Ex 4.3, 1 (iii) - Construct Parallelogram HEAR, HE = 5 cm, EA = 6 cm

Ex 4.3, 1 (iii) - Chapter 4 Class 8 Practical Geometry - Part 2
Ex 4.3, 1 (iii) - Chapter 4 Class 8 Practical Geometry - Part 3
Ex 4.3, 1 (iii) - Chapter 4 Class 8 Practical Geometry - Part 4
Ex 4.3, 1 (iii) - Chapter 4 Class 8 Practical Geometry - Part 5 Ex 4.3, 1 (iii) - Chapter 4 Class 8 Practical Geometry - Part 6 Ex 4.3, 1 (iii) - Chapter 4 Class 8 Practical Geometry - Part 7 Ex 4.3, 1 (iii) - Chapter 4 Class 8 Practical Geometry - Part 8

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Question 1 (iii) Construct the following quadrilaterals. Parallelogram HEAR HE = 5 cm EA = 6 cm ∠R = 85°cm Let’s draw a rough figure In parallelogram, Opposite angles are equal ∴ ∠ E = ∠ R = 85° Now, we need to find angles ∠ H and ∠ A Finding ∠ H and ∠ A Sum of angles of a quadrilateral = 360° ∠ H + ∠ E + ∠ A + ∠ R = 360° ∠ H + 85° + ∠ A + 85° = 360° ∠ H + ∠ A + 170° = 360° ∠ H + ∠ A = 360° – 170° ∠ H + ∠ A = 190° ∠ H + ∠ H = 190° 2∠ H = 190° ∠ H = (190°)/2 = 95° ∴ ∠ H = ∠ A = 95° Now, Let’s construct it As opposite angles of parallelogram are equal ∴ ∠ H = ∠ A Steps of construction 1. Draw side HE of length 5 cm 2. Now, we draw 95° from point H (using protractor) 3. Now, we draw 85° from point E (using protractor) Taking 6.5 cm as radius, and E as center, draw an arc. Where the arc intersects the line is point A 5. Now, we draw 95° from point A (using protractor) 6. Mark point R where the lines from point H and point A intersect Label the sides ∴ HEAR is the required parallelogram

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo