AP State Syllabus AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.2 Textbook Questions and Answers.

## AP State Syllabus 6th Class Maths Solutions 10th Lesson Practical Geometry Ex 10.2

Question 1.

Draw a line segment PQ = 5.8 cm and construct its perpendicular bisector using ruler and compasses.

Solution:

Steps of construction :

- Draw a line segment \(\overline{\mathrm{PQ}}\) = 5.8 cm.
- Set the compasses as radius with more than half of the length of \(\overline{\mathrm{PQ}}\).
- With P as centre, draw arcs below and above the line segment.
- With the same radius and Q as centre draw two arcs above and below the line segment to cut the previous arcs. Name the intersecting points of arcs as X and Y.
- Join the points X and Y. So, the line I is the perpendicular bisector of PQ.

Hence l is required perpendicular bisector of PQ which meets at A.

Question 2.

Ravi made a line segment of length 8.6 cm. He constructed a bisector of AB on C. Find the length of AC and BC.

Solution:

As it is a bisector, it divides the line segment into two equal parts.

Each equal part is half of AB (8.6 cm) = \(\frac{\mathrm{AB}}{2}=\frac{8.6}{2}\) = 4.3 cm

∴ AC = BC = 4.3 cm

Question 3.

Using ruler and compasses, draw AB = 6.4 cm. Locate its mid point by geometric construction.

Solution:

- Draw a line segment \(\overline{\mathrm{PQ}}\) = 6.4 cm
- Set the compasses as radius with more than half of the length of \(\overline{\mathrm{PQ}}\).
- With A as centre, draw arcs below and above the line segment.
- With the same radius and B as centre draw two arcs above and below the line segment to cut the previous arcs. Name the intersecting points of arcs as X and Y.
- Join the points X and Y. So, the line l is the perpendicular bisector of AB.

Hence l is the required perpendicular bisector of AB which meets at M.