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.