The DAV Class 6 Maths Book Solutions and **DAV Class 6 Maths Chapter 14 Worksheet 1 **Solutions of Constructions offer comprehensive answers to textbook questions.

## DAV Class 6 Maths Ch 14 WS 1 Solutions

Question 1.

Draw a line segment XY equal to 8 cm. Using compass and ruler, construct its perpendicular bisector.

Answer:

Step 1: Draw XY = 8 cm.

Step 2: Take X and Y as centres and draw two arc with a radius greater than \(\frac{1}{2}\) of XY above and below the segment XY.

Step 3: The two arcs intersect each other at C and D.

Step 4: Join CD and produce in both directions which meets XY at E. CD is the required perpendicular bisector of XY at E.

Question 2.

Draw a line segment PQ equal to 7 cm. Using ruler and compass, obtain a line segment of length i PQ. Measure each part.

Answer:

Step 1: Draw PQ = 7 cm.

Step 2: Take P and Q as centres and draw two arcs with radius more than \(\frac{1}{2}\) × PQ to intersect each other at R and S above and below the segment PQuestion

Step 3: Join R and S which intersects PQ at T.

Step 4: Measure PT and TQ, which is equal to 3.5 cm.

Hence, measure of each equal part is 3.5 = \(\frac{1}{2}\) × PQ

Question 3.

Draw a circle of radius 6 cm. Draw its diameter and name it AB. Using compasses and ruler, construct the perpendicular bisector of AB. Does it pass through the centre of the circle?

Answer:

Step 1 : Draw a circle with centre O and radius 6 cm.

Step 2: Draw any diameter AB.

Step 3: Taking A and B as centres, draw two arcs with the radius more than half of AB above and below the segment AB.

Step 4: The two arcs meet each other at C and D.

Step 5: Join CD and we observe that CD passes through the centre O of the circle.

### DAV Class 6 Maths Chapter 14 Worksheet 1 Notes

1. Construction of Angles:

For the construction of angles we generally use scale, pencil and protractor.

We use ruler, compass and pencil to construct the following angles:

15°, 30°, 45°, 105°, 120°, 135°, 22\(\frac{1}{2}\)°, 67\(\frac{1}{2}\)° etc.

2. Construction of the perpendicular bisector of a given line segment:

Step 1: Draw a line segment PQ.

Step 2: Take P as a centre and draw an arc equal to the radius more than half of PQ above and below the segment PQ.

Step 3: Draw another arc with centre Q and of same radius.

Step 4: Join the point of intersection of the two arcs which is the required perpendicular bisector of PQ.

3. Construction of an angle equal to the given angle:

Step 1: Draw an arc CD in the given angle AOB with O as centre.

Step 2: Draw a ray QR and draw an arc with centre Q and radius equal to OC which meets QR at S.

Step 3: With centre S cut an arc with a radius equal to CD which cut the first arc at T.

Step 4: Join QT and produce to P.

Step 5: ∠PQR is the required angle equal to ∠AOB.

4. Construction of the bisector of a given angle: ∠AOB is the given angle.

Step 1 : Take O as centre and draw an arc of suitable radius which meets the two arms of the angle at D and E.

Step 2: Draw two arcs with centres D and E with a radius more than half of arc DE.

Step 3: The two arcs intersect each other at F.

Step 4: Join OF and produce to C.

Step 5: OC is the required bisector of ∠AOB such that ∠AOC = ∠COB