Question
Draw an angle of $80^\circ$ using a protractor and divide it into four equal parts, using ruler and compasses. Check your construction by measurement.
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Answer
Here, to divide an angle of measure $80^\circ $ into four equal parts, we use the following steps of construction:
Step I: Draw a line segment $AB$ of any length. Place the centre of the protractor at A and the zero edge along $AB.$
Step II: Start with zero near $B$ and mark $C$ at $80^\circ .$
Step III: Join $AC,$ then $\angle\text{BAC}$ is an angle of measure $80^\circ .$
Step IV: With $A$ as centre and using compass, draw an arc that cuts both the rays of $\angle\text{A}$ at $P$ and $Q.$
Step V: With $P$ as centre, draw $($in the interior of $\angle\text{A})$ an arc, whose radius is more than half the length of $PQ.$
Step VI: With $Q$ as centre and the same radius, draw another arc in the interior of $A.$ Let the two arcs intersect at $D.$ Join $AD,$ cutting arc $PQ$ at $L.$ Then, $AD$ divides the $\angle\text{BCA}$ into two equal parts.
Step VII: Now, taking $P$ and $L$ as centre, having radius more than half of length $PL,$ draw two arcs respectively, which cut each other at $R.$
Step VIII: Join $AR,$ which divides $\angle\text{BAD}$ into two equal parts.
Step IX: Now, taking $Q$ and $L$ as centre, having radius more than half of length $QL,$ draw two arcs respectively, which cut each other at $M.$
Step X: Join $AM,$ which divide $\angle\text{CAD}$ into two equal parts. Thus, $AM, AD$ and $AR$ divide $\angle\text{BAC}$ into four equal parts.
Step I: Draw a line segment $AB$ of any length. Place the centre of the protractor at A and the zero edge along $AB.$
Step II: Start with zero near $B$ and mark $C$ at $80^\circ .$
Step III: Join $AC,$ then $\angle\text{BAC}$ is an angle of measure $80^\circ .$
Step IV: With $A$ as centre and using compass, draw an arc that cuts both the rays of $\angle\text{A}$ at $P$ and $Q.$
Step V: With $P$ as centre, draw $($in the interior of $\angle\text{A})$ an arc, whose radius is more than half the length of $PQ.$
Step VI: With $Q$ as centre and the same radius, draw another arc in the interior of $A.$ Let the two arcs intersect at $D.$ Join $AD,$ cutting arc $PQ$ at $L.$ Then, $AD$ divides the $\angle\text{BCA}$ into two equal parts.
Step VII: Now, taking $P$ and $L$ as centre, having radius more than half of length $PL,$ draw two arcs respectively, which cut each other at $R.$
Step VIII: Join $AR,$ which divides $\angle\text{BAD}$ into two equal parts.
Step IX: Now, taking $Q$ and $L$ as centre, having radius more than half of length $QL,$ draw two arcs respectively, which cut each other at $M.$
Step X: Join $AM,$ which divide $\angle\text{CAD}$ into two equal parts. Thus, $AM, AD$ and $AR$ divide $\angle\text{BAC}$ into four equal parts.
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No.
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Column A
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Column B
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$i$
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Points are collinear.
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$a.$
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May be parallel or intersecting.
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$ii$
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Line is completely known.
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$b.$
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Are undefined terms in geometry.
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$iii$
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Two lines in a plane.
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$c.$
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If they lie on the same line.
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$iv$
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Relations between points and lines.
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$d.$
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Can pall through a point.
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$v$
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Three non-collinear points.
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$e.$
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Determine a plane.
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$vi$
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A plane extends.
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$f.$
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Are called incidence porperties.
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$vii$
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Indefinite number of lines.
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$g.$
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It two points are givne.
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$viii$
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Point, line and plane are.
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$h.$
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Indefinitely in all directions.
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