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Yimin Math Centre
4 Unit Math Homework for Year 12 (Worked Answers)
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nt
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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
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Geometry of Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
in
1.1
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Harder 3 Unit Topics (Part1)
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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.
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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.
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• The line from the centre of a circle to the midpoint of the chord meets the chord at
right angles.
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• When two circles intersect, the line joining their centres bisects their common chord
at right angles.
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Exercise 1.1.1 Use trigonometry to find x in each diagram (correct to 1 decimals place.).
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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.
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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.
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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 =
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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
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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:
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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
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Solution:
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2. Hence, show that ∠P OC = ∠QOC.
Ce
∴ 4P OC ≡ 4QOC (three mattching side are equal).
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∴ ∠P OC = ∠QOC, matching angle of congruent traingles.
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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.
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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.
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∠AOB = ∠COD and OM = ON .
h
Exercise 1.1.5
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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.
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Year 12 Topic 1 Worked Answers
Angles in a Circle
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Definition: The angle at the centre of a a circle is twice the angle at the circumference
standing on the same arc.
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Definition: The angle in a semicircle is a right angle.
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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.
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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.
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Definition: The opposite angles of a cyclic quadrilateral are supplementary (sum to 180◦ ).
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Exercise 1.1.6 The diagram shows, AOB is the diameter of a circle, centre O. find angle ∠ADC.
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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
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• is called indirect, if the circles are on the opposite sides of the tangent.
The Alternate Segment Theorem
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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.
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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).
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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.
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AM × M B = P M × M Q
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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.
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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).
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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:
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1. Prove the ∠BOT = 2∠BT P .
Let ∠BT P be x◦ ⇒∴ ∠BAT = x◦ (angle in the alternate segment)
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∠BOT = 2x◦ (angle at centre is twice angle at circum. on the same arc)
∴ ∠BOT = 2∠BT P (Q.E.D)
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Q.E.D. quod erat demonstrandum ‘which was to be demonstrated’.
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2. Prove that ∠AT Q + ∠RBA + ∠P BT = 180◦ .
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Solution: ∠ABT + ∠RBA + ∠P BT = 180◦ (∠RBP is a straight angle)
But∠AT Q = ∠ABT (angle in the alternate segment)
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∴ ∠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)
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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.
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AX × BX = CX × DX and EY × F Y = GY × HY
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• The square of the length of a tangent is equal to the product of the intercepts of a
secant drawn from an external point.
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(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.
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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.
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Solution:
2. AB if BP = 10 cm and P T = 13 cm.
Since P T 2 = AP × BP BP = P T 2 ÷ BP
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Solution:
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BP = 132 ÷ 10 = 16.9 cm ⇒ AB = 16.9 − 10 = 6.9 cm.
Since CP × DP = AP × BP, CP = DP × AP ÷ BP
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Solution:
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3. CD if DP = 5 cm, AB = 8cm and BP = 6 cm.
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CP = 6 × (8 + 6) ÷ 5 = 16.8 cm ⇒ CD = 16.8 − 5 = 11.8 cm
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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.
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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:
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Solution:
AE × BE = CE × DE
Since AB bisects CD ∴ CE = DE
CE = 6 cm
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Ce
CE 2 = AE × BE
√
CE = AE × BE
√
CE = 4 × 9
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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
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Solution:
√
ab.
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1. Show that 4ADC|||4CDB, and hence show that DC =
consider the rectangular triangles ACB and CDB.
These triangle have the common angle ∠CAB.
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Hence 4ACB are similar 4CDB.
AD
CD
From here
=
⇒ CD2 = AD × BD;
CD
BD
√
AD = a, BD = b, ⇒ ∴ CD = ab.
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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 =
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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.
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2. Find the angel sum of ∠OAM and ∠AP M .
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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.
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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◦ .
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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.
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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.
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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).
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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.
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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.
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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.)
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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
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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.
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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.
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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.
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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◦ .
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Year 12 Topic 1 Worked Answers
Page 18 of 19
Exercise 1.2.8
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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.
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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.
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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.
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Year 12 Topic 1 Worked Answers
Page 19 of 19
Cyclic Quadrilaterals:
• Opposite angles of a cyclic quadrilateral are supplementary.
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• An exterior angle of a cyclic quadrilateral equals the opposite interior angle.
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Exercise 1.2.9 ABC is a triangle inscribed in a circle. AB = AC, point D lies on the circle,
produced BD to E.
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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 .
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