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Yimin Math Centre 4 Unit Math Homework for Year 12 (Worked Answers) Grade: Date: Score: nt re Student Name: Table of contents 1.2 1 1 1.1.1 Chord Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1.2 Angles in a Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.1.3 Circle Quadrilateral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.1.4 Tangents and Radii . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.1.5 Direct and indirect Common Tangents . . . . . . . . . . . . . . . . . . . . . . 7 1.1.6 The Alternate Segment Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 7 M at h Geometry of Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . in 1.1 Ce Harder 3 Unit Topics (Part1) Yi m 1 1.1.7 Intercepts on Intersecting Chords . . . . . . . . . . . . . . . . . . . . . . . . 8 1.1.8 Further Circle Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Pass Exam Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 This edition was printed on June 17, 2013. Camera ready copy was prepared with the LATEX2e typesetting system. Copyright © 2000 - 2013 Yimin Math Centre (www.yiminmathcentre.com) 4 Unit Math Homework for Year 12 Year 12 Topic 1 Worked Answers 1 Page 1 of 19 Harder 3 Unit Topics (Part1) 1.1 1.1.1 Geometry of Circle Chord Properties Chord Properties of Circle 1: • A perpendicular drawn to a chord from the centre of a circle bisects the chord, and the perpendicular bisector of a chord passes through the centre. nt re • The line from the centre of a circle to the midpoint of the chord meets the chord at right angles. Ce • When two circles intersect, the line joining their centres bisects their common chord at right angles. in M at h Exercise 1.1.1 Use trigonometry to find x in each diagram (correct to 1 decimals place.). Yi m 1. Given that ∠AOM = 50◦ , AB = 12 cm. Solution: AB = 12, cm ⇒ AM = 6 cm, ⇒ x = AM = 7.8 cm. sin 50◦ 2. Given that ∠P OQ = 150◦ , P Q = 15 cm. Solution: ∠P OQ = 150◦ , ⇒ ∠ROQ = 75◦ , ⇒ 7.5 = 7.8 cm. sin 75◦ 3. Given that ∠M ON = 130◦ , OM = 8.5 cm. Solution: ∠M ON = 130◦ , ⇒ ∠OM N = 25◦ , ⇒ x = 2 × 8.5 × cos 25◦ = 15.4 cm. Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers Page 2 of 19 Exercise 1.1.2 For diagram shown below AO = 26 cm, OM = 10 cm. Find the length of AB, giving reasons. Solution: OA2 = AM 2 + OM 2 (Phthagoras), ⇒ AM 2 = 262 − 102 √ ∴ AM = 576 = 24 cm. nt re N ow AB = 2 × AM (as OM is the perpendicular bisector of AB), ⇒ AB = 48 cm. Exercise 1.1.3 Ce 1. Find the radius of a circle in which a chord of length 14 cm subtends and angle of 70◦ at the centre (correct to 1 decimals place.). 7 7 ⇒ r= = 12.2 cm. r sin 35◦ h Solution: M at sin 35◦ = ◦ sin 55 = Yi m Solution: in 2. A chord subtends an angle of 110◦ at the centre of a circle of radius 5.6 cm. Find the length of the chord correct to one decimal place. l 2 5.6 ⇒ l = 2 × 5.6 × sin55◦ = 9.2 cm. 3. A chord of length 12 cm is drawn on a circle of radius 16 cm. How far this chord from the centre of the circle? Solution: d= √ 162 − 62 = 14.8, cm. 4. A chord of length 22 cm has a perpendicular distance of 8 cm from the centre of the circle. What is the radius of the circle? Solution: r= √ 112 + 82 = 13.6 cm Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers Page 3 of 19 Exercise 1.1.4 These two circles have as their centres points O and C. PQ is the common chord joining the points of intersection of the two circles. N is the point where PQ intersects the line OC which joins the centres. Solution: nt re 1. Prove that the triangles P OC and QOC are congruent. P C = CQ = r, P O = OQ = R and OC = OC (common) ∵ 4P OC ≡ 4QOC at Solution: h 2. Hence, show that ∠P OC = ∠QOC. Ce ∴ 4P OC ≡ 4QOC (three mattching side are equal). M ∴ ∠P OC = ∠QOC, matching angle of congruent traingles. Yi m in 3. Now, prove that the triangles P ON and QON are congruent. Solution: P O = QO, (matching sides of congruent traingles, prove above.) ∠P OC = ∠QOC, (matching angle of congruent traingles, prove above). ∠OP N = ∠OQN (base angles of an isosceles traingle). ∴ 4P ON ≡ 4QON .(ASA) 4. Hence, show that N bisects P Q and that P Q ⊥ OC. Solution: P N = N Q, (matching sides of congruent traingles, prove above) 180◦ and∠P N O = ∠ON Q = = 90◦ , (matching angles of congruent traingles, prove above) 2 ∴ N bisects P Q and that P Q ⊥ OC. Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers Page 4 of 19 Chord Properties of Circles 2: Equal chords of a circle are the same distance from the centre and subtend equal angles at the centre. Ce nt re ∠AOB = ∠COD and OM = ON . h Exercise 1.1.5 Yi m in M at 1. AB = BC = 25cm, OM = 5.8 cm. Find he length of ON giving reasons. Solution: ON = 5.8 cm as equal chords of a circle are the same distance from the centre. 2. OM = ON = 3.5 cm, P Q = 12.6 cm. Find the length of RS, giving reasons. Solution:RS = 12.6 cm as chords that are equidistant from the centre are equal in length. Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers Angles in a Circle Ce nt re Definition: The angle at the centre of a a circle is twice the angle at the circumference standing on the same arc. in M at h Definition: The angle in a semicircle is a right angle. Yi m 1.1.2 Page 5 of 19 Definition: Angles at the circumference of a circle standing on the same arc are equal. In another words, angles in the same segment of a circle are equal. Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers 1.1.3 Page 6 of 19 Circle Quadrilateral A cyclic quadrilateral is a quadrilateral whose vertices lie on the circumference of a circle. nt re Definition: The opposite angles of a cyclic quadrilateral are supplementary (sum to 180◦ ). at h Ce Exercise 1.1.6 The diagram shows, AOB is the diameter of a circle, centre O. find angle ∠ADC. Yi m in M Solution: Connect CB to form a cyclic quadrilateral ABCD, ∵ ∠BOC = 60◦ and CO = CB, ⇒ ∴ ∠OBC = ∠OCB = 60◦ . In the quadrilateral ABCD, ∠OBC + ∠ADC = 180◦ ∴ ∠ADC = 180◦ − 60◦ = 120◦ (Opposite angles of a cyclic quadrilateral are supplementary). Exercise 1.1.7 If O is the centre of the circle and ∠CAO = 40◦ , find the size of ∠ABC Solution: Connect AE and CE to form a cyclic quadrilateral ABCE, ∠OAC = ∠OCA, ⇒ ∠AOC = 180◦ − 2 × 40◦ = 100◦ , ∠AEC = ∠AOC ÷ 2 = 100◦ ÷ 2 = 50◦ ∴ ∠ABC = 180◦ − 50◦ = 130◦ (The opposite angles of a cyclic quadrilateral are supplementary). Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers 1.1.4 Page 7 of 19 Tangents and Radii Definition: The two tangents form a external point have equal lengths. 1.1.5 Direct and indirect Common Tangents Definition: A common tangent to a pair of circles: • is called direct, if both circles are on the same side of the tangent, 1.1.6 nt re • is called indirect, if the circles are on the opposite sides of the tangent. The Alternate Segment Theorem in M at h Ce Definition: The angle between a tangent to a circle and a chord at the point of contact is equal to any angle in the alternate segment. Yi m Exercise 1.1.8 In the figure the tangent TS is parallel to the chord PQ. If ∠T P Q = 40◦ , find the size of ∠P T Q. Solution: ∠RT P = ∠T P Q = 40◦ , and ∠ST Q = ∠T QP (Alternate angles, ST ||P Q) ∠ST Q = ∠T P Q = 40◦ (alternate segment theorem), ∴ ∠P T Q = 180◦ − 40◦ − 40◦ = 100◦ (angle sum of a straight line). Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers 1.1.7 Page 8 of 19 Intercepts on Intersecting Chords Definition: If two chords of a circle intersect, the product of the intercepts on the one chord is equal to the product of the intercepts on the other chord. nt re AM × M B = P M × M Q Ce Definition: Given a circle and two secants from an external point, the product of the two intervals from the point to the circle on the secant is equal to the product of these two intervals on the other secant. Yi m in M at h AM × M B = P M × M Q Exercise 1.1.9 O is the centre of the circle of radius 9 cm. If CD = 6 cm, DP = 4 cm, find P B. Solution: Since OA = OB = 9 cm, ⇒ AB = 18 cm and CP = CD + DP = 6 + 4 = 10 cm Let P B = x, AP = AB + x = 18 + x AP × BP = CP × DP, ⇒ (18 + x) × x = 10 × 4 Now 18x + x2 = 40 ⇒ x2 + 18x − 40 = 0 ⇒ (x − 2)(x + 20) = 0 ∴ x = 2 cm, x = −20 (invalid). Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers Page 9 of 19 Exercise 1.1.10 O is the centre of the circle. BT is a chord that subtends ∠BAT at the circumference and ∠T OB at the centre. P T and P B are tangents to the circle. Solution: nt re 1. Prove the ∠BOT = 2∠BT P . Let ∠BT P be x◦ ⇒∴ ∠BAT = x◦ (angle in the alternate segment) Ce ∠BOT = 2x◦ (angle at centre is twice angle at circum. on the same arc) ∴ ∠BOT = 2∠BT P (Q.E.D) at h Q.E.D. quod erat demonstrandum ‘which was to be demonstrated’. M 2. Prove that ∠AT Q + ∠RBA + ∠P BT = 180◦ . in Solution: ∠ABT + ∠RBA + ∠P BT = 180◦ (∠RBP is a straight angle) But∠AT Q = ∠ABT (angle in the alternate segment) Yi m ∴ ∠AT Q + ∠RBA + ∠P BT = 180◦ . 3. Prove that ∠BP T = 180◦ − 2∠BAT . Solution: Let ∠BAT be α ∴ ∠BT P = α◦ (angle in the alternate segment) ∠P BT = α◦ (angle in the alternate segment) ∴ ∠BP T + α◦ + α◦ = 180◦ (angle sum of a triangle) ∴ ∠BP T = 180◦ − 2α = 180◦ − 2∠BAT (Q.E.D) Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers 1.1.8 Page 10 of 19 Further Circle Properties • The products of intercepts of intersecting chords or secants are equal. nt re AX × BX = CX × DX and EY × F Y = GY × HY M at h Ce • The square of the length of a tangent is equal to the product of the intercepts of a secant drawn from an external point. Yi m in (P T )2 = AP × P B Exercise 1.1.11 Find the value of each pronumeral, giving reasons. Solution: T P 2 = AP × DP, ⇒ AP = T P 2 ÷ DP = 122 ÷ 8 = 18 cm ∴ x = 18 − 8 − 8 = 2 cm AC × CD = BC × CT, ⇒ CT = AC × CD ÷ CB ∴ y = 8 × 2 ÷ 2.5 = 6.4 cm. Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers Page 11 of 19 Exercise 1.1.12 For the diagram given below find: 1. P T if AB = 9 cm and BP = 3 cm AB = 9 and BP = 3 cm ⇒ AP = 3 + 9 = 12 cm √ AP × BP = T P 2 , ⇒ P T = 3 × 12 = 6 cm. nt re Solution: 2. AB if BP = 10 cm and P T = 13 cm. Since P T 2 = AP × BP BP = P T 2 ÷ BP Ce Solution: h BP = 132 ÷ 10 = 16.9 cm ⇒ AB = 16.9 − 10 = 6.9 cm. Since CP × DP = AP × BP, CP = DP × AP ÷ BP M Solution: at 3. CD if DP = 5 cm, AB = 8cm and BP = 6 cm. in CP = 6 × (8 + 6) ÷ 5 = 16.8 cm ⇒ CD = 16.8 − 5 = 11.8 cm Yi m 4. EG if GF = 20 cm, CG = 30 cm and DG = 25 cm. Solution: Since EG × GF = CG × DG ⇒ EG = GF × CG ÷ DG ∴ EG = 25 × 30 ÷ 20 = 37.5 cm. 5. CD if CG = 16 cm EF = 38 cm and EG = 22 cm. Solution: Since EG × GF = CG × DG ⇒ DG = GF × CG ÷ EG DG = (38 − 22) × 22 ÷ 16 = 22 cm ⇒ CD = 22 + 16 = 38 cm 6. CD if T P = 18 cm and DP = 9 cm. 2 2 Solution: T P = CP × DP, ⇒ CP = 18 ÷ 9 = 36, ⇒ CD = 36 − 9 = 27 cm. Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers Page 12 of 19 Exercise 1.1.13 1. AB is the diameter of a circle. AB bisects a chord CD at the point E. Find the length of CE if AE = 9 cm and BE = 4 cm. Because the product of intercepts are equal: nt re Solution: AE × BE = CE × DE Since AB bisects CD ∴ CE = DE CE = 6 cm at h Ce CE 2 = AE × BE √ CE = AE × BE √ CE = 4 × 9 Yi m in M 2. PT is a direct common tangent of the circles drawn. AB is a common chord that has been produced to meet the common tangent at C. Use the ‘quare of the tangent’of result to prove that CP = CT . Solution: By using the square of the tangent: CT 2 = AC × BC and P C 2 = AC × BC Since AB is a common chord ∴ CT 2 = P C 2 ∴ CP = CT. Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers 1.2 Page 13 of 19 Pass Exam Questions Exercise 1.2.1 ADB is a straight line with AD = a and DB = b . A circle is drawn on AB as diameter. DC is drawn perpendicular to AB to meet this circle at C. Let O be the center of the circle, and r be its radius Ce Solution: √ ab. nt re 1. Show that 4ADC|||4CDB, and hence show that DC = consider the rectangular triangles ACB and CDB. These triangle have the common angle ∠CAB. M at h Hence 4ACB are similar 4CDB. AD CD From here = ⇒ CD2 = AD × BD; CD BD √ AD = a, BD = b, ⇒ ∴ CD = ab. Yi m in 2. Deduce geometrically that if a > 0 and b > 0, then Solution: √ ab ≤ a+b . 2 a+b ; 2 OC = r and the triangle is rectangular √ ∴ ⇒ OC > CD ⇒ r > ab a+b √ > ab. ∴ 2 AB = 2r ⇒ r = Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers Page 14 of 19 Exercise 1.2.2 Let A is a point outside a circle, AP is the tangent of the circle, P is the point of contact. ABC is a secant. M is a midpoint of BC. 1. Prove that APOM is a cyclic quadrilateral. at h 2. Find the angel sum of ∠OAM and ∠AP M . Ce nt re Solution: As AP is a tangent ⇒ ∴ OP ⊥ AP. Since M is the midpoint of the chord BC, ⇒ ∴ OM ⊥ BC ∴ ∠OP A + ∠OM A = 180◦ . ∴ the opposite angles of the quadrilateral are supplementary. Therefore the quadrilateral APOM is a cyclic quadrilateral. Yi m in M Solution: As proven above the quadrilateral APOM is a cyclic quadrilateral. ∴ ∠OAM = ∠OP M Since OP ⊥ AP, ⇒ ∴ ∠OP M + ∠AP M = 90◦ , ∴ ∠OAM + ∠AP M = 90◦ . Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers Page 15 of 19 Exercise 1.2.3 1. ABC is a triangle. The internal bisectors of ∠ABC and ∠ACB meet at D. DP , DQ and DP are the perpendiculars from D to BC, CA and AB respectively. Show that DR = DQ and deduce that the internal bisectors of the three angles of a triangle are concurrent. at h Ce nt re Solution: The rectangular triangles BDR and BDP have a common side and ∠DBR = ∠DBP . Hence these triangles are congruent. Therefore DR = DP . Similarly, the rectangular triangles CDQ and CDP are congruent. Hence DQ = DP . We have DR = DP and DQ = DP . Consequently, DR = DQ. The rectangular triangles ADP and ADR are congruent, since they have a common side AD and DQ = DR. Hence ∠DAQ = ∠DAQ and so AD is the internal bisector of ∠BAC. Yi m in M 2. ABC is a triangle. E and F are the midpoints of AC and AB respectively. BE and CF intersects at D. Show that the triangle DEF and DBS are similar and hence deduce that the three medians of a triangle are concurrent. Solution: Since AE = 21 AC, AF = 12 AB and ∠CAB = ∠EAF , the triangles ABC and AEF are similar. Hence EF = 21 CB and these triangles have equal angles. From here ∠AEF = ∠ACB ⇒ EF ||CB. Hence ∠F EB = ∠CBE and ∠EF C = ∠BCF . DE FD EF Therefore 4DEF is similar to 4DCB ⇒ DB = DC = CB . EF 1 But it was shown that CB = 2 . Hence ED = 12 and F D = 12 DC , (i.e., the point D of intersection of the medians BE and CF divide each median in accordance with the relation 2 : 1 starting with the top. As it is true for any pair of the medians, all the medians intersect at a single point). Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers Page 16 of 19 Exercise 1.2.4 In the diagram, BD is the diameter of the circle, E is a point on the circle , Point A is the intersect of CE produced and and BD produced. BC ⊥ AC . and ∠CBE = ∠DBE. Prove that AC is a tangent of the circle. nt re Solution: Joint the points O, E, ⇒ OE = OB ∴ ∠OEB = ∠OBE. ∵ ∠CBE = ∠DBE ⇒ ∴ ∠CBE = ∠OEB, ⇒ ∴ OE||BC But BC ⊥ AE, ⇒ ∴ OE ⊥ AC, ⇒ ∴ AC is the tangent of the circle. M at h Ce Exercise 1.2.5 ABC is a triangle inscribed in a circle. P is a point on the minor arc AB. The points L, M, and N are the feet of the perpendiculars from P to CA produced, AB, and BC respectively. Show that L, M and N are collinear. (The line N L is called the Simpson line.) Yi m in Solution: In order to prove that L, M and N are collinear, it is sufficient to show that: ∠LM A = ∠N M B. For this purpose we hsow , that ∠N M B = ∠BP N = ∠SP A = ∠LM A. The first step: ∠N M B = ∠BP N . The 4P KM and 4BKN are rectangular and NK =M . ∠P KM = ∠BKN ⇒ 4P KM are similar 4BKN ⇒ KB PK K But ∠P KB = ∠M KN ⇒ 4P KB|||4M KN ⇒ ∠N M B = ∠BP N . The second step: ∠BP N = ∠SP A. The point P lies on the circle ⇒ P ACB is cyclic quadrilateral ∴ ⇒ ∠P AC + ∠P BC = 180◦ . But ∠P AC + ∠P AL = 180◦ . Hence ∠P BC = ∠P AL. From here as the 4P N B and 4OLA are rectangular, we have 4P N B|||4P LA ⇒ ∴ ∠BP N = ∠AP L. The third step: ∠SP A = ∠LM A it is obvious that 4ALS|||4P AL, As these rectangular triangles have the common angle ∠P SM . S Hence PASS = M ⇒ 4M LS|||4P AS ⇒ ∠SP A = ∠LM A. LS Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers Page 17 of 19 Exercise 1.2.6 The diagram shows that AB is a chord of the circle and CD is a tangent and meeting the circle at P . AC ⊥ CD, BD ⊥ CD and P Q ⊥ AB. Prove that P Q2 = AC × BD. Ce nt re Solution: Joint P A and P B as shown about. ∵ CD meets the circle at P ⇒ ∴ ∠α = ∠β. ∵ AC ⊥ CD, P Q ⊥ AB, ⇒ ∴ ∠ACP = ∠P QB = 90◦ , AP . ∴ 4ACP |||4P QB, ⇒ ∴ PACQ = BP PQ AP Similarly: BD = BP . PQ ∴ PACQ = BD , ⇒ ∴ P Q2 = AC × BD. in M at h Exercise 1.2.7 ABC is a triangle inscribed in a circle. AD bisects the ∠BAC and produced to E, and E is lies on the circle. Yi m 1. Prove that 4ABE|||4ADC. Solution: ∠BAE = ∠CAE = ∠CAD (angles on the same arc CE) ∠AEB = ∠ACB (angles on the same arc CE) ∴ 4ABE|||4ADC. 2. If the area of the triangle ABC is given by A = 21 AD × AE, find the size of the ∠BAC. Solution: As 4ABE|||4ADC ⇒ AB = AD , ⇒ AB × AC = AD × AE. AE AC 1 But A = 2 AB × AC sin ∠BAC and A = 12 AD × AE ∴ AB × AC sin ∠BAC = AD × AE. Hence sin ∠BAC = 1 ⇒ ∠BAC = 90◦ . Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers Page 18 of 19 Exercise 1.2.8 nt re 1. Given that E lies on the circle and AB is the diameter of the circle. M N is a tangent passes through E. AM ⊥ M N, N B ⊥ M N andED ⊥ AB. prove that M D ⊥ N D. M at h Ce Solution: Jointing AE and BE, ∵ AM ⊥ M N, ED ⊥ AB, ∴ ⇒ ∠AM E + ∠ADE = 180◦ ⇒ ∠M DA = ∠M EA, But M N is a tangent ∴ ⇒ ∠M EA = ∠ABE, ∠M DA = ∠ABE. Similarly ∠BDN = ∠BAE. ∵ AB is the diameter, ∴ ⇒ ∠AEB = 90◦ , ∴ ∠ABE + ∠BAE = 90◦ , ⇒ ∠ADM + ∠BDN = 90◦ , ∴ ∠M DN = 90◦ , ⇒ ∴ M D ⊥ N D. Yi m in 2. AB is the diameter of a circle, A tangent passes through D produced and meets the AB produced at C. If DA = DC, prove that AB = 2BC. Solution: Jointing OE, ⇒ OB = OE, ∵ ∠AED = ∠OBE ∵ BD is a diameter, ∴ ∠EBD = 90◦ , ⇒ ∠OEB + ∠OED = 90◦ , ∴ ∠AED + ∠OED = 90◦ , ⇒ ∴ OE ⊥ AE, Therefore AE is the tangent of the circle. Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com) Year 12 Topic 1 Worked Answers Page 19 of 19 Cyclic Quadrilaterals: • Opposite angles of a cyclic quadrilateral are supplementary. nt re • An exterior angle of a cyclic quadrilateral equals the opposite interior angle. at h Ce Exercise 1.2.9 ABC is a triangle inscribed in a circle. AB = AC, point D lies on the circle, produced BD to E. M 1. Prove that AD produced bisects the ∠CDE. in Solution: Produced AD to F, ⇒ ∠CDF = ∠ABC (An exterior angle of a cyclic quadrilateral) and AB = AC, ⇒ ∠ABC = ∠ACB, Yi m ∵ ∠ADB = ∠ACB, ∴ ∠ADB = ∠CDF Since ∠EDF = ∠ADB, ⇒ ∴ ∠EDF = ∠CDF ∴ AD produced bisects ∠CDE 2. If ∠BAC = 30◦ , It 4ABC, if the perpendicular height of base BC is 2 + of the circle inscribed the 4ABC. √ 3 cm, find the area Solution: Joint AO and produced to H, ⇒ AH ⊥ BC (height of an isosceles triangle ) Joint OC ⇒ ∠OAC = ∠OCA = 15◦ , ⇒ ∠ACB = 75◦ ∴ ∠OCH = 60◦ √ √ Let r be the radius of the circle, ⇒ r + 23 r = 2 + 3 ⇒ r = 2 units. Therefore the area of the circle is 2πr = 4π units2 . Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com)