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( i ) Using the angles shown in the diagram below, state at least two theorems.

1. ( i ) Using the angles shown in the diagram below, state at least two theorems. Theorem: The angles in any triangle add up to 180°. |  a | + |  b | + |  c | = 180°. Theorem: Each exterior angle in a triangle is equal to the sum of the interior opposite angles. |  a| + |  b| = |  d|.

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( i ) Using the angles shown in the diagram below, state at least two theorems.

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  1. 1. (i) Using the angles shown in the diagram below, state at least two theorems. Theorem: The angles in any triangle add up to 180° |a| + |b| + |c| = 180° Theorem: Each exterior angle in a triangle is equal to the sum of the interior opposite angles. |a| + |b| = |d|

  2. 1. (ii) Use the diagram and what you know about the sum of the angles of a triangle to explain why |a| + |b| = |d|. |a| + |b| + |c| = 180° 3 angles of triangle |c| + |d| = 180° Straight line 180° = 180° |a| + |b| + |c| = |c| + |d| Subtract |c| from both sides |a| + |b| = |d|

  3. 1. (iii) If |d| = 100° and |a| = 40°, find the measure of band c. |a| + |b| = |d| |a| = 40° Given |d| = 100° Given 40° + |b| = 100° 40° + |b| − 40° = 100° − 40° Subtract 40° from both sides

  4. 1. (iii) If |d| = 100° and |a| = 40°, find the measure of band c. |b| = 60° |a| + |b| + |c| = 180° 40° + 60° + |c| = 180° 100° + |c| = 180° 100° + |c| − 100° = 180° − 100° Subtract 100° from both sides |c| = 80°

  5. 2. QSbisects both of the angles PSRand PQR.Prove that QSis the diameter of the circle. (x + x) + (y + y) = 180° Opposite angles of a cyclic quadrilateral sum to 180° 2x + 2y = 180° 2(x + y) = 180° Take out common factor Divide both sides by 2 x + y = 90°

  6. 2. QSbisects both of the angles PSRand PQR.Prove that QSis the diameter of the circle. In PSQ |SQP| + |QPS| + |PSQ| = 180° 3 angles of triangle |SQP| = x Given |PSQ| = y Given x + |QPS| + y = 180° (x + y) + |QPS| = 180° 90° + |QPS| = 180° 90° + |QPS| − 90° = 180° − 90° Subtract 90° from both sides |QPS| = 90°

  7. 2. QSbisects both of the angles PSRand PQR.Prove that QSis the diameter of the circle. Similarly, for RSQ |QRS| = 90° Corollary: Each angle in a semicircle is a right angle PSQ and RSQ form semicircles of the same circle with QS as the diameter

  8. 3. |BE||| |CD|. ABCand AEDare straight lines. |AB| = 4 cm, |BC|= 6 cm, |BE|= 5 cm, |AE|= 4·8 cm. Calculate |CD|. (i)

  9. 3. |BE||| |CD|. ABCand AEDare straight lines. |AB| = 4 cm, |BC|= 6 cm, |BE|= 5 cm, |AE|= 4·8 cm. Calculate |ED|. (ii) |AD| = 12 

  10. 3. |BE||| |CD|. ABCand AEDare straight lines. |AB| = 4 cm, |BC|= 6 cm, |BE|= 5 cm, |AE|= 4·8 cm. Calculate |ED|. (ii) |AD| = |AE| + |ED| 12 = 4·8 + |ED| 12 − 4·8 = 4·8 + |ED| − 4·8 Subtract 4·8 from both sides 7·2 cm = |ED|

  11. 4. Ifn must be a whole number, what is the smallest possible value of n? n + n 15 2 sides of a triangle are together greater than the third 2n 15 n 7·5 If we change the angle between n and n so that it become a straight line n + n must be at least 15. If we choose n = 7, n + n = 7 + 7 = 14 and would be too short. The next whole number is eight so this is our answer. n= 8

  12. 5. The Jamaican flag is a parallelogram in which the yellow lines form the diagonals, as shown.Prove that the green triangles are congruent. Justify your answer. AEBDEC |BE| = |DE| Diagonals of parallelogram bisect each other |AEB| = |DEC | Vertically opposite angles |AE| = |EC| Diagonals of parallelogram bisect each other AEB = DEC by SAS

  13. 6. In the circle shown, Ois the centre and BCis a diameter.|ACP| = 40° and QPis a tangent at C. Find |ACO|. (i) |OCP| = 90° Tangent perpendicular to radius |OCP| = |ACP| + |ACO| |ACP| = 40° Given 90° = 40° + |ACO| 90° − 40° = 40° + |ACO| − 40° Subtract 40° from both sides 50° = |ACO|

  14. 6. In the circle shown, Ois the centre and BCis a diameter.|ACP| = 40° and QPis a tangent at C. Find |OAB|. (ii) Justify your answers. |OC| = |OA| Radii of circle OAC is isosceles |OAC| = |ACO| |OAC| = 50° |BAC| = 90° Semicircle corollary

  15. 6. In the circle shown, Ois the centre and BCis a diameter.|ACP| = 40° and QPis a tangent at C. Find |OAB|. (ii) Justify your answers. |OAB| + |OAC| = |BAC| |OAB| + 50° = 90° |OAB| + 50° − 50° = 90° − 50° Subtract 50° from both sides |OAB| = 40°

  16. 7. In the triangles PQRand TSQ, |QRP|= |QST|. Prove that the triangles RPQ and STQ are similar. (i) |QRP| = |QST| Given |PQR| = |SQT|Vertically opposite PQR |QRP| + |PQR| + |QPR| = 180° 3 angles of triangle Subtract |QRP| from both sides |QRP| + |PQR| + |QPR| − |QRP| = 180° − |QRP| PQR| + |QPR| = 180° − |QRP|

  17. 7. In the triangles PQRand TSQ, |QRP|= |QST|. Prove that the triangles RPQ and STQ are similar. (i) Subtract |PQR| from both sides |PQR| + |QPR| − |PQR| = 180° − |QRP| − |PQR| |QPR| = 180° − |QRP| − |PQR| SQT |QST| + |SQT| + |QTS| = 180° 3 angles of triangle Subtract |QST| from both sides |QST| + |SQT| + |QTS| − |QST| = 180° − |QST| |SQT| + |QTS| = 180° − |QST|

  18. 7. In the triangles PQRand TSQ, |QRP|= |QST|. Prove that the triangles RPQ and STQ are similar. (i) Subtract |SQT| from both sides |SQT| + |QTS| − |SQT| = 180° − |QST| − |SQT| |QTS| = 180° − |QST| − |SQT| But from above, |QTS| = 180° − |QRP| − |PQR| Therefore, |QTS| = |QPR| PQR and SQT are therefore equiangular and similar triangles

  19. 7. In the triangles PQRand TSQ, |QRP|= |QST|. Given that |QT| = 5 cm, |QR| = 16 cm and |PQ| = |QS|, find |PT|.Give your answer correct to the nearest cm. (ii)

  20. 8. (i) What is the specific name for the shape ABCD? A cyclic quadrilateral.

  21. 8. (ii) Give one property of the shape ABCD. Opposite angles in a cyclic quadrilateral sum to 180°.

  22. 8. (iii) Find the value of x. BCE |BCE| + |CBE| + |BEC| = 180° 3 angles of triangle |BEC| = 35° Given |CBE| = 67° Given |BCE| + 67° + 35° = 180° |BCE| + 102° = 180° |BCE| + 102° − 102° = 180° − 102° |BCE| = 78°

  23. 8. (iii) Find the value of x. |BCE| + |DCB| = 180° Straight line 78° + |DCB| = 180° 78° + |DCB| − 78° = 180° − 78° Subtract 78° from both sides |DCB| = 102° x + 102° = 180° Opposite angles sum to 180° x + 102° − 102° = 180° − 102° x = 78°

  24. 9. A circle has chord of length 10 cm.The shortest distance from the circle’s centre to the chord is 4 cm.Find the radius of the circle correct to one decimal place. OC is the perpendicular bisector of the chord AB |AC| = |CB| perpendicular from centre bisects the chord |AB| = 10 |AB| = |AC| + |CB| |AB| = |AC| + |AC| 10 = 2|AC| Divide both sides by 2 5 = |AC|

  25. 9. A circle has chord of length 10 cm.The shortest distance from the circle’s centre to the chord is 4 cm.Find the radius of the circle correct to one decimal place. OCA is a right angled triangle |OA| = Radius of the circle |OA| = Hypotenuse |OC| = 4 Given c2 = a2 + b2 Pythagoras c2 = 42 + 52 c2 = 16 + 25 c2 = 41 Square root both sides |OA| = radius = 6·4 cm c = 6·4 cm

  26. 10. During a total eclipse of the sun, the moon is directly in line with the sun and blocks the sun’s rays. The distance DA between the centre of the Earth and the centre of the sun is 149,600,000 km.The distance DE between the centre of Earth and the centre of the moon is 384,400 km. The radius AB of the sun is 696,300 km.Use the diagram and the given measurements to estimate the radius EC of the moon, to the nearest kilometre. |DA| = 149,600,000 km |DE| = 384,400 km |AB| = 696,300 km

  27. 10. During a total eclipse of the sun, the moon is directly in line with the sun and blocks the sun’s rays. The distance DA between the centre of the Earth and the centre of the sun is 149,600,000 km.The distance DE between the centre of Earth and the centre of the moon is 384,400 km. The radius AB of the sun is 696,300 km.Use the diagram and the given measurements to estimate the radius EC of the moon, to the nearest kilometre.

  28. 11. In the diagram, ABis parallel to CD. [AD] bisects CDB and [BC] bisects ABD.Show that ΔBODand ΔAOBare congruent. AB || CD Given [AD] bisects CDB Given [BC] bisects ABD Given |CDA| = |ADB| COD bisected by [AD] |CDA| = |BAD| Alternate angles |ABC| = |CBD| ABD bisected by [BC] |ABD| = |BDE| Alternate angles

  29. 11. In the diagram, ABis parallel to CD. [AD] bisects CDB and [BC] bisects ABD.Show that ΔBODand ΔAOBare congruent. At point D: x + x + 2y = 180° Straight line 2x + 2y = 180° 2(x + y) = 180° Take out common factors Divide both sides 2 x + y = 90° x + y + z = 180° 3 angles of triangle 90° + z = 180° 90° + z − 90° = 180° − 90° Subtract 90 from both sides z = 90°

  30. 11. In the diagram, ABis parallel to CD. [AD] bisects CDB and [BC] bisects ABD.Show that ΔBODand ΔAOBare congruent. BOD      AOB |BOD| = |AOB| Both 90°, shown common to both triangles |OB| = |OB| Given |DBO| = |ABO| BODAOB by ASA

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