RD Sharma Solutions: Geometrical Constructions (Exercise 19.2)

# Geometrical Constructions (Exercise 19.2) RD Sharma Solutions | Mathematics (Maths) Class 6 PDF Download

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``` Page 1

1. How many lines can be drawn which are perpendicular to a given line and pass through a given point
lying
(i) outside it?
(ii) on it?
Solution:

(i) We know that a line perpendicular from a given point to a given line is the shortest distance between them.
Here only one shortest distance is possible.
Therefore, only one perpendicular line is possible from a given point lying outside it.

(ii) We can draw only one perpendicular line from any point on the line.
Therefore, only one perpendicular line can be drawn from a given point lying on it.

2. Draw a line PQ. Take a point R on it. Draw a line perpendicular to PQ and passing through R.
(Using (i) ruler and a set-square (ii) ruler and compasses)
Solution:

(i) Construct a line PQ and mark a point R on it.
Now place the set square with its one arm of the right angle along the line PQ.
Place the ruler along its edge without disturbing the position of set square.
Remove the set square without disturbing the position of the ruler and construct a line MN through R.
Here, MN is the required line which is perpendicular to PQ through the point R.
Page 2

1. How many lines can be drawn which are perpendicular to a given line and pass through a given point
lying
(i) outside it?
(ii) on it?
Solution:

(i) We know that a line perpendicular from a given point to a given line is the shortest distance between them.
Here only one shortest distance is possible.
Therefore, only one perpendicular line is possible from a given point lying outside it.

(ii) We can draw only one perpendicular line from any point on the line.
Therefore, only one perpendicular line can be drawn from a given point lying on it.

2. Draw a line PQ. Take a point R on it. Draw a line perpendicular to PQ and passing through R.
(Using (i) ruler and a set-square (ii) ruler and compasses)
Solution:

(i) Construct a line PQ and mark a point R on it.
Now place the set square with its one arm of the right angle along the line PQ.
Place the ruler along its edge without disturbing the position of set square.
Remove the set square without disturbing the position of the ruler and construct a line MN through R.
Here, MN is the required line which is perpendicular to PQ through the point R.

(ii) Construct a line PQ and mark R on it.
Considering R as the centre and measuring convenient radius, draw an arc which touches the line PQ at A and B.
Considering A and B as centres and radius which is greater than AR, draw two arcs which cuts each other at the
point S.
Now join the points RS and extend in both directions.
Here, RS is the required line which is perpendicular to PQ through the point R.
3. Draw a line l. Take a point A, not lying on l. Draw a line m such that m ? l and passing through A.
(Using (i) ruler and a set-square (ii) ruler and compasses)
Solution:
(i) Construct a line L and mark a point A outside it.
Now place the set square PQR with its one arm PQ of the right angle along the line L.
Place the ruler along the edge PR without disturbing the position of set square.
Slide the set square along the ruler until its arm QR reaches A without disturbing the position of the ruler.
Draw a line m.
Page 3

1. How many lines can be drawn which are perpendicular to a given line and pass through a given point
lying
(i) outside it?
(ii) on it?
Solution:

(i) We know that a line perpendicular from a given point to a given line is the shortest distance between them.
Here only one shortest distance is possible.
Therefore, only one perpendicular line is possible from a given point lying outside it.

(ii) We can draw only one perpendicular line from any point on the line.
Therefore, only one perpendicular line can be drawn from a given point lying on it.

2. Draw a line PQ. Take a point R on it. Draw a line perpendicular to PQ and passing through R.
(Using (i) ruler and a set-square (ii) ruler and compasses)
Solution:

(i) Construct a line PQ and mark a point R on it.
Now place the set square with its one arm of the right angle along the line PQ.
Place the ruler along its edge without disturbing the position of set square.
Remove the set square without disturbing the position of the ruler and construct a line MN through R.
Here, MN is the required line which is perpendicular to PQ through the point R.

(ii) Construct a line PQ and mark R on it.
Considering R as the centre and measuring convenient radius, draw an arc which touches the line PQ at A and B.
Considering A and B as centres and radius which is greater than AR, draw two arcs which cuts each other at the
point S.
Now join the points RS and extend in both directions.
Here, RS is the required line which is perpendicular to PQ through the point R.
3. Draw a line l. Take a point A, not lying on l. Draw a line m such that m ? l and passing through A.
(Using (i) ruler and a set-square (ii) ruler and compasses)
Solution:
(i) Construct a line L and mark a point A outside it.
Now place the set square PQR with its one arm PQ of the right angle along the line L.
Place the ruler along the edge PR without disturbing the position of set square.
Slide the set square along the ruler until its arm QR reaches A without disturbing the position of the ruler.
Draw a line m.

Here, line m is the required line which is perpendicular to the line L.
(ii) Considering A as centre, construct an arc PQ which intersects the line L at P and Q.
By taking P and Q as centres, draw two arcs which intersects each other at the point B.
Now join the points A and B and extend in both the directions.
Here, AB is the required line.
4. Draw a line AB and take two points C and E on opposite sides of AB. Through C, draw CD ? AB and
through E draw EF ? AB.
(Using (i) ruler and a set-square (ii) ruler and compasses)
Solution:
Q
Page 4

1. How many lines can be drawn which are perpendicular to a given line and pass through a given point
lying
(i) outside it?
(ii) on it?
Solution:

(i) We know that a line perpendicular from a given point to a given line is the shortest distance between them.
Here only one shortest distance is possible.
Therefore, only one perpendicular line is possible from a given point lying outside it.

(ii) We can draw only one perpendicular line from any point on the line.
Therefore, only one perpendicular line can be drawn from a given point lying on it.

2. Draw a line PQ. Take a point R on it. Draw a line perpendicular to PQ and passing through R.
(Using (i) ruler and a set-square (ii) ruler and compasses)
Solution:

(i) Construct a line PQ and mark a point R on it.
Now place the set square with its one arm of the right angle along the line PQ.
Place the ruler along its edge without disturbing the position of set square.
Remove the set square without disturbing the position of the ruler and construct a line MN through R.
Here, MN is the required line which is perpendicular to PQ through the point R.

(ii) Construct a line PQ and mark R on it.
Considering R as the centre and measuring convenient radius, draw an arc which touches the line PQ at A and B.
Considering A and B as centres and radius which is greater than AR, draw two arcs which cuts each other at the
point S.
Now join the points RS and extend in both directions.
Here, RS is the required line which is perpendicular to PQ through the point R.
3. Draw a line l. Take a point A, not lying on l. Draw a line m such that m ? l and passing through A.
(Using (i) ruler and a set-square (ii) ruler and compasses)
Solution:
(i) Construct a line L and mark a point A outside it.
Now place the set square PQR with its one arm PQ of the right angle along the line L.
Place the ruler along the edge PR without disturbing the position of set square.
Slide the set square along the ruler until its arm QR reaches A without disturbing the position of the ruler.
Draw a line m.

Here, line m is the required line which is perpendicular to the line L.
(ii) Considering A as centre, construct an arc PQ which intersects the line L at P and Q.
By taking P and Q as centres, draw two arcs which intersects each other at the point B.
Now join the points A and B and extend in both the directions.
Here, AB is the required line.
4. Draw a line AB and take two points C and E on opposite sides of AB. Through C, draw CD ? AB and
through E draw EF ? AB.
(Using (i) ruler and a set-square (ii) ruler and compasses)
Solution:
Q

(i) Construct a line AB and mark two points C and E on the opposite sides of AB.
Place the set square PQR on the side E with its one arm PQ of the right angle along the line AB. Place the ruler
along the edge PR.
Slide the set square along the ruler until the arm QR reaches C.
Construct a line CD where D is a point on the line AB.
Here CD is the required line where CD ? AB
Now follow the same steps using a set square on the side E, we construct a line EF ? AB.

Page 5

1. How many lines can be drawn which are perpendicular to a given line and pass through a given point
lying
(i) outside it?
(ii) on it?
Solution:

(i) We know that a line perpendicular from a given point to a given line is the shortest distance between them.
Here only one shortest distance is possible.
Therefore, only one perpendicular line is possible from a given point lying outside it.

(ii) We can draw only one perpendicular line from any point on the line.
Therefore, only one perpendicular line can be drawn from a given point lying on it.

2. Draw a line PQ. Take a point R on it. Draw a line perpendicular to PQ and passing through R.
(Using (i) ruler and a set-square (ii) ruler and compasses)
Solution:

(i) Construct a line PQ and mark a point R on it.
Now place the set square with its one arm of the right angle along the line PQ.
Place the ruler along its edge without disturbing the position of set square.
Remove the set square without disturbing the position of the ruler and construct a line MN through R.
Here, MN is the required line which is perpendicular to PQ through the point R.

(ii) Construct a line PQ and mark R on it.
Considering R as the centre and measuring convenient radius, draw an arc which touches the line PQ at A and B.
Considering A and B as centres and radius which is greater than AR, draw two arcs which cuts each other at the
point S.
Now join the points RS and extend in both directions.
Here, RS is the required line which is perpendicular to PQ through the point R.
3. Draw a line l. Take a point A, not lying on l. Draw a line m such that m ? l and passing through A.
(Using (i) ruler and a set-square (ii) ruler and compasses)
Solution:
(i) Construct a line L and mark a point A outside it.
Now place the set square PQR with its one arm PQ of the right angle along the line L.
Place the ruler along the edge PR without disturbing the position of set square.
Slide the set square along the ruler until its arm QR reaches A without disturbing the position of the ruler.
Draw a line m.

Here, line m is the required line which is perpendicular to the line L.
(ii) Considering A as centre, construct an arc PQ which intersects the line L at P and Q.
By taking P and Q as centres, draw two arcs which intersects each other at the point B.
Now join the points A and B and extend in both the directions.
Here, AB is the required line.
4. Draw a line AB and take two points C and E on opposite sides of AB. Through C, draw CD ? AB and
through E draw EF ? AB.
(Using (i) ruler and a set-square (ii) ruler and compasses)
Solution:
Q

(i) Construct a line AB and mark two points C and E on the opposite sides of AB.
Place the set square PQR on the side E with its one arm PQ of the right angle along the line AB. Place the ruler
along the edge PR.
Slide the set square along the ruler until the arm QR reaches C.
Construct a line CD where D is a point on the line AB.
Here CD is the required line where CD ? AB
Now follow the same steps using a set square on the side E, we construct a line EF ? AB.

(ii) Construct a line AB and mark two points C and E on the opposite sides.
Considering C as centre, construct an arc PQ which intersects AB at P and Q.
Considering P and Q as centres, draw two arcs which intersect each other at the point H.
Now join the points H and C where HC crosses the line AB at the point D.
We know that CD ? AB
In the same way, consider E as centre and construct an arc RS.
Considering R and S as centres, construct two arcs which intersect each other at the point G.
Now join the point G and E such that GE crosses the line AB at the point F
We know that EF ? AB.

5. Draw a line segment AB of length 10 cm. Mark a point P on AB such that AP = 4 cm. Draw a line
through P perpendicular to AB.
Solution:

Construct a line L and mark a point A on it.
With the help of a ruler and a compass mark a point B which is at 10 cm from the point A on the line L.
Here, AB is the required line segment of length 10 cm.
Now, mark a point P which is 4 cm from the point A in the direction of B.
Taking P as centre and radius 4 cm draw an arc which intersects the line L at A and E.
Taking A and E as centres and radius 6 cm draw two arcs which intersects each other at the point R.
Now join P and R and extend it.
Here, PR is the required line which is perpendicular to the line AB.
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## Mathematics (Maths) Class 6

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## Mathematics (Maths) Class 6

94 videos|347 docs|54 tests

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