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Practical Geometry: Constructing Parallel Lines and Triangles in Class VII

This presentation by N.K. Srivastava from KV NTPC Shaktinagar covers essential concepts in practical geometry for Class VII students. It includes the step-by-step construction of a line parallel to a given line through an external point and detailed methods for constructing triangles using different criteria (SSS, SAS, ASA, RHS). Students will learn to apply geometric techniques, measure accurately, and understand the relationships between angles and sides. This knowledge not only reinforces geometry skills but also enhances spatial reasoning and problem-solving capabilities.

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Practical Geometry: Constructing Parallel Lines and Triangles in Class VII

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  1. Power Point Prepared By N K Srivastava KV NTPC Shaktinagar

  2. Class : VII Topic : Practical Geometry

  3. Construction of a line parallel to a given line , through a point not on the line. Step 1 : Take a line ‘l’ and a point ‘A’ not on ‘l’ .A l

  4. Step 2: Take any point B on ‘l’ and join B to A . A l B

  5. Step 3 : With B as centre and a convenient radius , draw an arc cutting l at C and BA at D. A D l C B

  6. Step 4 : Now with A as centre and the same radius as in Step 3,draw an arc EF cutting AB at G. Place the pointed tip of the compass at C and adjust the opening so that the pencil tip is at D F A E G D B l C

  7. Step 5 : With the same opening as in step 4 and with G as centre draw an arc cutting the arc EF at H. F A H E G D B l C

  8. Step 6 : Now , join AH to draw a line m . This line m is the required parallel line parallel to line ‘l’ and passing through A . F A m H E G D B l C

  9. Construction Of Triangles 1.Construction of a triangle when the measurements of all the three sides are given. ( SSS Criterion ) Example : Construct a triangle ABC , given that AB = 5 cm , BC = 6 cm and AC = 7 cm

  10. SolutionStep 1 : First , we draw a rough sketch with given measures. A 7 cm 5 cm B C 6 cm

  11. Step 2 : Draw a line segment BC of length 6 cm . B C 6 cm

  12. Step 3 : From point B , point A is at a distance of 5 cm . So , with B as centre , draw an arc of radius 5 cm. C B 6 cm

  13. Step 4 : From point C , point A is at a distance of 7 cm . So , with C as centre , draw an arc of radius 7 cm. C B 6 cm

  14. Step 5 : Mark the point of intersection of arcs as A . Join AB and AC .Δ ABC is the required triangle. A 7 cm 5 cm C B 6 cm

  15. 2.Construction of a triangle when the lengths of two sides and the measurement of angle between them are given. ( SAS Criterion ) Example : Construct a triangle PQR , given that PQ = 3 cm , QR = 5.5 cm and ˂ PQR = 60⁰

  16. SolutionStep 1 : First , we draw a rough sketch with given measures. P 3 cm 60⁰ Q R 5.5 cm

  17. Step 2 : Draw a line segment QR of length 5.5 cm . Q R 5.5 cm

  18. Step 3 : AT point Q , Draw QX making 60⁰ with QR. X 60⁰ R Q 5.5 cm

  19. Step 4 : With Q as centre draw an arc of radius 3 cm. IT cuts QX at point P. X P 60⁰ R Q 5.5 cm

  20. Step 5 : Join PR .ΔPQR is the required triangle. X P 3 cm 60⁰ R Q 5.5 cm

  21. 3.Construction of a triangle when the measures of two of its angles and the length of side between them is given. ( ASA Criterion ) Example : Construct a triangle PQR , given that XY = 6 cm ,m YXZ = 30⁰ and m XYZ = 100⁰

  22. SolutionStep 1 : First , we draw a rough sketch with given measures. Z 100⁰ 30⁰ X Y 6 cm

  23. Step 2 : Draw a line segment XY of length 6 cm . X Y 6 cm

  24. Step 3 : AT point X , Draw XP making 30⁰ with XY and at point Y draw YQ making an angle 100⁰ with YX. P Q 100⁰ 30⁰ Y X 6 cm

  25. Step 4 : Z has to lie on both the rays XP and YQ. So the point of Intersection of these two rays is Z. Δ XYZ is the required triangle. Q P Z 100⁰ 30⁰ Y X 6 cm

  26. 4.Construction of a Right -angled triangle when the length of its one leg and its hypotenuse are given. ( RHS Criterion ) Example : Construct a triangle LMN , right –angled at M , given that LN = 5 cm and MN = 3 cm.

  27. SolutionStep 1 : First , we draw a rough sketch with given measures. L 5 cm 90⁰ N M 3 cm

  28. Step 2 : Draw a line segment MN of length 3 cm . M N 3 cm

  29. Step 3 : AT point M , Draw MX perpendicular to MN. X 90⁰ N M 3 cm

  30. Step 4 : With N as centre , draw an arc of radius 5 cm. X L 90⁰ 3 cm M N

  31. Step 5 : Now join LN.Δ MNL is the required right – angled triangle. X L 90⁰ N M 3 cm

  32. HOME WORK Take any line m and draw a line perpendicular to passing through it and not on it. Construct a triangle ABC in which AB = 6 cm , BC = 7 cm and AC = 4 cm Construct a triangle PQR in which PQ = 8 cm Q = 60⁰ and QR = 6 cm Construct a triangle XYZ in which YZ = 7 cm , Y = 30⁰ and Z = 100⁰ .

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