RS Aggarwal Solutions: Construction (Using Ruler and a Pair of Compasses)

# RS Aggarwal Solutions: Construction (Using Ruler and a Pair of Compasses) | Mathematics (Maths) Class 6 PDF Download

``` Page 1

EXERCISE 14 A
Q. 1. Draw a line segment PQ = 6·2 cm. Draw
the perpendicular bisector of PQ.
Sol. Steps of construction :
(i) Draw a line segment PQ = 6·2 cm
(ii) With centre P and Q and radius more than
half of PQ, draw arcs on each side
intersecting each other at L and M.
(iii) Join LM intersecting PQ at N.
Then, LM is the perpendicular bisector
of PQ.
Q. 2. Draw a line segment AB = 5.6 cm. Draw
the perpendicular bisector of AB.
Sol. Steps of Construction :
1. Draw a line segment AB = 5.6 cm.
2. With A as centre and radius more than
half AB, draw arcs, one one each side
of AB.
3. With B as centre and same radius as
before, draw arcs, cutting the previous
arcs at P and Q respectively.
4. Join P and Q, meeting AB at M. Then
PQ is the required perpendicular
bisector of AB.
Verification : Measure AMP. We see
that AMP = 90º. So, PQ is the
perpendicular bisector of AB.
Q. 3. Draw an angle equal to AOB given in
Page 2

EXERCISE 14 A
Q. 1. Draw a line segment PQ = 6·2 cm. Draw
the perpendicular bisector of PQ.
Sol. Steps of construction :
(i) Draw a line segment PQ = 6·2 cm
(ii) With centre P and Q and radius more than
half of PQ, draw arcs on each side
intersecting each other at L and M.
(iii) Join LM intersecting PQ at N.
Then, LM is the perpendicular bisector
of PQ.
Q. 2. Draw a line segment AB = 5.6 cm. Draw
the perpendicular bisector of AB.
Sol. Steps of Construction :
1. Draw a line segment AB = 5.6 cm.
2. With A as centre and radius more than
half AB, draw arcs, one one each side
of AB.
3. With B as centre and same radius as
before, draw arcs, cutting the previous
arcs at P and Q respectively.
4. Join P and Q, meeting AB at M. Then
PQ is the required perpendicular
bisector of AB.
Verification : Measure AMP. We see
that AMP = 90º. So, PQ is the
perpendicular bisector of AB.
Q. 3. Draw an angle equal to AOB given in
Sol. Steps of Contruction :
1. Draw a ray RX.
2. With O as centre and any radius draw
an arc cutting OA and OB at P and Q
respectively.
3. With R as centre and same radius draw
an arc cutting RX at S.
4. With S as centre and radius PQ cut the
arc through S at T.
5. Join RT and produce it to Y. Then XRY
is the required angle equal to AOB.
Verification : Measuring angle AOB and
XRY, we observe that XRY =
AOB.
Q. 4. Draw an angle of 50º with the help of a
protractor. Draw a ray bisecting this
angle.
Sol. Steps of constructions :
(i) Draw an angle ABC = 50º with the help
of a protractor.
(ii) With centre B and C and a suitable radius,
draw an arc meeting AB at Q and BC at
P.
(iii) With centres P and Q and with a suitable
radius draw two arcs intersecting each
other at R inside the angle ABC.
(iv) Join RB.
Then ray BR is the bisector of ABC.
Q. 5. Construct AOB = 85º with the help of
a protractor. Draw a ray OX bisecting
AOB.
Sol. Steps of construction :
(i) Draw an angle AOB = 85º with the help
of the protractor.
(ii) With centre O, draw an arc with a suitable
radius meeting OB at E and OA at F.
(iii) With centre E and F and with a suitable
radius draw arcs intersecting each other
at X inside the angle AOB.
Then ray OX is the bisector of AOB.
Q. 6. Draw a line AB. Take a point P on it.
Draw a line passing through P and
perpendicular to AB.
Page 3

EXERCISE 14 A
Q. 1. Draw a line segment PQ = 6·2 cm. Draw
the perpendicular bisector of PQ.
Sol. Steps of construction :
(i) Draw a line segment PQ = 6·2 cm
(ii) With centre P and Q and radius more than
half of PQ, draw arcs on each side
intersecting each other at L and M.
(iii) Join LM intersecting PQ at N.
Then, LM is the perpendicular bisector
of PQ.
Q. 2. Draw a line segment AB = 5.6 cm. Draw
the perpendicular bisector of AB.
Sol. Steps of Construction :
1. Draw a line segment AB = 5.6 cm.
2. With A as centre and radius more than
half AB, draw arcs, one one each side
of AB.
3. With B as centre and same radius as
before, draw arcs, cutting the previous
arcs at P and Q respectively.
4. Join P and Q, meeting AB at M. Then
PQ is the required perpendicular
bisector of AB.
Verification : Measure AMP. We see
that AMP = 90º. So, PQ is the
perpendicular bisector of AB.
Q. 3. Draw an angle equal to AOB given in
Sol. Steps of Contruction :
1. Draw a ray RX.
2. With O as centre and any radius draw
an arc cutting OA and OB at P and Q
respectively.
3. With R as centre and same radius draw
an arc cutting RX at S.
4. With S as centre and radius PQ cut the
arc through S at T.
5. Join RT and produce it to Y. Then XRY
is the required angle equal to AOB.
Verification : Measuring angle AOB and
XRY, we observe that XRY =
AOB.
Q. 4. Draw an angle of 50º with the help of a
protractor. Draw a ray bisecting this
angle.
Sol. Steps of constructions :
(i) Draw an angle ABC = 50º with the help
of a protractor.
(ii) With centre B and C and a suitable radius,
draw an arc meeting AB at Q and BC at
P.
(iii) With centres P and Q and with a suitable
radius draw two arcs intersecting each
other at R inside the angle ABC.
(iv) Join RB.
Then ray BR is the bisector of ABC.
Q. 5. Construct AOB = 85º with the help of
a protractor. Draw a ray OX bisecting
AOB.
Sol. Steps of construction :
(i) Draw an angle AOB = 85º with the help
of the protractor.
(ii) With centre O, draw an arc with a suitable
radius meeting OB at E and OA at F.
(iii) With centre E and F and with a suitable
radius draw arcs intersecting each other
at X inside the angle AOB.
Then ray OX is the bisector of AOB.
Q. 6. Draw a line AB. Take a point P on it.
Draw a line passing through P and
perpendicular to AB.
Sol. Steps of Construction :
1. Draw the given line AB and take a point
P on it.
2. With P as centre and any suitable radius
draw a semi-circle to cut the line AB at
X and Y.
3. With centre X and radius more than XP
draw an arc.
4. With centre Y and same radius draw
another arc to cut the previous arc at Q.
5. Join PQ. Then, PQ is the required line
passing through P and perpendicular to
AB.
Verification : Measure APQ, we see
that APQ = 90º
Q. 7. Draw a line AB. Take a point P outside
it. Draw a line passing through P and
perpendicular to AB.
Sol. Steps of Construction :
1. Draw the given line AB and take a point
P outside it.
2. With P as centre and suitable radius,
draw an arc intersecting AB at C and D.
3. With C as centre and radius more than
half CD, draw an arc.
4. With D as centre and same radius, draw
another arc to cut the previous arc at Q.
5. Join PQ, meeting AB at L. Then PL is
the required line passing thrugh P and
perpendicular to AB.
Verification : Measure PLB. We see
that PLB = 90º.
Q. 8. Draw a line AB. Take a point P outside
it. Draw a line passing through P and
parallel to AB.
Sol. Steps of Construction :
1. Draw a given line AB and take a point P
outside it.
2. Take a point R on AB.
3. Join PR.
4. Draw RPQ such that RPQ = PRB
as shown in the figure.
5. Produce PQ on both sides to form a line.
Then, PQ is the required line passing
through P and parallel to AB.
Verification : Since RPQ = PRB and
these are alternate interior angles, it
follows that PQ | | AB.
Q. 9. Draw an BAC of measure 60º such
that AB = 4.5 cm and AC = 5 cm.
Through C draw a line parallel to AB
and through B draw a line parallel to AC,
intersecting each other at D. Measure
BD and CD.
Sol. Steps of Construction :
1. Draw a ray AX and cut of AC = 5 cm.
2. With A as centre and suitable radius draw
an arc above AX and cutting it at P.
Page 4

EXERCISE 14 A
Q. 1. Draw a line segment PQ = 6·2 cm. Draw
the perpendicular bisector of PQ.
Sol. Steps of construction :
(i) Draw a line segment PQ = 6·2 cm
(ii) With centre P and Q and radius more than
half of PQ, draw arcs on each side
intersecting each other at L and M.
(iii) Join LM intersecting PQ at N.
Then, LM is the perpendicular bisector
of PQ.
Q. 2. Draw a line segment AB = 5.6 cm. Draw
the perpendicular bisector of AB.
Sol. Steps of Construction :
1. Draw a line segment AB = 5.6 cm.
2. With A as centre and radius more than
half AB, draw arcs, one one each side
of AB.
3. With B as centre and same radius as
before, draw arcs, cutting the previous
arcs at P and Q respectively.
4. Join P and Q, meeting AB at M. Then
PQ is the required perpendicular
bisector of AB.
Verification : Measure AMP. We see
that AMP = 90º. So, PQ is the
perpendicular bisector of AB.
Q. 3. Draw an angle equal to AOB given in
Sol. Steps of Contruction :
1. Draw a ray RX.
2. With O as centre and any radius draw
an arc cutting OA and OB at P and Q
respectively.
3. With R as centre and same radius draw
an arc cutting RX at S.
4. With S as centre and radius PQ cut the
arc through S at T.
5. Join RT and produce it to Y. Then XRY
is the required angle equal to AOB.
Verification : Measuring angle AOB and
XRY, we observe that XRY =
AOB.
Q. 4. Draw an angle of 50º with the help of a
protractor. Draw a ray bisecting this
angle.
Sol. Steps of constructions :
(i) Draw an angle ABC = 50º with the help
of a protractor.
(ii) With centre B and C and a suitable radius,
draw an arc meeting AB at Q and BC at
P.
(iii) With centres P and Q and with a suitable
radius draw two arcs intersecting each
other at R inside the angle ABC.
(iv) Join RB.
Then ray BR is the bisector of ABC.
Q. 5. Construct AOB = 85º with the help of
a protractor. Draw a ray OX bisecting
AOB.
Sol. Steps of construction :
(i) Draw an angle AOB = 85º with the help
of the protractor.
(ii) With centre O, draw an arc with a suitable
radius meeting OB at E and OA at F.
(iii) With centre E and F and with a suitable
radius draw arcs intersecting each other
at X inside the angle AOB.
Then ray OX is the bisector of AOB.
Q. 6. Draw a line AB. Take a point P on it.
Draw a line passing through P and
perpendicular to AB.
Sol. Steps of Construction :
1. Draw the given line AB and take a point
P on it.
2. With P as centre and any suitable radius
draw a semi-circle to cut the line AB at
X and Y.
3. With centre X and radius more than XP
draw an arc.
4. With centre Y and same radius draw
another arc to cut the previous arc at Q.
5. Join PQ. Then, PQ is the required line
passing through P and perpendicular to
AB.
Verification : Measure APQ, we see
that APQ = 90º
Q. 7. Draw a line AB. Take a point P outside
it. Draw a line passing through P and
perpendicular to AB.
Sol. Steps of Construction :
1. Draw the given line AB and take a point
P outside it.
2. With P as centre and suitable radius,
draw an arc intersecting AB at C and D.
3. With C as centre and radius more than
half CD, draw an arc.
4. With D as centre and same radius, draw
another arc to cut the previous arc at Q.
5. Join PQ, meeting AB at L. Then PL is
the required line passing thrugh P and
perpendicular to AB.
Verification : Measure PLB. We see
that PLB = 90º.
Q. 8. Draw a line AB. Take a point P outside
it. Draw a line passing through P and
parallel to AB.
Sol. Steps of Construction :
1. Draw a given line AB and take a point P
outside it.
2. Take a point R on AB.
3. Join PR.
4. Draw RPQ such that RPQ = PRB
as shown in the figure.
5. Produce PQ on both sides to form a line.
Then, PQ is the required line passing
through P and parallel to AB.
Verification : Since RPQ = PRB and
these are alternate interior angles, it
follows that PQ | | AB.
Q. 9. Draw an BAC of measure 60º such
that AB = 4.5 cm and AC = 5 cm.
Through C draw a line parallel to AB
and through B draw a line parallel to AC,
intersecting each other at D. Measure
BD and CD.
Sol. Steps of Construction :
1. Draw a ray AX and cut of AC = 5 cm.
2. With A as centre and suitable radius draw
an arc above AX and cutting it at P.
3. With P as centre and the same radius as
before draw another arc to cut the
previous arc at Q.
4. Join PQ and produce it to the point B
such that. AB = 4.5 cm. Then BAC =
60º is the required angle.
5. Draw ÐRBA such that ÐRBA = ÐBAC.
6. Produce RB on both sides to form a line.
Then, RY is the line parallel to AC and
passing through B.
7. Now, draw ÐSCX = ÐBAC at the point
C.
8. Produce CS to intersect the line RY at
D. Then CD is the required line thrugh
C and parallel to AB.
9. Measure BD and CD. We see that BD =
5 cm. and CD = 4.5 cm.
Verification. Since RBA = BAC and
these are alternate angles, it follows that
RY | | AC.
Also SCX = BAC and these are
corresponding  angles, it follows that
CD | | AB.
Q. 10. Draw a line segment AB = 6 cm. Take a
point C on AB such that AC = 2.5 cm.
Draw CD perpendicular to AB.
Sol. Steps of Construction :
1. With the help of a rular, draw a line
segment AB = 6 cm. and off AC = 2.5
cm such that the point C is on AB.
2. With C as centre and any suitable radius
draw a semi-circle to cut AB at P and
Q.
3. With P as centre and any radius more
than PC draw an arc.
4. With Q as centre and same radius draw
another arc to cut the previous arc at
D.
5. Join CD. Then CD is the required line
perpendicular to AB.
Verification : Measure ACD. We see
that ACD = 90º.
Q. 11. Draw a line segment AB = 5.6 cm. Draw
the right bisector of AB.
Sol. Steps of Construction :
1. With the help of rular, draw a line segment
AB = 5.6 cm.
2. With A as centre and radius more than
half AB, draw arcs, one on each side of
AB.
3. With B as centre and the same radius as
before draw arcs, cutting the previous
arcs at P and Q respectively.
4. Join PQ, meeting AB at M. Then, PQ is
the required right bisector of AB.
Verification : On measuring AM and
BM and AMP, we see that AM = BM
and AMP = 90º.
So, PQ is the right bisector of AB.
Q. 12. Using a rular and a pair of compasses,
construct AOB = 60º and draw the
bisector of this angle.
Sol. Steps of Construction :
1. With the help of a rular, draw a ray OA.
2. With O as centre and suitable radius
draw an arc to cut OA at P.
Page 5

EXERCISE 14 A
Q. 1. Draw a line segment PQ = 6·2 cm. Draw
the perpendicular bisector of PQ.
Sol. Steps of construction :
(i) Draw a line segment PQ = 6·2 cm
(ii) With centre P and Q and radius more than
half of PQ, draw arcs on each side
intersecting each other at L and M.
(iii) Join LM intersecting PQ at N.
Then, LM is the perpendicular bisector
of PQ.
Q. 2. Draw a line segment AB = 5.6 cm. Draw
the perpendicular bisector of AB.
Sol. Steps of Construction :
1. Draw a line segment AB = 5.6 cm.
2. With A as centre and radius more than
half AB, draw arcs, one one each side
of AB.
3. With B as centre and same radius as
before, draw arcs, cutting the previous
arcs at P and Q respectively.
4. Join P and Q, meeting AB at M. Then
PQ is the required perpendicular
bisector of AB.
Verification : Measure AMP. We see
that AMP = 90º. So, PQ is the
perpendicular bisector of AB.
Q. 3. Draw an angle equal to AOB given in
Sol. Steps of Contruction :
1. Draw a ray RX.
2. With O as centre and any radius draw
an arc cutting OA and OB at P and Q
respectively.
3. With R as centre and same radius draw
an arc cutting RX at S.
4. With S as centre and radius PQ cut the
arc through S at T.
5. Join RT and produce it to Y. Then XRY
is the required angle equal to AOB.
Verification : Measuring angle AOB and
XRY, we observe that XRY =
AOB.
Q. 4. Draw an angle of 50º with the help of a
protractor. Draw a ray bisecting this
angle.
Sol. Steps of constructions :
(i) Draw an angle ABC = 50º with the help
of a protractor.
(ii) With centre B and C and a suitable radius,
draw an arc meeting AB at Q and BC at
P.
(iii) With centres P and Q and with a suitable
radius draw two arcs intersecting each
other at R inside the angle ABC.
(iv) Join RB.
Then ray BR is the bisector of ABC.
Q. 5. Construct AOB = 85º with the help of
a protractor. Draw a ray OX bisecting
AOB.
Sol. Steps of construction :
(i) Draw an angle AOB = 85º with the help
of the protractor.
(ii) With centre O, draw an arc with a suitable
radius meeting OB at E and OA at F.
(iii) With centre E and F and with a suitable
radius draw arcs intersecting each other
at X inside the angle AOB.
Then ray OX is the bisector of AOB.
Q. 6. Draw a line AB. Take a point P on it.
Draw a line passing through P and
perpendicular to AB.
Sol. Steps of Construction :
1. Draw the given line AB and take a point
P on it.
2. With P as centre and any suitable radius
draw a semi-circle to cut the line AB at
X and Y.
3. With centre X and radius more than XP
draw an arc.
4. With centre Y and same radius draw
another arc to cut the previous arc at Q.
5. Join PQ. Then, PQ is the required line
passing through P and perpendicular to
AB.
Verification : Measure APQ, we see
that APQ = 90º
Q. 7. Draw a line AB. Take a point P outside
it. Draw a line passing through P and
perpendicular to AB.
Sol. Steps of Construction :
1. Draw the given line AB and take a point
P outside it.
2. With P as centre and suitable radius,
draw an arc intersecting AB at C and D.
3. With C as centre and radius more than
half CD, draw an arc.
4. With D as centre and same radius, draw
another arc to cut the previous arc at Q.
5. Join PQ, meeting AB at L. Then PL is
the required line passing thrugh P and
perpendicular to AB.
Verification : Measure PLB. We see
that PLB = 90º.
Q. 8. Draw a line AB. Take a point P outside
it. Draw a line passing through P and
parallel to AB.
Sol. Steps of Construction :
1. Draw a given line AB and take a point P
outside it.
2. Take a point R on AB.
3. Join PR.
4. Draw RPQ such that RPQ = PRB
as shown in the figure.
5. Produce PQ on both sides to form a line.
Then, PQ is the required line passing
through P and parallel to AB.
Verification : Since RPQ = PRB and
these are alternate interior angles, it
follows that PQ | | AB.
Q. 9. Draw an BAC of measure 60º such
that AB = 4.5 cm and AC = 5 cm.
Through C draw a line parallel to AB
and through B draw a line parallel to AC,
intersecting each other at D. Measure
BD and CD.
Sol. Steps of Construction :
1. Draw a ray AX and cut of AC = 5 cm.
2. With A as centre and suitable radius draw
an arc above AX and cutting it at P.
3. With P as centre and the same radius as
before draw another arc to cut the
previous arc at Q.
4. Join PQ and produce it to the point B
such that. AB = 4.5 cm. Then BAC =
60º is the required angle.
5. Draw ÐRBA such that ÐRBA = ÐBAC.
6. Produce RB on both sides to form a line.
Then, RY is the line parallel to AC and
passing through B.
7. Now, draw ÐSCX = ÐBAC at the point
C.
8. Produce CS to intersect the line RY at
D. Then CD is the required line thrugh
C and parallel to AB.
9. Measure BD and CD. We see that BD =
5 cm. and CD = 4.5 cm.
Verification. Since RBA = BAC and
these are alternate angles, it follows that
RY | | AC.
Also SCX = BAC and these are
corresponding  angles, it follows that
CD | | AB.
Q. 10. Draw a line segment AB = 6 cm. Take a
point C on AB such that AC = 2.5 cm.
Draw CD perpendicular to AB.
Sol. Steps of Construction :
1. With the help of a rular, draw a line
segment AB = 6 cm. and off AC = 2.5
cm such that the point C is on AB.
2. With C as centre and any suitable radius
draw a semi-circle to cut AB at P and
Q.
3. With P as centre and any radius more
than PC draw an arc.
4. With Q as centre and same radius draw
another arc to cut the previous arc at
D.
5. Join CD. Then CD is the required line
perpendicular to AB.
Verification : Measure ACD. We see
that ACD = 90º.
Q. 11. Draw a line segment AB = 5.6 cm. Draw
the right bisector of AB.
Sol. Steps of Construction :
1. With the help of rular, draw a line segment
AB = 5.6 cm.
2. With A as centre and radius more than
half AB, draw arcs, one on each side of
AB.
3. With B as centre and the same radius as
before draw arcs, cutting the previous
arcs at P and Q respectively.
4. Join PQ, meeting AB at M. Then, PQ is
the required right bisector of AB.
Verification : On measuring AM and
BM and AMP, we see that AM = BM
and AMP = 90º.
So, PQ is the right bisector of AB.
Q. 12. Using a rular and a pair of compasses,
construct AOB = 60º and draw the
bisector of this angle.
Sol. Steps of Construction :
1. With the help of a rular, draw a ray OA.
2. With O as centre and suitable radius
draw an arc to cut OA at P.
3. With P as centre and the same radius,
draw another are to cut the previous arc
at Q.
4. Join OQ and produce it to any point B,
then AOB = 60º is the required angle.
5. With P as centre and radius more than
half PQ, draw an arc.
6. With Q as centre and the same radius,
draw another arc to cut the previous arc
at R.
7. Join OR and produce it to the point C.
Then OC is the required bisector of
AOB.
Verification : Measure AOC and
BOC. We see that AOC = BOC.
So, OC is the bisector of AOB.
Q. 13. Construct an angle of 135º, using a rular
and pair of compasses.
Sol. Steps of construction :
1. Draw a ray OA with the help of a rular.
2. With O as centre and suitable radius
draw an arc above OA to cut it at P .
3. With P as centre and same radius, cut
the arc at Q and again with Q as centre
and same radius cut the arc at R. With
R as centre and same radius, again cut
the arc at S.
4. Join OR and produce it to B and join OS
and produce it to C.
5. Draw the bisector OD of BOC.
6. Draw the bisector OE of BOD. Then,
AOE = 135º is the required angle.
EXERCISE 14B
Q. 1. Using a pair of compasses construct the
following angles :
(i) 60º (ii) 120º (iii) 90º
Sol. (i) 60º
Steps of construction :
(i) Draw a ray OA.
(ii) With centre O and with a suitable radius
drawn an arc meeting OA at E.
(iii) With centre E and with same radius, draw
another arc cutting the first arc at F.
(iv) Join OF and produce it to B
Then AOB = 60º
(ii) 120º
Steps of construction :
(i) Draw a ray OA
(ii) With centre O and with a suitable radius
draw an arc meeting OA at E.
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## Mathematics (Maths) Class 6

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## FAQs on RS Aggarwal Solutions: Construction (Using Ruler and a Pair of Compasses) - Mathematics (Maths) Class 6

 1. How can I construct an equilateral triangle using a ruler and a pair of compasses?
Ans. To construct an equilateral triangle using a ruler and a pair of compasses, follow these steps: 1. Draw a line segment using a ruler. 2. Place the compass at one end of the line segment and draw an arc that intersects the line segment. 3. Without changing the compass width, place the compass at the other end of the line segment and draw another arc that intersects the line segment. 4. The intersection point of the two arcs is the third vertex of the equilateral triangle. 5. Draw line segments connecting each endpoint of the original line segment to the third vertex to complete the equilateral triangle.
 2. How can I construct a perpendicular bisector using a ruler and a pair of compasses?
Ans. To construct a perpendicular bisector using a ruler and a pair of compasses, follow these steps: 1. Draw a line segment using a ruler. 2. Place the compass at one endpoint of the line segment and draw an arc that intersects the line segment. 3. Without changing the compass width, place the compass at the other endpoint of the line segment and draw another arc that intersects the line segment. 4. The intersection point of the two arcs is the midpoint of the line segment. 5. Use a ruler to draw a line passing through the midpoint and perpendicular to the line segment. This line is the perpendicular bisector.
 3. How can I construct a triangle with given sides using a ruler and a pair of compasses?
Ans. To construct a triangle with given sides using a ruler and a pair of compasses, follow these steps: 1. Draw a line segment using a ruler, representing one of the given sides of the triangle. 2. With the compass width set to the length of the second given side, place the compass at one endpoint of the line segment and draw an arc. 3. Without changing the compass width, place the compass at the other endpoint of the line segment and draw another arc that intersects the first arc. 4. The intersection points of the two arcs are the possible locations for the third vertex of the triangle. 5. Use a ruler to connect each endpoint of the line segment to each of the possible third vertex locations. These lines will form the triangle with the given sides.
 4. How can I construct an isosceles triangle using a ruler and a pair of compasses?
Ans. To construct an isosceles triangle using a ruler and a pair of compasses, follow these steps: 1. Draw a line segment using a ruler, representing the base of the triangle. 2. With the compass width set to the desired length of the equal sides, place the compass at one endpoint of the line segment and draw an arc. 3. Without changing the compass width, place the compass at the other endpoint of the line segment and draw another arc that intersects the first arc. 4. The intersection point of the two arcs is the third vertex of the isosceles triangle. 5. Use a ruler to draw line segments connecting each endpoint of the base to the third vertex. These lines will form the isosceles triangle.
 5. How can I construct an angle bisector using a ruler and a pair of compasses?
Ans. To construct an angle bisector using a ruler and a pair of compasses, follow these steps: 1. Draw the angle using a ruler, with its vertex at the center. 2. Place the compass at the vertex of the angle and draw an arc that intersects both sides of the angle. 3. Without changing the compass width, place the compass at one of the points where the arc intersects a side of the angle and draw another arc that intersects the first arc. 4. Repeat step 3 from the other point where the arc intersects the other side of the angle. 5. The intersection point of the two arcs is the desired angle bisector. 6. Use a ruler to draw a line passing through the vertex and the intersection point of the arcs. This line is the angle bisector.

## Mathematics (Maths) Class 6

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