + {tex}\angle{/tex}C
{tex}\therefore{/tex} {tex}\angle{/tex}B > {tex}\angle{/tex}A
and {tex}\angle{/tex}B > {tex}\angle{/tex}C
{tex}\therefore{/tex} AC > BC . . . [Side opposite to greater angle is longer]
and AC > AB
{tex}\therefore{/tex} AC is the longest side, i.e. hypotenuse is the longest side.
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Yuvika Mourya 5 years, 5 months ago
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