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Chapter 5 Class 6 Understanding Elementary Shapes
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Ex 5.5, 4 - The line l is perpendicular to line m (a) Is CE = EG?

Ex 5.5, 4 - Chapter 5 Class 6 Understanding Elementary Shapes - Part 2
Ex 5.5, 4 - Chapter 5 Class 6 Understanding Elementary Shapes - Part 3 Ex 5.5, 4 - Chapter 5 Class 6 Understanding Elementary Shapes - Part 4 Ex 5.5, 4 - Chapter 5 Class 6 Understanding Elementary Shapes - Part 5 Ex 5.5, 4 - Chapter 5 Class 6 Understanding Elementary Shapes - Part 6

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Transcript

Ex 5.5, 4 Study the diagram. The line l is perpendicular to line m (a) Is CE = EG? CE = CD + DE = 1 + 1 = 2 EG = EF + FG = 1 + 1 = 2 ∴ CE = EG Ex 5.5, 4 Study the diagram. The line l is perpendicular to line m (b) Does PE bisect CG? PE & CG Intersect at point E & CE = EG ∴ PE is the bisector of CG Ex 5.5, 4 Study the diagram. The line l is perpendicular to line m (c) Identify any two line segments for which PE is the perpendicular bisector. PE is perpendicular bisector for (𝑪𝑮) ̅ As (𝐶𝐸) ̅ = (𝐸𝐺) ̅ = 2 & (𝑃𝐸) ⃡ ⊥ (𝐶𝐺) ̅ (𝑩𝑯) ̅ As (𝐵𝐸) ̅ = (𝐸𝐴) ̅ = 3 & (𝑃𝐸) ⃡ ⊥(𝐵𝐻) ̅ Ex 5.5, 4 Study the diagram. The line l is perpendicular to line m perpendicular bisector. (d) Are these true? (i) AC > FG True AC = AB + BC = 1 + 1 =2 FG = 1 ∴ AC > FG Ex 5.5, 4 Study the diagram. The line l is perpendicular to line m perpendicular bisector. (d) Are these true? (ii) CD = GH True CD = 1 GH = 1 ∴ CD = GH Ex 5.5, 4 Study the diagram. The line l is perpendicular to line m perpendicular bisector. (d) Are these true? (iii) BC < EH True BC = 1 EH = EF + FG + GH = 1 + 1 + 1 = 3 ∴ BC < EH

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.