1 / 48

( i ) Construct a rectangle ABCD in which | AB | = 8 cm and | BC | = 6 cm .

1. ( i ) Construct a rectangle ABCD in which | AB | = 8 cm and | BC | = 6 cm . Mark the point O , the point of intersection of the diagonals. 1. (ii) Draw the image of the rectangle ABCD under the enlargement of scale factor 1·75 and centre O. Steps:. 1. Measure | OA |.

robertgoodman
Download Presentation

( i ) Construct a rectangle ABCD in which | AB | = 8 cm and | BC | = 6 cm .

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 1. (i) Construct a rectangle ABCD in which |AB| = 8 cm and |BC| = 6 cm. Mark the point O, the point of intersection of the diagonals.

  2. 1. (ii) Draw the image of the rectangle ABCDunder the enlargement of scale factor 1·75 and centreO. Steps: 1. Measure |OA|. 2. Multiply |OA| by 1·75 3. Draw a line 1·75 |OA| cm long from O through A. 4. Repeat for B, C and D 5. Join points to form image.

  3. 1. (iii) Find the area of the image rectangle. Area of rectangle ABCD = length  width = 8  6 = 48 cm2 (Scale factor)2 = Scale factor = 1·75 3·0625  48 = Multiply both sides by 48 147 cm2 = Image area

  4. 1. (iv) Under another enlargement the area of the image of the rectangle ABCD is 27 cm2. What is the scale factor of this enlargement? Give your answer in the form , where a, b ℕ. Object area = 48 cm2 Image area = 27 cm2 (Scale factor)2 = (Scale factor)2 = (Scale factor)2 = Scale factor = Square root both sides Scale factor =

  5. 2. (i) Construct an equilateral triangle, PQR, of side 8 cm. Steps: 1. Draw [PQ] 8 cm long. 2. With compass point on P and compass width 8 cm, draw an arc. 3. With compass point on Q and compass width 8 cm, draw an arc. 4. Mark the intersection of the arcs R. 5. Join P to R and Q to R.

  6. 2. (ii) Construct the image of the triangle PQR under enlargement with centreP and scale factor 2. Steps: 1. Measure |RP|. 2. Multiply |RP| by 2. 3. Draw a line 2 |RP| in length from P through R. 4. Repeat for Q 5. Join points to form a image. Note: P = P′

  7. 2. (iii) Given that the height of PQR is cm, find the area of the image triangle, correct to the nearest cm2. Area ∆ = base  perpendicular height base = 8 cm height = Area ∆ = object area = = 27·71 cm2 (Scale factor)2 = Scale factor = 2

  8. 2. (iii) Given that the height of PQR is cm, find the area of the image triangle, correct to the nearest cm2. (2)2 = 4  27·71 = multiply both sides by 27·71 110·84 = Image area Image ∆ = 111 cm2 rounded to nearest cm2

  9. 2. (iv) Another triangle, XYZ, is the image of triangle PQR under a different enlargement with centreP. The area of XYZ is cm2. Calculate the scale factor of this enlargement. Object area = from part (iii) Image area = (Scale factor)2 = (Scale factor)2 = Scale factor = Square root both sides Scale factor = 0·43

  10. 3. The triangle CDE is the image of the triangle CAB under an enlargement with centreC. |CA| = 12, |AD| = 9 and |CB|= 8. Object = ∆CAB Image = ∆CDE Find the scale factor of the enlargement. (i) Scale factor = Scale factor = Scale factor = Scale factor = = 1·75

  11. 3. The triangle CDE is the image of the triangle CAB under an Enlargement with centreC. |CA| = 12, |AD| = 9 and |CB|= 8. Object = ∆CAB Image = ∆CDE Find |BE|. (ii) Scale factor = Scale factor = Scale factor =

  12. 3. The triangle CDE is the image of the triangle CAB under an Enlargement with centreC. |CA| = 12, |AD| = 9 and |CB|= 8. (ii) Find |BE|. Multiply both sides by 8

  13. 3. The triangle CDE is the image of the triangle CAB under an Enlargement with centreC. |CA| = 12, |AD| = 9 and |CB|= 8. (ii) Find |BE|. 8 + = 14 8 + − 8 = 14 − 8 Subtract 8 from both sides = 6

  14. 3. The triangle CDE is the image of the triangle CAB under an Enlargement with centreC. |CA| = 12, |AD| = 9 and |CB|= 8. The area of the triangle CDE is 98 square units. Find the area of the triangle CAB. (iii) Area ∆CDE = 98 square units ∆ ∆ ∆ Multiply both sides by area ∆CAB ∆ 3·0625  area ∆CAB = ∆

  15. 3. The triangle CDE is the image of the triangle CAB under an Enlargement with centreC. |CA| = 12, |AD| = 9 and |CB|= 8. The area of the triangle CDE is 98 square units. Find the area of the triangle CAB. (iii) 3·0625  area ∆CAB = 98 ∆ Divide both sides by 3·0625 Area ∆CAB = 32 square units

  16. 4. The triangle AʹBʹCʹ is the image of the triangle ABC under an enlargement. Find, by measurement, the scale factor of the enlargement. (i) |AB| = 1·3 cm (object length) |A′B′| = 2·6 cm (image length) Scale factor = 2

  17. 4. The triangle AʹBʹCʹ is the image of the triangle ABC under an enlargement. Find, by measurement, the scale factor of the enlargement. (i) Similarly, |BC| = 2 cm |AC| = 2·3 cm |B′C′| = 4 cm |A′C′| = 4·6 cm

  18. 4. The triangle AʹBʹCʹ is the image of the triangle ABC under an enlargement. Copy the diagram and show how to find the centre of the enlargement. (ii) To find the centre of enlargement: Join A′ to A and extend Join B′ to B and extend Join C′ to C and extend The point where these 3 lines intersect is the centre of enlargement.

  19. 4. The triangle AʹBʹCʹ is the image of the triangle ABC under an enlargement. Units are chosen so that |BC| = 10 units. How many of these units is BʹCʹ? (iii) |BC| = 10 units |B′C′| = 10 × 2 = 20 units

  20. 4. The triangle AʹBʹCʹ is the image of the triangle ABC under an enlargement. Find the area of triangle ABC, given that the area of AʹBʹCʹ is 104 square units. (iv) image area = k2 (object area) Image area = 104 square units 26 square units = object area. k = 2

  21. 5. (i) Construct a triangle ABC in which |AB| = 10 cm, |BC| = 8 cm and |AC| = 6 cm. Steps: 1. Draw [AB] = 10 cm. 2. With compass point on A and width = 8 cm draw an arc. 3. With compass point on B and width = 6 cm draw an arc. 4. Mark the intersection of the arcs, C. 5. Join A to C and B to C.

  22. 5. (ii) Choose any point P that is outside the triangle and construct the image of ABC under the enlargement of scale factor 0·6 and centreP. Answers will vary depending on where you put your centre of enlargement. Steps: 1. Pick a point P outside the triangle. 2. Measure |AP| 3. Multiply |AP| by 0·6 4. Draw a line equal in length to 0·6 |AP| from P through A. 5. Repeat for B and C. 6. Join points to form the image.

  23. 5. (iii) Given that the area of this original triangle is 24 cm2, calculate the area of the image triangle.

  24. 6. (i) Draw a square, OPQR, with sides 12 cm.

  25. 6. (ii) Draw the image of this square under the enlargement with centreO and scale factor 0·5. Steps: 1. Measure |OP|. 2. Multiply |OP| by 0·5. 3. Draw a line 0·5|OP| from O through P. 4. Repeat for other points. Join to form image.

  26. 6. (iii) Calculate the area of this image square. Area OPQR = length  width = 12  12 = 144

  27. 6. (iv) Under another enlargement the area of the image of the square OPQR is 120 cm2. What is the scale factor of this enlargement?

  28. 7. The diagram shows two identical squares ABCD and DCEF. Their diagonals intersect at X and Y. Under the translation what is the image of: (i) (a) D F (b) [DC] [FE] (c) ΔBXC ΔCYE

  29. 7. The diagram shows two identical squares ABCD and DCEF. Their diagonals intersect at X and Y. Under the translation what is the image of: (i) (d) [AX] [DY] (e) [XD] [YF] (f) XBA YCD

  30. 7. The diagram shows two identical squares ABCD and DCEF. Their diagonals intersect at X and Y. Name another square. (ii) XDYC

  31. 7. The diagram shows two identical squares ABCD and DCEF. Their diagonals intersect at X and Y. If the area of ΔABX is 6 cm2, find the area of (iii) (a) ΔABD Area ΔABX = 6 cm2 Area ΔABX = Area ΔADX Congruent triangles Area ΔABD = Area ΔABX + Area ΔADX Area ΔABD = 6 + 6 Area ΔABD = 12 cm2

  32. 7. The diagram shows two identical squares ABCD and DCEF. Their diagonals intersect at X and Y. If the area of ΔABX is 6 cm2, find the area of (iii) (iii) (b) ΔACF Area ΔACF = Area ΔADX + Area ΔDCX + Area ΔDCY + Area ΔDYF Area ΔADX = Area ΔDCX = Area ΔDCY = Area ΔDYF Congruent triangles Area ΔADX = 6 cm2 Area ΔACF = 6 + 6 + 6 + 6 Area ΔACF = 24 cm2

  33. 7. The diagram shows two identical squares ABCD and DCEF. Their diagonals intersect at X and Y. If the area of ΔABX is 6 cm2, find the area of (iii) (iii) (c) the rectangle ABEF Area ABEF = Area ΔABX + Area ΔADX + Area ΔDCX + Area ΔBCX + Area ΔDCY + Area ΔDYF + Area ΔFEY + Area ΔECY All Δ’s are congruent ABEF = 8  ΔABX ABEF = 8  6 ABEF = 48 cm2

  34. 8. The diagram shows four equilateral triangles. Under the translation what is the image of: (i) (c) ΔRUV (a) R STU S (d) VQR (b) [QV] URS [RU]

  35. 8. The diagram shows four equilateral triangles. Name a translationthat maps ΔQRV onto ΔRSU. (ii)

  36. 8. The diagram shows four equilateral triangles. If the area of the parallelogram QSTV is 20 cm2, what is the area of RSUV? (iii) Area QSTV = 20 cm2 Area QSTV = Area QVR + Area VRU + Area RUS + Area UST Area QVR = Area VRU = Area RUS = Area UST 4 equilateral triangles Area QSTV = 4  Area VRU 20 = 4  Area VRU

  37. 8. The diagram shows four equilateral triangles. If the area of the parallelogram QSTV is 20 cm2, what is the area of RSUV? (iii) 5 = Area VRU Area RSUV = Area VRU + Area RUS Area VRU = Area RUS equilateral triangles Area RSUV = 2 VRU

  38. 8. The diagram shows four equilateral triangles. If the area of the parallelogram QSTV is 20 cm2, what is the area of RSUV? (iii) Area RSUV = 2 VRU Area RSUV = 2  5 Area RSUV = 10 cm2

  39. 8. The diagram shows four equilateral triangles. What is the measure of RUT? (iv) Equilateral triangle has 3 equal angles Sum of angles in a triangle = 180° |RUT| = |RUS| + |SUT| |RUT| = 60° + 60° |RUT| = 120°

  40. 9. A triangle, A, is drawn on graph paper as shown. Calculate the area of triangle A. (i)

  41. 9. A triangle, A, is drawn on graph paper as shown. Copy the triangle A and enlarge it by a scale factor of 3 with centre of enlargement (0, 0). Call this triangle B. (ii) Steps: 1. Measure O to one of the vertices of A 2. Multiply this by 3 3. Draw a line equal in length to the answer above from O through the choosen vertex. 4. Repeat for each vertex. 5. Join points and label triangle B.

  42. 9. A triangle, A, is drawn on graph paper as shown. Find the area of triangle B. (iii) Multiply both sides by 15

  43. 9. A triangle, A, is drawn on graph paper as shown. Find the ratio of the area of triangle B to triangle A. (iv) ΔB : Δ A 13·5 : 1·5 How many times does 1·5 go into 13·5? 9 : 1

  44. 9. A triangle, A, is drawn on graph paper as shown. Comment on the significance of your answer to part (iv). (v) Multiply both sides by Δ4

  45. 9. A triangle, A, is drawn on graph paper as shown. Comment on the significance of your answer to part (iv). (v) With a scale factor (k) of 3 we find that the area of triangle A must be multiplied by 9 (k2) to get the area of the image, triangle B. The ratio of the areas of the triangles is also 9:1, where triangle A has 1 unit of area and triangle B has 9 units of area. ΔB = k2ΔA

  46. 10. The 2 litre and 500 ml bottles of cola shown in the diagram are similar. The height of the label on bottle A is 8 cm. The curved surface area of the label on bottle A is 60 cm2. The curved surface area of the label on bottle B is 375 cm2. Find the height of the label on bottle B.

  47. 10. The 2 litre and 500 ml bottles of cola shown in the diagram are similar. The height of the label on bottle A is 8 cm. The curved surface area of the label on bottle A is 60 cm2. The curved surface area of the label on bottle B is 375 cm2. Find the height of the label on bottle B.

More Related