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Sia ? 6 years, 5 months ago
- Draw OQ of any length.
- Place the centre of the protractor at O and the zero edge along OQ.
- Start with 0 near Q. Mark point P at 90°.
- Join {tex}\overline{OP}{/tex}. Then, ∠POQ = 90°.
- With O as centre and using compasses, draw an arc that cuts both rays of ∠POQ. Label the points of intersection as P' and Q'.
- With Q' as centre, draw (in the interior of ∠POQ) an arc whose radius is more than half the length Q'P'.
- With the same radius and with P' as centre, draw another arc in the interior of ∠POQ. Let the two arcs intersect at R. Then, {tex}\overline{OR}{/tex} is the bisector of ∠POQ.

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- Draw a line segment {tex}\overline {AB} {/tex} of length 7.3cm.
- With A as centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of {tex}\overline {AB} {/tex}.
- With the same radius and with B as centre, draw another circle using compasses. Let it cut the previous circle at C and D.
- Join CD. Then, {tex}\overline {CD} {/tex} is the axis of symmetry of {tex}\overline {AB} {/tex}.
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Gurpreet Kaur 8 years ago
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