Class 6 Exam  >  Class 6 Notes  >  RD Sharma Solutions for Class 6 Mathematics  >  RD Sharma Solutions -Ex-19.3, Geometrical Constructions, Class 6, Maths

Ex-19.3, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics PDF Download

Q.1 Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.

Sol.1 : Draw a line segment AB of length 8.6 cm.

With A as centre and radius more than half of AB, draw arcs on both sides of AB.

With the same radius and B as centre, draw arcs on the both sides of AB, cutting the previous two arcs at E and F.

Draw a line segment from E to F intersecting AB at C.

On measuring AC and BC, we get: AC = BC = 4.3 cm.

Ex-19.3, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

 

Q.2 Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.

Sol.2 : Draw a line segment AB of length 5.8 cm using a ruler.

With A as centre and radius more than half of AB, draw arcs on both sides of AB.

With the same radius and B as centre, draw arcs on both sides of AB, intersecting the previous arcs at L and M.

Draw the line segment LM with L and M as end-points.

LM is the required perpendicular bisector of AB.

Ex-19.3, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

 

Q.3 Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment Ab. Does it pass through the centre of the circle?

Sol.3 : Draw a point O. With O as centre and radius equal to 5 cm, draw a circle.

Take any two points A and B on the circumference of the circle and draw a line segment with A and B as its end points.

AB is the chord of the circle.

With A as centre and radius more than half of AB, draw arcs on both sides of AB.

With the same radius and B as a centre, draw arcs on both sides of AB, cutting the previous two arcs at E and F.

Draw a line passing through E and F.

Line EF passes through the centre of the circle O.

Ex-19.3, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

 

Q.4 Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisector of AB and CD. At what point do they intersect?

Sol.4 : Draw a circle with centre at 0. We draw two chords AB and CD as shown in the figure.

(i) With A as centre and radius more than half of AB, draw arcs on both sides of AB.

(ii) With the same radius and B as centre, draw arcs cutting the arcs of step (i) at P and Q.

(iii) Join P and Q.

(iv) With C as centre and radius more than half of CD, draw arcs on both sides of CD.

(v) With the same radius and D as centre, draw arcs cutting the arcs of step (iv) at R and S.

(vi) Join R and S.

We draw the line segments of perpendicular bisector of AB and CD.

We see that the perpendicular bisector of AB and CD meet at 0, the centre of the circle.

Ex-19.3, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

 

Q.5 Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.

Sol.5 : Draw a line segment AB of length 10 cm and bisect it.

(i) With A as centre and radius more than half of AB, draw arcs on both sides of AB.

(ii) With the same radius and B as centre, draw arcs cutting the arcs of step (i) at P and Q, respectively.

(iii) Join P and Q. Line PQ intersects line AB at C.

(iv) With A as centre and radius more than half of AC, draw arcs on both sides of AB.

(v) With the same radius and C as centre, draw arcs cutting the arcs of step (iv) at R and S, respectively.

(vi) Join R and S.

Line RS intersects AC at D.

If we measure AD with the ruler, we have AD = 2.5 cm

Ex-19.3, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

 

Q.6 Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length 1/2(AB).

Sol.6 : Draw a line segment AB.

(i) With A as centre and radius more than half of AB , draw arcs on both sides of AB.

(ii) With the same radius and B as centre, draw arcs cutting the arcs drawn in step (i) at P and Q.

(iii) Join P and Q. PQ intersects AB at C.

(iv) With A as centre and radius more than half of AC, draw arcs on both sides of AC.

(v) With the same radius and C as centre, draw arcs cutting the arcs drawn in step (iv) at R and S.

(vi) Join R and S. RS intersects AB at D.

Now, AC and CB are equal.

Both are 1/2(AB). Again, divide AC at D.

So, AD and AC are of same length, i.e., 1/4(AB).

Ex-19.3, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

 

Q.7 Draw a line segment AB and by ruler and compasses, obtain a line segment of length 3/4(AB).

Sol.7 : Draw a line segment AB using the ruler.

(i) With A as centre and radius more than half of AB, draw arcs on both sides of AB.

(ii) With the same radius and B as centre, draw arcs cutting the arcs drawn in step (i) at P and Q.

(iii) Join P and Q. PQ intersects AB at C.

(i) With A as centre and radius more than half of AB, draw arcs on both sides of AC.

(ii) With the same radius and C as centre, draw arcs cutting the arcs drawn in step (iv) at R and S.

(iii) Join R and S. RS intersects AB at D.

Bisect AC again and mark the point of bisection as D.

So, we have: AD = 1/4(AB),

DC = 1/4 (AB) and CB = 1/2(AB)

Therefore, DB = 1/4 (AB) + 1/2(AB) = 3/4(AB)

Thus, DB is the required line segment of length 3/4(AB).

Ex-19.3, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

The document Ex-19.3, Geometrical Constructions, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics is a part of the Class 6 Course RD Sharma Solutions for Class 6 Mathematics.
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FAQs on Ex-19.3, Geometrical Constructions, Class 6, Maths RD Sharma Solutions - RD Sharma Solutions for Class 6 Mathematics

1. How can I construct a perpendicular bisector of a line segment?
Ans. To construct a perpendicular bisector of a line segment, follow these steps: 1. Draw a line segment AB. 2. With A and B as centers, draw arcs of the same radius that intersect each other at points P and Q. 3. Draw the line passing through P and Q, which is the perpendicular bisector of the line segment AB.
2. How can I construct an angle bisector?
Ans. To construct an angle bisector, follow these steps: 1. Draw an angle with vertex O and sides OA and OB. 2. With O as the center, draw an arc that intersects the sides OA and OB at points P and Q, respectively. 3. Draw the line passing through O and the point of intersection between the arc and the angle, which is the angle bisector.
3. What is the construction for drawing a triangle when the lengths of its three sides are given?
Ans. To construct a triangle when the lengths of its three sides are given, follow these steps: 1. Draw a line segment AB of the given length. 2. With A as the center, draw an arc of the length of the second side, intersecting AB at point C. 3. With B as the center, draw an arc of the length of the third side, intersecting AB at point D. 4. Draw the lines joining points C and D to form the required triangle.
4. How can I construct a line parallel to a given line through a point not on the line?
Ans. To construct a line parallel to a given line through a point not on the line, follow these steps: 1. Draw the given line and mark a point P outside the line. 2. Place the compass on point P and draw an arc that intersects the given line at points A and B. 3. With the same radius, place the compass on point A and draw an arc above and below point P. 4. With the same radius, place the compass on point B and draw arcs intersecting the previous arcs. 5. Draw a line passing through point P and the point of intersection of the arcs, which is the required line parallel to the given line.
5. How can I construct a triangle given its base, one angle, and the difference of the other two sides?
Ans. To construct a triangle given its base, one angle, and the difference of the other two sides, follow these steps: 1. Draw a line segment AB of the given base length. 2. At point A, draw an angle of the given angle measure. 3. From point B, measure the difference of the other two sides and mark a point C on the same side of AB. 4. Draw the line segment AC and extend it to point D such that AD is equal to the difference of the other two sides. 5. Join points B and D to form the required triangle.
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