F.3 MathematicsM.C. Revision Exercise – Ch.3 Special Lines and Centre on a Triangle

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Multiple-choice Questions13073001

In the figure, GB = EC, ∠GBC =∠ECB and ∠GAB =∠EDC = 90. Which of the following triangles is congruent to △GBC? State the reason.A. △GAB, RHSB. △ECB, SASC. △EBC, SASD. △EBC, AAS-- ans --Solution:The answer is B.In △GBC and △ECB, GB = EC given BC = CB common side ∠GBC =∠ECB given

△GBC △ECB SAS △ ECB is congruent to △ GBC , and the reason is SAS .

-- ans end --

13073002

In the figure, AD // BC, if we need to prove △ADE △CBE, which of the followingconditions is needed?A. ∠AED =∠CBEB. ∠DAE =∠ADEC. AB = DCD. AD = BC

P. 1

-- ans --Solution:The answer is D.Consider AD = BC.In △ADE and △CBE, ∠DAE =∠BCE alt. ∠s, AD // BC ∠ADE =∠CBE alt. ∠s, AD // BC AD = BC given

△ ADE △ CBE ASA

-- ans end --

13073003

In the figure, ∠BAC =∠AED = 42, ∠ABC = 60, ∠BAE = 142, AB = EA and AC = ED. Write down the reason of △BAC △AED, and find ∠CAD.A. SAS, 40B. SAS, 50C. AAS, 40D. AAS, 50-- ans --Solution:The answer is A.In △BAC and △AED, AB = EA given AC = ED given ∠BAC =∠AED = 42 given

△BAC △AED SAS ∠EAD =∠ABC = 60

∠BAC +∠CAD +∠EAD =∠BAE 42 +∠CAD + 60 = 142

∠CAD =

-- ans end --

P. 2

13073004Determine which pair(s) of triangles in the following is/are congruent.I. II.

△ABC and △EDB △ABC and △CDA

III.

△ABE and △CDE

A. I onlyB. II onlyC. I and II onlyD. II and III only-- ans --Solution:The answer is C.I: △ABC △EDB (SSS)II: △ABC △CDA (RHS)III: △ABE and △CDE are not congruent.

Only the pairs of triangles in I and II must be congruent.-- ans end --

13073005

P. 3

In the figure, D is a point of BC. ∠BAC =∠EDF = 90, BA // ED, BC // EF, BA = ED = 4 cm and DF = 5 cm. Write down the reason of △ABC △DEF, and find the area of the figure.A. RHS, 15 cm2

B. RHS, 10 cm2

C. ASA, 15 cm2

D. ASA, 20 cm2

-- ans --Solution:The answer is D.In △ABC and △DEF, ∠BDE =∠DEF alt. ∠s, BC // EF ∠ABC =∠BDE alt. ∠s, BA // ED

∠ABC =∠DEF

∠BAC =∠EDF = 90 given BA = ED = 4 cm given

△ABC △DEF ASAArea of the figure

= area of △ABC + area of △DEF= 2 × (area of △DEF)

-- ans end --

13073006

P. 4

In the figure, AC = AD, BC = ED, ∠ABC = 58 and ∠DAE = 18. Write down thereason of △ABC △AED, and find ∠CAD.A. SAS, 24B. SAS, 28C. ASA, 24D. SSA, 28-- ans --Solution:The answer is B.In △ABC and △AED, BC = ED given AC = AD given

∠ACD =∠ADC base ∠s, isos. △∠ACB = 180 –∠ACD and∠ADE = 180 –∠ADC adj. ∠s on st. line

∠ACB =∠ADE△ABC △AED SAS

∠BAC =∠EAD = 18　∠BAC +∠ABC =∠ACD　 18 + 58 =∠ACD ∠ACD = 76

In △ACD,∠CAD +∠ACD +∠ADC = 180

∠CAD + 76 + 76 = 180 ∠CAD =

-- ans end --

13073007

P. 5

In the figure, BECF is a straight line. AB // DE, AC // DF, AB = DE, BE = 2 cm and EC = 2.5 cm. Write down the reason of △ABC △DEF, and find BF.A. AAS, 6.5 cmB. AAS, 7 cmC. SAS, 6.5 cmD. SAS, 7 cm-- ans --Solution:The answer is A.In △ABC and △DEF, ∠ABC =∠DEF corr. ∠s, AB // DE ∠ACB =∠DFE corr. ∠s, AC // DF

AB = DE given△ABC △DEF AAS

BC = EF

BF = BC + EF – EC = 2BC – EC = 2(BE + EC) – EC = [2(2 + 2.5) – 2.5] cm =

-- ans end --

13073008

In the figure, APCR is a straight line. ∠BAC =∠QPR = 90, BC // QR and BC = QR. If the area of △ABC is 24 m2 and AB = 4 m, write down the reason of △ABC △PQR, and find PR.A. AAS, 12 mB. AAS, 14 mC. RHS, 16 mD. SAS, 18 m-- ans --Solution:The answer is A.In △ABC and △PQR, ∠BAC =∠QPR = 90 given

P. 6

∠BCA =∠QRP corr. ∠s, BC // QR BC = QR given

△ABC △PQR AAS

Area of △ABC

PR = AC

=

-- ans end --

13073009

In the figure, ADB, AFC and BEC are straight lines. ∠DAE =∠FAE, ∠BDE =∠CFE and BD = CF. Which of the following may not be true?A. BE = CEB. AD = AFC. ∠AED =∠AEFD. ∠DBE =∠DEA-- ans --Solution:The answer is D.In △ADE and △AFE,∠DAE =∠FAE given∠BDE =∠CFE given∠ADE = 180 –∠BDE and∠AFE = 180 –∠CFE adj. ∠s on st. line

∠ADE =∠AFE

AE = AE common side △ADE △AFE AAS

AD = AF corr. sides, △s

∠ AED = ∠ AEF corr. ∠s, △s

In △BDE and △CFE,

P. 7

∠BDE =∠CFE given BD = CF given

△ADE △AFE proved DE = FE corr. sides, △s

△BDE △CFE SAS BE = CE corr. sides, △s

Only ∠ DBE = ∠ DEA may not be true.

-- ans end --

13073010

In the figure, BCEFG and AED are straight lines. ∠ABC = 40, ∠DCE = 48, ∠AEC = 90 and ∠GAF = 21. If △ABC △AGF, which of the following must be true?A. ∠CAD = 29B. ∠BAC = 30C. ∠EDF = 40D. ∠AFE = 48-- ans --Solution:The answer is A.

△ABC △AGF given∠BAC =∠GAF = 21 corr. ∠s, △s

In △ABC,∠BAC +∠ABC =∠ACE ext. ∠ of △

21 + 40 =∠ACEi.e. ∠ACE = 61In △ACE,∠CAD +∠AEC +∠ACE = 180 ∠ sum of △

∠CAD + 90 + 61 = 180 ∠ CAD = 29

-- ans end --

P. 8

13073011

In the figure, PQ = 4 cm, QR = 14 cm, RS = 16 cm and PS = 7 cm. If △SPQ ~ △RQS, which of the following must be true?A. SQ = 7 cmB. SQ = 8 cmC. ∠PQS = 2∠SPQ D. ∠PQS = ∠SQR +∠SRQ-- ans --Solution:The answer is B.

△SPQ △RQS given

corr. sides, ~ △s

-- ans end --

13073012

In the figure, ∠BAC =∠CAD, AB = 4.5 m, AC = 3 m, AD = 2 m and CD = 2.6 m. Write down the reason of △ABC △ACD, and find BC.A. ratio of 2 sides, inc. ∠, 3.9 mB. ratio of 2 sides, inc. ∠, 4.1 mC. 3 sides proportional, 3.9 mD. 3 sides proportional, 4.1 m-- ans --Solution:The answer is A.In △ABC and △ACD,

P. 9

∠BAC =∠CAD given

△ABC △ACD ratio of 2 sides, inc. ∠

　

-- ans end --

13073013

In the figure, if we need to prove △CDE △CAB, which of the following conditions is needed?A. CE = DEB. CB = AB

C.

D. AB // DE-- ans --Solution:The answer is D.Consider AB // DE.In △CDE and △CAB, ∠DCE =∠ACB common angle ∠CDE =∠CAB corr. ∠s, AB // DE ∠CED =∠CBA corr. ∠s, AB // DE

△ CDE △ CAB AAA

-- ans end --

P. 10

13073014Determine which pair(s) of triangles in the following is/are similar.I. II.

△ABC and △DEF △ABC and △ADE

III.

△ABC and △ADE

A. I onlyB. I and II onlyC. II and III onlyD. I, II and III-- ans --Solution:The answer is D.I: △ABC △DEF (3 sides proportional)II: △ABC △ADE (AAA)III: △ABC △ADE (ratio of 2 sides, inc. ∠)

The pairs of triangles in I, II and III are similar.-- ans end --

13073015

P. 11

In the figure, ∠BAC =∠QPR, AB = 12 m, BC = 16 m, AC = 20 m, PQ = 3 m andPR = 5 m. Write down the reason of △ABC △PQR, and find the area of the shaded region.A. ratio of 2 sides, inc. ∠, 90 m2

B. ratio of 2 sides, inc. ∠, 96 m2

C. AAA, 114 m2

D. AAA, 120 m2

-- ans --Solution:The answer is A.In △ABC and △PQR,

∠BAC =∠QPR given

△ABC △PQR ratio of 2 sides, inc. ∠

Area of the shaded region

= area of △ABC – area of △PQR

-- ans end --

13073016

P. 12

Given that △PQR △STP, SP = ST and PQ = PR, which of the following must betrue?I. PT = TRII. ST // QRIII. PT = QRA. I onlyB. II onlyC. III onlyD. I and II only-- ans --Solution:The answer is B.

With the notation in the figure, in △STP,SP = ST given∠SPT = a base ∠s, isos. △

In △STP and △PQR,∠PRQ =∠SPT = a corr. ∠s, ~ △s∠STP =∠PRQ = a

i.e. ST // QR corr. ∠s equalOnly II is true.

-- ans end --

13073017

P. 13

In the figure, ABC is a right-angled triangle and DEFG is a rectangle. Given that ∠BAC =∠GFC = 90, AG = 10 cm, GC = 6 cm and GF = 4 cm. Write down the reason of △DAG △GFC, and find DG, correct to 3 significant figures.A. AAA, 13.4 cmB. AAA, 15.2 cmC. ratio of 2 sides, inc. ∠, 13.4 cmD. ratio of 2 sides, inc. ∠, 15.2 cm-- ans --Solution:The answer is A.In △DAG and △GFC,∠DAG =∠GFC = 90 given∠AGD =∠FCG corr. ∠s, DG // BC∠ADG = 180 –∠DAG –∠AGD and∠FGC = 180 –∠GFC –∠FCG ∠ sum of △

∠ADG =∠FGC△DAG △GFC AAA

　

cor. to 3 sig. fig.-- ans end --

13073018

How many pairs of similar triangles are there in the figure?A. 4

P. 14

B. 5C. 6D. 7-- ans --Solution:The answer is C.The following shows all pairs of similar triangles:△ AEF ~ △ ABC , △ AEF ~ △ CGF , △ CGF ~ △ ABC , △ ABC ~ △ CDA , △ CGF ~ △ CDA , △ AEF ~ △ CDA . -- ans end --

P. 15

13073019

In △PQR, ∠QPR = 40, ∠PRS = 110, PQ = 10 cm and QR = 6 cm. What kind oftriangle is △PQR?A. Scalene triangleB. Right-angled triangleC. Equilateral triangleD. Isosceles triangle-- ans --Solution:The answer is D.In △PQR,∠QPR +∠PQR =∠PRS ext. ∠ of △ 40 +∠PQR = 110

∠PQR = 70∠PRQ +∠PRS = 180 adj. ∠s on st. line ∠PRQ + 110 = 180

∠PRQ = 70∠PQR =∠PRQ = 70

i.e. PR = PQ sides opp. eq. ∠s △ PQR is an isosceles triangle.

-- ans end --

P. 16

13073020

In the figure, PQR and PTS are straight lines. ∠PQU =∠PTU and ∠URS =∠USR. Which of the following must be true?I. UR = USII. PR = PSIII. PQ = PTA. II onlyB. I and II onlyC. I and III onlyD. I, II and III-- ans --Solution:The answer is D.I: In △URS,

∠URS =∠USR givenUR = US sides opp. eq. ∠s

II: In △QRU and △TSU,∠PQU = ∠PTU given∠UQR = 180 –∠PQU and∠UTS = 180 –∠PTU adj. ∠s on st. line

∠UQR =∠UTS

∠QUR =∠TUS vert. opp. ∠sUR = US proved

△QRU △TSU AAS ∠QRU =∠TSU corr. ∠s, △s

∠PRS =∠QRU +∠URS ∠PSR =∠TSU +∠USR

∠PRS =∠PSRPR = PS 　 sides opp. eq. ∠s

III: △QRU △TSU provedQR = TS corr. sides, △s

PQ = PR – QR

P. 17

PT = PS – TSPQ = PT I, II and III must be true.

-- ans end --

13073021

In the figure, ABCD and DEFG are identical squares, and ∠DAG = 40. Which of the following must be true?A. ∠DCE = 40B. ∠DCE = 50C. ∠CDE = 90D. ∠CDE = 100-- ans --Solution:The answer is B.In △DAG,

AD = GD given∠DGA =∠DAG = 40 base ∠s, isos. △

∠DAG +∠DGA +∠ADG = 180 ∠ sum of △ 40 + 40 +∠ADG = 180

∠ADG = 100∠ADC =∠GDE = 90 given∠ADG +∠ADC +∠GDE +∠CDE = 360 ∠s at a pt.

100 + 90 + 90 +∠CDE = 360 ∠CDE = 80

In △CDE,DC = DE given∠DCE =∠DEC base ∠s, isos. △

∠CDE +∠DCE +∠DEC = 180 ∠ sum of △ 80 + 2∠DCE = 180

∠ DCE = 50 -- ans end --

P. 18

13073022

Given that ABCDE is a regular pentagon, how many isosceles triangles are there in the figure?A. 1B. 2C. 3D. 4-- ans --Solution:The answer is C.

ABCDE is a regular pentagon.AB = BC = AE = ED

i.e. △ ABC and △ AED are isosceles triangles. In △ABC and △AED, AB = AE given BC = ED given ∠ABC =∠AED given

△ABC △AED SAS AC = AD corr. sides, △s

△ ACD is an isosceles triangle. There are 3 isosceles triangles in the figure.

-- ans end --

13073023

In the figure, CE = CF, BA // CF and BC // DF. Which of the following must be true?I. △ABC is isosceles.II. AE = ECIII. △ADE △CFE

P. 19

A. I onlyB. II onlyC. III onlyD. I and II only-- ans --Solution:The answer is A.

CE = CF given∠CEF =∠CFE base ∠s, isos. △

∠ADF =∠CFE alt. ∠s, BA // CF ∠AED =∠CEF vert. opp. ∠s

∠ADE =∠AED

∠ADE =∠ABC corr. ∠s, BC // DF∠AED =∠ACB corr. ∠s, BC // DF∠ABC =∠ACB

i.e. AB = AC sides opp. eq. ∠s△ ABC is isosceles.

-- ans end --

13073024

Referring to the figure, which of the following must be true?A. PS is an angle bisector of ∠RPT.B. PS is an angle bisector of ∠PST.C. PS is a median.D. PS is not any special line.-- ans --Solution:The answer is A.

∠RPS =∠SPTPS is an angle bisector of ∠ RPT .

-- ans end --

P. 20

13073025

Referring to the figure, which of the following must be true?I. PR is the only altitude of △PRT.II. QU is a perpendicular bisector.III. QU is an altitude of △PRU.A. I onlyB. II onlyC. I and II onlyD. II and III only-- ans --Solution:The answer is D.I: RT is also an altitude of △PRT.

I is not true.II: QU PR and PQ = QR

QU is a perpendicular bisector. II must be true.

III: QU PRQU is an altitude of △PRU.

III must be true.Only II and III must be true.

-- ans end --

13073026

In the figure, if PR is an angle bisector of ∠QPS, QT is an angle bisector of ∠PQS, ∠RPS = 35 and ∠TQS = 24, find ∠PRS.A. 81B. 83

P. 21

C. 85D. 87-- ans --Solution:The answer is B.

PR is an angle bisector of ∠QPS. ∠QPR =∠RPS = 35QT is an angle bisector of ∠PQS. ∠PQT =∠TQS = 24

In △PQR, ∠PQR +∠QPR =∠PRS　

∠PQT +∠TQS +∠QPR =∠PRS 24 + 24 + 35 =∠PRS ∠PRS =

-- ans end --

13073027

In the figure, ∠BAD =∠DAC, ∠ADB = 90 and BD = DC. Determine which of thefollowing must be true.I. AD is an angle bisector of ∠BAC.II. AD is a perpendicular bisector of BC.III. AD is an altitude of △ABC.A. I and II onlyB. I and III onlyC. II and III onlyD. I, II and III-- ans --Solution:The answer is D.-- ans end --

P. 22

13073028

In the figure, ∠RPT = 60, ∠RVT = 120, ∠RUT = 83 and ∠VTR = 37. Whichof the following is/are angle bisector(s)?I. RUII. QTIII. PSA. I onlyB. II onlyC. I and II onlyD. I, II and III-- ans --Solution:The answer is C.I: In △PUR,

∠PRU +∠RPU =∠RUT ext. ∠ of △ ∠PRU + 60 = 83

∠PRU = 23In △RVT,∠VRT +∠RVT +∠VTR = 180 ∠ sum of △

∠VRT + 120 + 37 = 180 ∠VRT = 23

∠PRU =∠URT = 23

i.e. RU is an angle bisector of ∠PRT.

II: In △UVT,∠VTU +∠VUT =∠RVT ext. ∠ of △

∠VTU + 83 = 120 ∠VTU = 37

∠QTR =∠QTP = 37QT is an angle bisector of ∠PTR.

Only RU and QT are angle bisectors.

-- ans end --

P. 23

13073029

In the figure, ABC and AED are straight lines. AB = BC, EC = ED, BE // CD and ∠ACD = 90. Which of the following must be true?I. EB is a perpendicular bisector of AC.II. EB is a perpendicular bisector of AD.III. EC is a median of AD.A. I onlyB. II onlyC. I and II onlyD. I and III only-- ans --Solution:The answer is D.I: ∠ABE =∠ACD = 90 corr. ∠s, BE // CD

AB = BC givenEB is a perpendicular bisector of AC.

III: In △ABE and △CBE, EB = EB　 common side

∠ABE =∠CBE = 90 proved AB = CB given

△ABE △CBE　 SAS AE = CE corr. sides, △s

EC = EDAE = ED

i.e. EC is a median of AD.Only I and III must be true.

-- ans end --

13073030It is known that the lengths of two sides of a triangle are 7 and 11 respectively, find the possible length of the remaining side.A. 2 B. 3 C. 4 D. 5-- ans --Solution:The answer is D.

P. 24

Since 5 + 7 > 11, 5 + 11 > 7 and 7 + 5 > 11,the three line segments of lengths 5, 7, 11 can form a triangle.The possible length of the remaining side is 5.

-- ans end --

13073031Which of the following descriptions about the circumcentre is true?I. Circumcentre is the point of intersection of angle bisectors.II. Circumcentre only forms outside the triangle.III. The centre of a circumcircle is circumcentre.A. I onlyB. II onlyC. III onlyD. None of the above-- ans --Solution:The answer is C.I: Circumcentre is the point of intersection of perpendicular bisectors.II: Circumcentre may lie outside the triangle, inside the triangle or on one side of the triangle.

Only III is true.-- ans end --

13073032

In the figure, ADB, AFC and BEC are straight lines. BE = EC, CF = FA, BD = DA and AG = 4 cm. Find GE.A. 1 cm B. 2 cm C. 3 cm D. 4 cm-- ans --Solution:The answer is B.

G is the centroid.

AG : GE

P. 25

GE

P. 26