Geometric Constructions – Maharashtra Board Class 6 Solutions for Mathematics

Geometric Constructions – Maharashtra Board Class 6 Solutions for Mathematics (English Medium)

MathematicsGeneral ScienceMaharashtra Board Solutions

Exercise-78

Solution 1:

1. seg MR ⊥ ray ST
2. line LM ⊥ seg PQ
3. line HP ⊥ ray OK
4. seg KG ⊥ seg VJ
5. line AD ⊥ line EF

Exercise-79

Solution 1:

Steps of construction:

1. Draw line XY.
2. Take a point S outside the line.
3. Place one side of the right angle of the set square on the line XY and let the other side of the right angle pass through point S.
4. Draw a line along the side of the set square which passes through S.
5. Name the point where this line intersects line XY as M.

Solution 2:

Steps of construction:

1. Draw line AB.
2. Take a point C outside the line.
3. Adjust the compass to a convenient radius.
4. Place the point of the compass on C and draw two arcs cutting AB at points E and F.
5. Using the same radius and placing the point of compass on E and F draw two arcs of circle intersecting each other on the side of line AB opposite to point C.
6. Name the point of intersection as D.
7. Draw the line CD passing through points C and D.

Hence, CD ⊥ AB.

Exercise-80

Solution 1:
Steps of construction:

1. Draw line PQ.
2. Take a point R anywhere on line PQ.
3. Place one side of the right angle of the set square along the line PQ. Move the set square so that the vertex of the right angle falls on point R.
4. Draw a line SR along the other side of the set square’s right angle, through R.

Hence, line RS ⊥ line PQ.

Solution 2:
Steps of construction:

1. Draw line MN.
2. Take a point L anywhere on the line MN.
3. Adjust the compass to a convenient span.
4. Place the point of the compass on L and draw two arcs, one on each side of point L, to intersect line MN at points E and F respectively.
5. Open the compass to a span a little greater than half the distance between points E and F.
6. Place the point of the compass first on point E and then on point F and draw two arcs on one side of the line MN, intersecting each other. Name the point of intersection as D.
7. Draw line LD.

Hence, line CD ⊥ line MN.

Exercise-81

Solution 1(1):
Steps of construction:

1. Draw segment MN of length 8 cm.
2. Open the compass to a span greater than half the length of seg MN.
3. Taking M as the centre draw an arc on each side of seg MN.
4. Now, keeping the same span and taking N as the centre, draw an arc on each side of seg MN to intersect the previous arcs.
5. Name the points of intersection as C and B.
6. Draw the line BC.

Hence, BC is the perpendicular bisector of seg MN.

Solution 1(2):
Steps of construction:

1. Draw segment AB of length 6.5 cm.
2. Open the compass to a span greater than half the length of seg AB.
3. Taking A as the centre draw an arc on each side of seg AB.
4. Now, keeping the same span and taking B as the centre, draw an arc on each side of seg AB to intersect the previous arcs.
5. Name the points of intersection as M and N.
6. Draw the line MN.

Hence, MN is the perpendicular bisector of seg AB.

Solution 1(3):
Steps of construction:

1. Draw segment PQ of length 7 cm.
2. Open the compass to a span greater than half the length of seg PQ.
3. Taking P as the centre draw an arc on each side of seg PQ.
4. Now, keeping the same span and taking Q as the centre, draw an arc on each side of seg PQ to intersect the previous arcs.
5. Name the points of intersection as S and T.
6. Draw the line ST.

Hence, ST is the perpendicular bisector of seg PQ.

Solution 1(4):
Steps of construction:

1. Draw segment XY of length 5.7 cm.
2. Open the compass to a span greater than half the length of seg XY.
3. Taking X as the centre draw an arc on each side of seg XY.
4. Now, keeping the same span and taking Y as the centre, draw an arc on each side of seg XY to intersect the previous arcs.
5. Name the points of intersection as V and W.
6. Draw the line VW.

Hence, VW is the perpendicular bisector of seg XY.

Solution 1(5):
Steps of construction:

1. Draw segment AB of length 9.2 cm.
2. Open the compass to a span greater than half the length of seg AB.
3. Taking A as the centre, draw an arc on each side of seg AB.
4. Now, keeping the same span and taking B as the centre, draw an arc on each side of seg AB to intersect the previous arcs.
5. Name the points of intersection as L and M.
6. Draw the line LM.

Hence, LM is the perpendicular bisector of seg AB.

Exercise-82

Solution 1(1):
Steps of construction:

1. Draw ∠ABC of measure 40°.
2. Place the point of the compass on point B.
3. Taking a suitable span, draw an arc intersecting ray BA and ray BC at the points M and N respectively.
4. Keeping the same span, draw arcs, first with centre M and then another with centre N, to intersect the first arc. Let the point of intersection be O.
5. Draw ray BO.

Thus, ray BO is the bisector of ∠ABC.

Solution 1(2):
Steps of construction:

1. Draw ∠ABC of measure 90°.
2. Place the point of the compass on point B.
3. Taking a suitable span, draw an arc intersecting ray BA and ray BC at the points M and N respectively.
4. Keeping the same span, draw arcs, first with centre M and then another with centre N, to intersect the first arc. Let the point of intersection be O.
5. Draw ray BO.

Thus, ray BO is the bisector of ∠ABC.

Solution 1(3):
Steps of construction:

1. Draw ∠ LMN of measure 120°.
2. Place the point of the compass on point M.
3. Taking a suitable span, draw an arc intersecting ray ML and ray MN at the points P and Q respectively.
4. Keeping the same span, draw arcs, first with centre P and then another with centre Q, to intersect the first arc. Let the point of intersection be O.
5. Draw ray MO.

Thus, ray MO is the bisector of ∠LMN.

Solution 1(4):
Steps of construction:

1. Draw ∠ LMN of measure 55°.
2. Place the point of the compass on point M.
3. Taking a suitable span, draw an arc intersecting ray ML and ray MN at the points P and Q respectively.
4. Keeping the same span, draw arcs, first with centre P and then another with centre Q, to intersect the first arc. Let the point of intersection be O.
5. Draw ray MO.

Thus, ray MO is the bisector of ∠LMN.

Solution 1(5):
Steps of construction:

1. Draw ∠LMN of measure 105°.
2. Place the point of the compass on point M.
3. Taking a suitable span, draw an arc intersectingray ML and ray MN at the points P and Q respectively.
4. Keeping the same span, draw arcs, first with centre P and then another with centre Q, to intersect the first arc. Let the point of intersection be O.
5. Draw ray MO.

Thus, ray MO is the bisector of ∠LMN.

Solution 1(6):
Steps of construction:

1. Draw ∠ LMN of measure 146°.
2. Place the point of the compass on point M.
3. Taking a suitable span, draw an arc intersecting ray ML and ray MN at the points P and Q respectively.
4. Keeping the same span, draw arcs, first with centre P and then another with centre Q, to intersect the first arc. Let the point of intersection be O.
5. Draw ray MO.

Thus, ray MO is the bisector of ∠LMN.

Exercise-83

Solution 1(1):
Steps of construction:

1. Draw ray QR.
2. Open the compass to a convenient span.
3. Place the point of the compass on point B of ∠ABC and draw an arc to intersect ray BA and ray BC at points D and E respectively.
4. Keeping the same span, place the point of the compass on point Q and draw an arc. Let it intersect the ray QR at point T.
5. Now, place the point of the compass on point E and adjust the span so that the tip of the pencil falls on point D.
6. Place the point of the compass on point T and draw an arc to intersect the arc drawn before. Name the point of intersection as S.
7. Draw ray QS. Take any point P on this ray.

Thus, ∠PQR is equal in measure to the given angle.

Solution 1(2):
Steps of construction:

1. Draw ray QR.
2. Open the compass to a convenient span.
3. Place the point of the compass on point C of ∠BCD and draw an arc to intersect ray CB and ray CD at points F and E respectively.
4. Keeping the same span, place the point of the compass on point Q and draw an arc. Let it intersect the ray QR at point T.
5. Now, place the point of the compass on point E and adjust the span so that the tip of the pencil falls on point F.
6. Place the point of the compass on point T and draw an arc to intersect the arc drawn before. Name the point of intersection as S.
7. Draw ray QS.Take any point P on this ray.

Thus, ∠ PQR is equal in measure to the given angle.

Solution 1(3):
Steps of construction:

1. Draw ray QR.
2. Open the compass to a convenient span.
3. Place the point of the compass on point Y of ∠XYZ and draw an arc to intersect ray YX and ray YZ at points F and E respectively.
4. Keeping the same span, place the point of the compass on point Q and draw an arc. Let it intersect the ray QR at point T.
5. Now, place the point of the compass on point E and adjust the span so that the tip of the pencil falls on point F.
6. Place the point of the compass on point T and draw an arc to intersect the arc drawn before. Name the point of intersection as S.
7. Draw ray QS. Take any point P on this ray.

Thus, ∠PQR is equal in measure to the given angle.

Solution 1(4):
Steps of construction:

1. Draw ray QR.
2. Open the compass to a convenient span.
3. Place the point of the compass on point B of ∠ ABC and draw an arc to intersect ray BA and ray BC at points D and E respectively.
4. Keeping the same span, place the point of the compass on point Q and draw an arc. Let it intersect the ray QR at point T.
5. Now, place the point of the compass on point E and adjust the span so that the tip of the pencil falls on point D.
6. Place the point of the compass on point T and draw an arc to intersect the arc drawn before. Name the point of intersection as S.
7. Draw ray QS. Take any point P on this ray.

Thus, ∠PQR is equal in measure to the given angle.

Solution 1(5):
Steps of construction:

1. Draw ray QR.
2. Open the compass to a convenient span.
3. Place the point of the compass on point B of ∠ABC and draw an arc to intersect ray BA and ray BC at points D and E respectively.
4. Keeping the same span, place the point of the compass on point Q and draw an arc. Let it intersect the ray QR at point T.
5. Now, place the point of the compass on point E and adjust the span so that the tip of the pencil falls on point D.
6. Place the point of the compass on point T and draw an arc to intersect the arc drawn before. Name the point of intersection as S.
7. Draw ray QS. Take any point P on this ray.

Thus, ∠ PQR is equal in measure to the given angle.

Exercise-84

Solution 1:
Steps of construction:

1. Draw a line XY.
2. Take a point R outside line XY.
3. Place one side of the right angle of the set square along the line XY and let the other side of the right angle pass through point R.
4. Take another set square. Place one side of the right angle of the second set square along the side of the first set square which is perpendicular to line XY with the other side parallel to line XY.
5. Now, keeping the first set square in its place, move the second one so that the vertex of its right angle falls on point R.
6. Draw a line ST through point R along the free side of the right angle of the second set square.

Thus, line ST passes through point R and is parallel to line XY.

Solution 2:
Steps of construction:

1. First draw segments ST and KL. We have to draw a segment equal in length to the sum of their lengths.
2. Draw a line l and mark any point D on it.
3. Measure a distance equal to the length of seg ST between the tips of the divider.
4. Without changing the span of the divider, place one of its tips on D and mark F on the line where its other tip falls.
5. Now measure a distance equal to the length of seg KL.
6. Without changing the span of the divider, place one of its tips on point F, and the other on line l on the side opposite to point D. Name this point E.

Now seg DE has a length equal to the sum of the lengths of seg ST and seg KL.

Solution 3:
Steps of construction:

1. First draw two segments ST and PQ of unequal

lengths.

1. Draw another line m.
2. Take a point M on line m.
3. Open the divider to a span equal to the length of the longer of the two segments ST and PQ. Hence, first take a span equal to length of seg ST.
4. Without changing the span, place one tip of the divider on point M and mark O, the point on the line m where its other tip falls.
5. Now, take a distacne equal to the length of seg PQ between the tips of the divider.
6. Place one tip of the divider on point O and bring the other tip down on the line m on the same side as point M. Mark this as point N.

Now segment MN is equal in length to the difference between the lengths of seg ST and seg PQ.