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Tree height and triangle measurements, Circle segment and parallelogram angles and dimensions

Calculate the height of a tree using angle measurements, find missing angles in triangles and parallelograms, determine area and perimeter of a shape with a segment removed, and calculate dimensions of a field.

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Tree height and triangle measurements, Circle segment and parallelogram angles and dimensions

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  1. 1. Nia wants to measure the height of a tree. She stands at the point A and measures the angle of elevation to the point C, at the top of the tree, to be 65°.She then walks 3 m away from the tree until she gets to the point B. She measures the angle of elevation from B to C to be 42°. Find: (i) |AC|. Give your answer to the nearest centimetre. Find the missing angles in triangle ABC: |CAB| = 180° − 65° = 115° |ACB| = 180° – (42° + 115°) = 180° – 157° = 23°

  2. 1. Nia wants to measure the height of a tree. She stands at the point A and measures the angle of elevation to the point C, at the top of the tree, to be 65°.She then walks 3 m away from the tree until she gets to the point B. She measures the angle of elevation from B to C to be 42°. Find: (i) |AC|. Give your answer to the nearest centimetre.

  3. 1. Nia wants to measure the height of a tree. She stands at the point A and measures the angle of elevation to the point C, at the top of the tree, to be 65°.She then walks 3 m away from the tree until she gets to the point B. She measures the angle of elevation from B to C to be 42°. Find: (ii) the height of the tree, |CD|. Give your answer to the nearest centimetre.

  4. 2. The triangle XYZ has dimensions |XY| = 54 cm, |XZ| = 62 cm, |∠XZY| = 40°. (i) Find two possible values for |∠XYZ|, correct to the nearest degree. |XY| = z = 54, |XZ| = y = 62, |XZY| = Z = 40°

  5. 2. The triangle XYZ has dimensions |XY| = 54 cm, |XZ| = 62 cm, |∠XZY| = 40°. (i) Find two possible values for |∠XYZ|, correct to the nearest degree. 1st Quadrant = 48° 2nd Quadrant = 180° − 48° = 132°

  6. 2. The triangle XYZ has dimensions |XY| = 54 cm, |XZ| = 62 cm, |∠XZY| = 40°. (ii) Draw a sketch of the two possible triangles XYZ. For |∠XYZ| = 132° For |∠XYZ| = 48°

  7. 3. The diagram shows a circle, of radius 40 cm, with a segment removed. Find: (i) the area of this shape, correct to one decimal place. Area of shape = Area of sector + Area of triangle = 3351·032 + 692·82 = 4043·852 = 4043·9 cm2

  8. 3. The diagram shows a circle, of radius 40 cm, with a segment removed. Find: (ii) the perimeter of this shape, correct to one decimal place. A |AB|2 = 402 + 40 − 2(40)(40) cos 120° |AB|2 = 1600 + 1600 − 3200(−0·5) |AB|2 = 3200 + 1600 B |AB|2 = 4800 |AB| = 69·282

  9. 3. The diagram shows a circle, of radius 40 cm, with a segment removed. Find: (ii) the perimeter of this shape, correct to one decimal place. A Perimeter of shape = Length of arc + |AB| 69·282 B = 167·5516 + 69·282 = 236·8336 = 236·8 cm

  10. 4. Two adjacent sides of a parallelogram are of length 7·5 cm and 9·3 cm, and the longer diagonal is oflength 12 cm. (i) Draw a diagram of the parallelogram, showing all known dimensions.

  11. 4. Two adjacent sides of a parallelogram are of length 7·5 cm and 9·3 cm, and the longer diagonal is oflength 12 cm. (ii) Find the angles of the parallelogram. Give your answer to the nearest degree. a2 = b2 + c2 – 2bccosA 122 = (9·3)2 + (7·5)2 – 2(9·3)(7·5) cos |DCB| 144 = 86·49 + 56·25 − 139·5 cos |DCB| 144 = 142·74 − 139·5 cos |DCB| 139·5 cos |DCB| = 142·74 – 144 cos|DCB| = −0·00903 |DCB| = cos–1 (−0·00903) = 90·517

  12. 4. Two adjacent sides of a parallelogram are of length 7·5 cm and 9·3 cm, and the longer diagonal is oflength 12 cm. (ii) Find the angles of the parallelogram. Give your answer to the nearest degree. 4 angles in parallelogram: |DCB| = 91º (to nearest degree) |DAB| = 91º |ADC| and |ABC| = 89º

  13. 4. Two adjacent sides of a parallelogram are of length 7·5 cm and 9·3 cm, and the longer diagonal is oflength 12 cm. (iii) Find the length of the other diagonal of the parallelogram, correct to one decimal place. a2 = b2 + c2 – 2bccosA |AC|2 = (7·5)2 + (9·3)2 – 2(7·5)(9·3) cos 89º AC|2 = 56·25 + 86·49 – 139·5(0·0175) |AC|2 = 142·74 – 2·44125 AC|2 = 140·29875 AC| = 11·844 Other diagonal = 11·8 cm

  14. 5. The diagram shows a sketch of a field ABCD. A farmer needs to buy fertiliser for this field and to put a new fence around the perimeter of the field. A bag of fertiliser costs €25 and can cover 18 m2.Fencing comes in 2 m lengths, each costing €45. Find: (i) |BD|, correct to one decimal place. a2 = b2 + c2 – 2bccosA |BD|2 = 302 + 502 – 2(30)(50) cos 80º |BD|2 = 900 + 2500 – 3000(0·1736) |BD|2 = 3400 – 520·8 |BD|2 = 2879·2 |BD| = 2879·2 |BD| = 53·658 |BD| = 53·7 m

  15. 5. The diagram shows a sketch of a field ABCD. A farmer needs to buy fertiliser for this field and to put a new fence around the perimeter of the field. A bag of fertiliser costs €25 and can cover 18 m2.Fencing comes in 2 m lengths, each costing €45. Find: (ii) |AD|, correct to one decimal place |BAD| = 180º − (25º + 37º) = 180º − 62º = 118º

  16. 5. The diagram shows a sketch of a field ABCD. A farmer needs to buy fertiliser for this field and to put a new fence around the perimeter of the field. A bag of fertiliser costs €25 and can cover 18 m2.Fencing comes in 2 m lengths, each costing €45. Find: (ii) |AD|, correct to one decimal place 118º |AD| = 25·7 m

  17. 5. The diagram shows a sketch of a field ABCD. A farmer needs to buy fertiliser for this field and to put a new fence around the perimeter of the field. A bag of fertiliser costs €25 and can cover 18 m2.Fencing comes in 2 m lengths, each costing €45. Find: (iii) the perimeter of the field, to the nearest whole number. Find |AB|: 118º = 36·6 m

  18. 5. The diagram shows a sketch of a field ABCD. A farmer needs to buy fertiliser for this field and to put a new fence around the perimeter of the field. A bag of fertiliser costs €25 and can cover 18 m2.Fencing comes in 2 m lengths, each costing €45. Find: (iii) the perimeter of the field, to the nearest whole number. Perimeter = |AB| + |BC| + |CD| + |DA| 25·7 m = 36·6 + 50 + 30 + 25·7 36·6 m = 142·3 m = 142 m

  19. 5. The diagram shows a sketch of a field ABCD. A farmer needs to buy fertiliser for this field and to put a new fence around the perimeter of the field. A bag of fertiliser costs €25 and can cover 18 m2.Fencing comes in 2 m lengths, each costing €45. Find: (iv) the area of the field, to the nearest whole number. Area of field = Area of 2 triangles = Area of ΔABD + Area of ΔBCD 25·7 m 36·6 m

  20. 5. The diagram shows a sketch of a field ABCD. A farmer needs to buy fertiliser for this field and to put a new fence around the perimeter of the field. A bag of fertiliser costs €25 and can cover 18 m2.Fencing comes in 2 m lengths, each costing €45. Find: (v) the total amount the farmer will pay for the fertiliser and the fencing. cost = 71  €45 = €3195 = 65 Bags cost = 65  €25 = €1625 Total cost = €1625 + €3195 = €4820

  21. 6. A circle, of radius 8 cm, is divided into six equal sectors, as shown. (i) Find |∠AOB| = 60°

  22. 6. A circle, of radius 8 cm, is divided into six equal sectors, as shown. (ii) Find |FD|, correct to one decimal place. a2 = b2 + c2 – 2bccosA |FD|2 = 82 + 82 − 2(8)(8) cos 120° |FD|2 = 64 + 64 – 128(−0·5) |FD|2 = 128 + 64 |FD|2 = 192 |FD| = 13·856 |FD| = 13·9 cm

  23. 6. A circle, of radius 8 cm, is divided into six equal sectors, as shown. (iii) Find |DC| ΔODC is an equilateral triangle  All sides are equal length  |DC| = 8 cm

  24. 6. A circle, of radius 8 cm, is divided into six equal sectors, as shown. (iv) Given |∠FDC| = |∠FAC| = 90°, find the area of the rectangle FDCA. Area of rectangle = LW = 13·9  8 = 111·2 cm2

  25. 6. A circle, of radius 8 cm, is divided into six equal sectors, as shown. (v) What percentage of the circle is occupied by the rectangle FDCA? Give your answer to one decimal place. Area of circle = πr2 = π(8)2 = 64π cm2 = 55·3%

  26. 7. Three ships are situated in a straight line at points A, B and C. P is a port such that: |∠BAP| = 55°, |∠ABP| = 110° |AB| = 10 km and |BC| = 22 km. Find: (i) |BP|, correct to the nearest metre. First find |APB| = 180º − (110º + 55º) = 180º − 165º |APB| = 15º

  27. 7. Three ships are situated in a straight line at points A, B and C. P is a port such that: |∠BAP| = 55°, |∠ABP| = 110° |AB| = 10 km and |BC| = 22 km. Find: (i) |BP|, correct to the nearest metre. = 31·6496 km |BP| = 31,650 m

  28. 7. Three ships are situated in a straight line at points A, B and C. P is a port such that: |∠BAP| = 55°, |∠ABP| = 110° |AB| = 10 km and |BC| = 22 km. Find: (ii) |CP|, correct to the nearest metre. |PBC| = 180º − 110º = 70º a2 = b2 + c2 − 2bccosA |CP|2 = 222 + (31·65)2 − 2(22)(31·65) cos 70º |CP|2 = 484 + 1001·7225 − 476·3 |CP|2 = 1009·4223 |CP| = 31·771 km = 31,771 m

  29. 7. Three ships are situated in a straight line at points A, B and C. P is a port such that: |∠BAP| = 55°, |∠ABP| = 110° |AB| = 10 km and |BC| = 22 km. Find: (iii) the area of the triangle APC, correct to one decimal place. Find |PCB| 31·771 sin |PCB| = 0·9361 |PCB| = sin−1 (0·9361) |PCB| = 69·4º

  30. 7. Three ships are situated in a straight line at points A, B and C. P is a port such that: |∠BAP| = 55°, |∠ABP| = 110° |AB| = 10 km and |BC| = 22 km. Find: (iii) the area of the triangle APC, correct to one decimal place. = (508·336) (0·9361) = 475·853 km2 Area of ∆APC = 475·9 km2

  31. 8. In the quadrilateral ABCD, |AC |= 5 cm, |BC| = 4 cm, |∠ACD| = 35° and |∠CDA| = 22°. Find: (i) |AB|, correct to two decimal places. a2 = b2 + c2 − 2bccosA |AB|2 = 42 + 52 − 2(4)(5) cos 110º |AB|2 = 16 + 25 − 40 (−0·342) |AB|2 = 41 + 13·68 |AB|2 = 54·68 |AB| = 7·394 |AB| = 7·39 cm

  32. 8. In the quadrilateral ABCD, |AC |= 5 cm, |BC| = 4 cm, |∠ACD| = 35° and |∠CDA| = 22°. Find: (ii) |CD|, correct to one decimal place. |CAD| = 180° − (35° + 22°) = 123° |CD| = 11·2 cm

  33. 8. In the quadrilateral ABCD, |AC |= 5 cm, |BC| = 4 cm, |∠ACD| = 35° and |∠CDA| = 22°. Find: (iii) the area of the quadrilateral ABCD, correct to one decimal place Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD = 9·39692 + 16·06 = 25·457 = 25·5 cm2

  34. 9. The diagram shows a sector of a circle with centreO. The radius of the circle is 10.4 cm. AC is a chord of the circle. Calculate: (i) the area of the shaded region ABC, correct to one decimal place. Area of shaded region = Area of sector − Area of triangle = 36·053π − 46·8346 = 66·429 = 66·4 cm2

  35. 9. The diagram shows a sector of a circle with centreO. The radius of the circle is 10.4 cm. AC is a chord of the circle. Calculate: (ii) the perimeter of the segment ABC, correct to one decimal place. Find |AC|: a2 = b2 + c2 − 2bccosA |AC|2 = (10·4)2 + (10·4)2 − 2(10·4)(10·4) cos 120° |AC|2 = 108·16 + 108·16 − 216·32 (−0·5) |AC|2 = 216·32 + 108·16 |AC|2 = 324·48

  36. 9. The diagram shows a sector of a circle with centreO. The radius of the circle is 10.4 cm. AC is a chord of the circle. Calculate: (ii) the perimeter of the segment ABC, correct to one decimal place. Perimeter of segment ABC = Length of arc + |AC| = 6·93 (π) + 18 = 21·78 + 18 = 39·78 = 39·8 cm

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