Certificate Mathematics in Action Full Solutions 4B

Certificate Mathematics in Action Full Solutions 4B

7 Basic Properties of Circles (II)

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Certificate Mathematics in Action Full Solutions 4B

Activity

Activity 7.1 (p. 63)

2. 90°. IfF the angle between ST and PQA is not equal
to 90°, a right-angled triangle with ST as the hypothenuse can be drawn. In siuch case, the opposite side perpendicular to PQ is shorter than ST. So, the line segment ST is shortest when ST ^ PQ.

3. (a) OC

(b) No

4. 90°

Activity 7.2 (p. 82)

1. (a)

(b) (i) Yes

(ii) Yes

3. Yes

Follow-up Exercise

p. 66

1. (alt. Ðs, TC // DE)

ÐOTB = 90° (tangent ^ radius)

(tangent ^ radius)

2. OT = OC (radii)

(base Ðs, isos. △)

(Ð sum of △)

(tangent ^ radius)

3. ∵

4. TG = TE (given)

(base Ðs, isos. △)

(alt. Ðs, AB // CD)

(tangent ^ radius)

p. 69

1. ÐPTO = ÐQTO (tangent properties)

TQ = TP (tangent properties)

2. (tangent ^ radius)

(ext. Ð of △)

(tangent properties))

3. (tangent ^ radius)

4. (Ð at centre twice Ð at ⊙ce)

(tangent ^ radius)

Consider quadrilateral OQTP.

∴ ÐPQR = ÐAQP

∴ PQ bisects ÐRQA.

p. 75

1. (Ð in alt. segment)

(Ð in alt. segment)

2. (Ð in alt. segment)

(Ð in alt. segment)

3. (Ð sum of △)

(Ð in alt. segment)

4. (Ð in alt. segment)

(Ð sum of △)

p. 77

1. (Ð in alt. segment)

(Ð in alt. segment)

2. (Ð in alt. segment)

(Ð at centre twice Ð at ⊙ce)

4. (Ð in alt. segment)

p. 87

1. ∵ ÐADB = ÐACB = 50° (given)

∴ A, B, C and D are concyclic. (converse of Ðs in

the same segment)

2. ∵

∴ A, B, C and D are not concyclic.

3. (adj. Ðs on st. line)

∴ A, B, C and D are not concyclic.

4. (Ð sum of △)

∴ A, B, C and D are not concyclic. (opp. Ðs supp.)

∵

∴ A, B, C and D are concyclic. (ext. Ð = int. opp. Ð)

Exercise

Exercise 7A (p. 70)

Level 1

1. (radii)

Consider △OPB.

ÐOPB = 90° (tangen(tangent ^
radius)t

4. (tangent

properties)

(tangent ^
radius)

5. radii

∴ PA2 + OA2 = PO2

∴ OA ^ PA converse of Pyth. theorem

∴ PA is the tangent to the converse of tangent
circle at A. converse of tangent ^
radius ^ radius

converse of tangent ^ radius

6. ∵ AP = AR (tangent properties)

∴ AR = 3 cm

∴

∵ RC = QC (tangent properties)

∴ QC = 5 cm

∵ BP = BQ (tangent properties)

∴ BQ = 2 cm

7. (tangent properties)

(Ð sum of △)

( (tangent properties) (tangent properties) (tangent properties) (tangent properties)

∴

8. OA = OB (radii)

9. OT = OP (radii)

∴

(ext. Ð of △)

10. ÐABC = 90° (tangent ^ radius)

(Ð sum of △)

(Ð at centre twice Ð at ⊙ce)

Level 2

11. (a) (Ð at centre twice Ð at ⊙ce)

(b) (tangent ^ radius)

Consider quadrilateral AOBT.

∵ Sum of the interior angles of quadrilateral

∴

12.

(tangent ^ radius)

(Ð sum of )

13. Let the radius of the circle be r cm.

ÐOTA = 90° (tangent ^ radius)

(Pyth. theorem)

∴ The radius of the circle is 5 cm.

14.

15. SMEFSU07EX@F01

Join OA, OB and OD.

(tangent ^ radius)

∵

and AE // BF

∴ AOB is a straight line.

16. (a) Reflex (Ðs at a pt.)

(b) (int. Ðs, QP // TO)

OT ^ AB (tangent ^ radius)

∴

17. (a) (ext. Ð, cyclic quad.)

(Ð sum of △)

(b) OT ^ PT (tangent ^ radius)

(radii)

(base Ðs, isos. △)

∴ (Ð sum of △)

18. SMEFSU07EX@F02

Join AB.

Let

∴ AC // TO alt. Ðs equal

19. Let AP = x cm.

AP = AR (tangent properties)

BP = BQ (tangent properties)

CR = CQ (tangent properties)

∵ BP = (12 - x) cm

∴ BQ = (12 - x) cm

∵

∴ CR = (x - 5) cm

∵ AP = AR

∴ x = 15 - x

∴ x = 7.5

∴ AP =

20. BP = BR (tangent properties)

CR = CQ (tangent properties)

∵

From (1), we have

∴

∴ AP =

21. AP = AS, BP = BQ, CQ = CR

and DR = DS (tangent properties)

Perimeter of ABCD

22. SMEFSU07EX@F03

Join OP.

Let

Consider △QOB.

23. (a) AB = BD tangent properties

BD = BC tangent properties

∴ AB = BC

(b) ÐBAD = ÐBDA base Ðs, isos. △

ext. Ð of △

ÐBDC = ÐBCD base Ðs, isos. △

ext. Ð of △

∴

Exercise 7B (p. 78)

Level 1

1. (Ð in alt. segment)

(Ð sum of △)

(Ð in alt. segment)

(ext. Ð of △)

5. TP = TA (given)

(base Ðs, isos. △)

6. (Ð in alt. segment)

(Ð in alt. segment)

(alt. Ðs, CA // TB)

(opp. Ðs, cyclic quad.)

7. (a) Consider △BCD and △CAD.

∴ ÐDBC = ÐDCA

∴ △BCD ~ △CAD AAA

(b) ∵ △BCD ~ △CAD

∴

8. (a) Consider △BCT and △CAT.

∴ ÐCBT = ÐACT

∴ △BCT ~ △CAT AAA

(b) ∵ △BCT ~ △CAT

∴

10. (Ð sum of △)

11. SMEFSU07EX@F04

Join BD.

(Ð in alt. segment)

(opp. Ðs, cyclic quad.)

12. (Ð in alt. segment)

(adj. Ðs, on st. line)

(opp. Ðs, cyclic quad.)

13. ∴

∴ DE is the tangent to the

circle at A. converse of Ð in
alt. segment

14. ∴

∴ PA is the tangent to

the circle at T. converse of Ð in
alt. segment

15.

Let ÐBCD = t.

Take ,

(or any other reasonable answers)

Level 2

16.

Consider △BCD.

17. (adj. Ðs on st. line)

(Ð in alt. segment)

(ext. Ð of △)

(Ð sum of △)

(opp. Ðs, cyclic quad.)

18. Let ÐABC = q.

∴ (Ð in semi-circle)

Consider △BPA.

(Ð sum of △)

19. (a) (Ð in alt. segment)

(Ð sum of △)

(b) (Ð in alt. segment)

(alt. Ðs, PD // BC)

(ext. Ð of △)

20. (a) Consider △ABC and △BTC.

∴ △ABC ~ △BTC AAA

(b) (Pyth. theorem)

∴

∵ △ABC ~ △BTC

∴

21. SMEFSUO7EX@F05

Join AM, AN, AB, BM and BN.

(arcs prop. to Ðs at ⊙ce)

(opp. Ðs, cyclic quad.)

(Ð sum of △)

22.

Consider △ATB.

(Ð sum of △)

23. (a) Consider △PAT and △PTB.

∴ △PAT ~ △PTB AAA

(b) Let PA = x cm.

∵ △PAT ~ △PTB

∴

24. (a)

(Ð sum of △)

(Ð in alt. segment)

(opp. Ðs, cyclic quad.)

∴

(b) (Ð in alt. segment)

(ext. Ð of △)

25. SMEFSU07EX@F06

Join AB.

Consider △ABT.

(Ð sum of △)

Consider △ABC.

(Ð sum of △)

26. (a) (Ð in alt. segment) (adj. Ðs on st. line)

(Ðs in the same segment)

Consider △APT.

(ext. Ð of △)

(b) Consider △ABT.

(Ð sum of △)

Exercise 7C (p. 88)

Level 1

1. (a) 

\ A, B, C and D are concyclic. opp. Ðs supp.

(b) (Ðs in the same segment)

2. (a)  ÐDBC

Ð sum of △

\ A, B, C and D are

concyclic. converse of Ðs

in the same

segment

(b) (Ðs in the same segment)

3. (a)  ÐCAD = ÐDBC = 90°

\ A, B, C and D are

concyclic. converse of Ðs

in the same

segment

(b) ÐBAC

(adj. Ðs on st. line)

\ (Ðs in the same segment)

4. (a)

Mark the point F as shown in the figure.

Consider △ABF.

Ð sum of △

\ ÐABC = ÐADE

\ A, B, C and D are

concyclic. ext. Ð = int.

opp. Ð

(b) \ (Ðs in the same segment)

5. Consider △PBC.

ext. Ð of △

\ A, B, C and D are

concyclic. converse of Ðs in

the same segment

6. ÐABC = ÐADC opp. Ðs of // gram

ÐPQB = ÐPDC ext. Ð, cyclic quad.

ÐDPQ = ÐPQB alt. Ðs, AD // BC

\ ÐDPQ = ÐABC

\ A, B, Q and P are

concyclic. ext. Ð = int. opp. Ð

7. AD = AE given

ÐADE = ÐAED base Ðs, isos. △

ÐADE = ÐABC opp. Ðs of // gram

\ ÐAED = ÐABC

\ A, B, C and E are

concyclic. ext. Ð = int. opp. Ð

8. ÐBPT = 90° Ð in semi-circle

ÐSQC = 90° Ð in semi-circle

adj. Ðs on st. line

\ ÐAQS = ÐBPT = 90°

\ A, P, R and Q are

concyclic. ext. Ð = int. opp. Ð

9. ÐATQ = ÐABT Ð in alt. segment

ÐATQ = ÐCDT alt. Ðs, PQ // CD

\ ÐABT = ÐCDT

\ A, B, C and D are

concyclic. ext. Ð = int. opp. Ð

\ ABCD is a cyclic

quadrilateral.

10. ÐBAD = 180° – (x + y) (Ð sum of △)

(opp. Ðs, cyclic quad.)

Take x = 30° and y = 50°, we have

\ x = 30°, y = 50°, z = 80°

(or any other reasonable answers)

11.

D is a point on the other side of AB such that
AD ^ BD.

Level 2

12. (a) ÐOAT = 90° tangent ^ radius

ÐOBT = 90° tangent ^ radius

\ ÐOAT + ÐOBT = 180°

\ O, B, T and A are

concyclic. opp. Ðs supp.

(b)  O, B, T and A are

concyclic. (proved in (a))

\ (Ðs in the same segment)

13. (a) ÐAPB = 90° Ð in semi-circle

adj. Ðs on st. line

ÐAOM = 90° given

\ ÐAPM = ÐAOM

\ O, P, M and A are

concyclic. converse of Ðs in

the same segment

(b) OA = OP radii

ÐOAP = ÐOPA base Ðs, isos. △

ÐOAP = ÐOMP Ðs in the same segment

\ ÐOPA = ÐOMB

14. ÐBRS = ÐSQA ext. Ð, cyclic quad.

ÐSRC = 180° – ÐBRS adj. Ðs on st. line

ÐSRC = ÐSPA ext. Ð, cyclic quad.

\ 180° – ÐSQA = ÐSPA

\ ÐSQA + ÐSPA = 180°

\ A, Q, S and P are

concyclic. opp. Ðs supp.

\ AQSP is a cyclic

quadrilateral.

15. (a)

Join PB and let ÐARP = q.

ÐPBA = ÐARP = q Ðs in the

same segment

ÐAPB = 90° Ð in

semi-circle

In △APB,

Ð sum of △

In △AQP,

ext. Ð of △

(b)

Join RB.

ÐTQB = 90° proved in (a)

ÐTRB = 90° Ð in semi-circle

ÐTQB + ÐTRB = 180°

\ R, T, Q and B are

concyclic. opp. Ðs supp.

\ RTQB is a cyclic

quadrilateral.

16. Consider △ACB and △DBC.

AC = DB given

ÐACB = ÐDBC given

BC = CB common side

\ △ACB @ △DBC SAS

ÐBAC = ÐCDB corr. Ðs, @ △s

\ A, B, C and D are concyclic. converse of Ðs in

the same segment

17. (a) ÐACB = 90° Ð in semi-circle

adj. Ðs on st. line

ÐADB = 90° Ð in semi-circle

\ ÐECP = ÐADB

\ P, D, E and C are

concyclic. ext. Ð = int. opp. Ð

(b) ÐCOD = 2ÐCAD Ð at centre twice

Ð at ⊙ce

Ð sum of △

18. (a) (i) ÐBQP = x Ð in alt.

segment

ÐCQP = y Ð in alt.

segment

 ÐBQC = ÐBQP + ÐCQP

\ ÐBQC = x + y

(ii) Consider △APD.

Ð sum of △

ÐBQC + ÐBPC

\ B, P, C and Q are

concyclic. opp. Ðs supp.

(b)

Join BC.

ÐCQP = y Ð in alt. segment

 B, P, C and Q are

concyclic.

\ ÐCBP = ÐCQP = y Ðs in the same

segment

\ ÐCDA = ÐCBP = y

\ A, B, C and D are

concyclic. ext. Ð = int. opp. Ð

Revision Exercise 7 (p. 93)

Level 1

1. TB = TA (tangent properties)

ÐTBA = ÐTAB = x (base Ðs, isos. △)

(Ð sum of △)

ÐTBA = ÐACB (Ð in alt. segment)

ÐACB = ÐCBF = y (alt. Ðs, AC // TF)

\ ÐTBA = y

But ÐTBA = x = 61°

2. ÐOAB = 90° (tangent ^ radius)

ÐODC = 90° (tangent ^ radius)

Sum of interior angles of pentagon = 180° (5 – 2)

= 540°

3. (a) ÐBAP = 90° (tangent ^ radius)

(Ð sum of △)

(b)

Join OC.

(Ð at centre twice Ð at ⊙ce)

ÐOCP = 90° (tangent ^ radius)

(Ð sum of △)

4. (Ð in alt. segment)

(adj. Ðs on st. line)

ÐABC = ÐTAC (Ð in alt. segment)

TA = TC (tangent properties)

ÐTAC = ÐTCA (base Ðs, isos. △)