# Ex 14.5, 5 - Chapter 14 Class 6 Practical Geometry

Last updated at Sept. 28, 2018 by Teachoo

Last updated at Sept. 28, 2018 by Teachoo

Transcript

Ex 14.5, 5 With (ππ) Μ of length 6.1 cm as diameter, draw a circle. Letβs take (ππ) Μ of length 6.1 cm To draw a circle, we need to find its center. If (ππ) Μ is the diameter, its mid-point will be the center of the circle. So, making perpendicular bisector of (ππ) Μ Letβs follow these steps 1. With P as center, and radius more than half PQ, draw an arc on top and bottom of PQ 3. Now with Q as center, and same radius, draw an arc on top and bottom of PQ 4. Where the two arcs intersect above PQ is point C and where the two arcs intersect below PQ is point D Join CD Let CD intersect PQ at point O β΄ O is the mid-point of PQ Now, we need to draw a circle with PQ as diameter So, we draw a circle with O as center Letβs follow these steps 1. Measure OQ using compass. 2. Now keeping compass opened the same length. We keep pointed end at the center, and draw a circle using the pencil end of the compass. So, this is the circle with PQ as diamter

Chapter 14 Class 6 Practical Geometry

Concept wise

- Basics
- Construction of a circle when its radius is known
- Construction of a line segment of a given length
- Constructing a copy of a given line segment
- Perpendicular to a line through a point on it
- Perpendicular to a line through a point not on it
- Perpendicular bisector of a line segment
- Constructing an angle using protractor
- Bisector of an angle
- Constructing angles using compass
- Constructing a copy of a given angle

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.