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

3. Class : VII Topic : Practical Geometry

4. 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

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

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

7. 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 B D C F A l E G

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

9. 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. B D C F A l E G H m

10. 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

11. Solution Step 1 : First, we draw a rough sketch with given measures. BC A 6 cm 7 cm 5 cm

12. Step 2 : Draw a line segment BC of length 6 cm. BC 6 cm

13. 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. B C 6 cm

14. 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. B C 6 cm

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

16. 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

17. Solution Step 1 : First, we draw a rough sketch with given measures. QR P 5.5 cm 3 cm 60⁰

18. Step 2 : Draw a line segment QR of length 5.5 cm. QR 5.5 cm

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

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

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

22. 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

23. Solution Step 1 : First, we draw a rough sketch with given measures. XY Z 6 cm 30⁰ 100⁰

24. Step 2 : Draw a line segment XY of length 6 cm. XY 6 cm

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

26. 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. X Y 6 cm 30⁰ 100⁰ P Q Z

27. 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.

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

29. Step 2 : Draw a line segment MN of length 3 cm. MN 3 cm

30. Step 3 : AT point M, Draw MX perpendicular to MN. M3 cm N X 90⁰

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

32. Step 5 : Now join LN.Δ MNL is the required right – angled triangle. M3 cm N X L 90⁰

33. HOME WORK 1.Take any line m and draw a line perpendicular to passing through it and not on it. 2.Construct a triangle ABC in which AB = 6 cm, BC = 7 cm and AC = 4 cm 3.Construct a triangle PQR in which PQ = 8 cm Q = 60⁰ and QR = 6 cm 4.Construct a triangle XYZ in which YZ = 7 cm, Y = 30⁰ and Z = 100⁰.