Question 1: State True or False:

(i) Two similar polygons are necessarily congruent – False

(ii) Two congruent polygons are necessarily similar – True

(iii) All equi-angular triangles are similar – True

(iv) All isosceles triangles are similar – False

(v) Two isosceles right angles triangles are similar – True

(vi) Two isosceles triangles are similar, if an angle of one is congruent to the corresponding angle of the other – True

(vii) The diagonals of the trapezium, divide each into proportional segments – True

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Question 2: In \triangle ABC, DE \parallel BC , where D and E  are points on AB and AC  respectively. Prove that \triangle ADE \sim \triangle ABC . Also find the length of DE, \ if \ AD = 12 \ cm,  BD = 24 \ cm \ and \ BC = 8 \ cm .

Answer:s11.jpg

In \triangle ABC, DE \parallel BC

Consider \triangle ABC \ and\  \triangle ADE

\angle ABC = \angle ADE (alternate angles)

\angle ACB =\angle AED  (alternate angles)

Therefore \triangle ADE \sim \triangle ABC    (AAA postulate)

Hence \frac{AD}{AB} = \frac{DE}{BC} = \frac{AE}{AC} 

\frac{12}{12+24} = \frac{DE}{8}

\Rightarrow DE = \frac{12}{36} \times 8 = \frac{8}{3}

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Question 3: s1Given \angle GHE = \angle DFE= 90^o , DH= 8, DF=12, DG=3x-1  and DE=4x+2 . Find the length of the segments DG \ and \ DE

Answer:

Consider \triangle EFD and \triangle GHD

\angle GHE = \angle DFE= 90^o (Given)

\angle EDG = \angle HDG \ (\angle D \ is \ common) 

Therefore \triangle EFD \sim \triangle GHD    (AAA postulate)

\frac{DH}{DF} = \frac{DG}{DE} 

\frac{8}{12} = \frac{3x-1}{4x+2} 

32x+16 = 36x-12 

4x=28 \Rightarrow x = 7 

Substituting DG = 3 \times 7 -1 = 20 

DE = 4 \times 7 +2 = 30 

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Question 4: D is a point of side BC of  \triangle ABC such that \angle ADC = \angle BAC . Prove that CA^2=CB \times CD .

Answer:s12

Consider \triangle ABC \ and \  \triangle ADC

\angle BAC = \angle ADC (Given)

\angle ACB = \angle ACD \ (\angle C \ is \ common) 

Therefore \triangle ABC \ and \  \triangle ADC   (AAA postulate)

Therefore \frac{CA}{CB} = \frac{CD}{CA} 

\Rightarrow CA^2=CB \times CD

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Question 5:  sm10In the given figure, \triangle ABC and \triangle AMP are right angled at B and M  respectively.  Given AC=10 \ cm, AP = 15 \ cm  and PM= 12 \ cm . (i) Prove \triangle ABC \sim \triangle AMP (ii) Find AB and AC . [2012]

Answer:

In  \triangle ABC \ and \  \triangle AMP

\angle BAC = \angle PAM (common angle)

\angle ABC =\angle PMA

Therefore \triangle ABC \sim \triangle AMP (AAA postulate)

Since \triangle ABC \sim \triangle AMP

\frac{AP}{AM} = \frac{BC}{PM} = \frac{AC}{AP}

Given AC=10 cm, AP = 15 cm and PM= 12 cm

\Rightarrow AB = \frac{10}{15} \times 11 = 7.33 \ cm

\Rightarrow BC = \frac{10}{15} \times 12 = 8 \ cm

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Question 6: E \ and \  F are points in sides DC \ and \ AB respectively of parallelogram ABCD . If diagonal AC and segment EF intersect at G ; prove that: AG \times EG = FG \times CG .

Answer:s10

In  \triangle AGF \ and \  \triangle EGC

\angle AGF = \angle EGC (opposite angles)

\angle ECG =\angle FAG   (alternate angles)

Therefore \triangle AGF \sim \triangle ECG (AAA postulate)

Therefore   \frac{EG}{FG} = \frac{CG}{AG} 

\Rightarrow AG \times EG = FG \times CG

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Question 7: Given RS \ and \  PT  are altitudes of  \triangle PQR  . Prove that (i) \triangle PQT \sim \triangle QRS  (ii) PQ \times QS = RQ \times QT 

Answer:s9

(i)   Consider \triangle PQT \ and \  \triangle QRS

\angle QTP = \angle QSR= 90^o (Given, altitudes)

\angle PQT = \angle SQR \ (\angle Q \ is \ common) 

Therefore \triangle PQT \sim \triangle QRS    (AAA postulate)

(ii)  \frac{PQ}{RQ} = \frac{QT}{QS} 

PQ \times QS = RQ \times QT 

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Question 8: s3Given ABCD  is a rhombus, DPR \ and \  CBR  are straight lines. Prove that DP \times CR = DC \times PR  .

Answer:

In  \triangle DPA \ and \  \triangle RPC

\angle DPA = \angle RPC (vertically opposite angles)

\angle PAD =\angle PCR   (alternate angles)

Therefore \triangle DPA \sim \triangle RPC (AAA postulate)

Therefore   \frac{DB}{PR} = \frac{AD}{CR} 

\frac{DP}{PR} = \frac{DC}{CR}  ( AD = DC \ as \ ABCD is a Rhombus)

\Rightarrow DP \times CR = DC \times PR

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Question 9: Given FB=FD, AE \perp FD \ and \  FC \perp AD   . Prove \frac{GB}{AD} = \frac{BC}{ED}   s4

Answer:

FB=FD, AE \perp FD   and FC \perp AD   .

Therefore \angle FDB=\angle FBD 

Consider \triangle AED \ and \  \triangle FCB

\angle AED = \angle FCB= 90^o (Given)

\angle ADE = \angle FBC 

Therefore \triangle AED \sim \triangle FBC    (AAA postulate)

\frac{AD}{FB} = \frac{ED}{BC} 

\frac{FB}{AD} = \frac{BC}{ED} 

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Question 10: In \triangle ABC, \angle B = 2 \angle C   and the bisector of \angle B  meets CA  at point D  . Prove that  (i) \triangle ABC \sim \triangle ABD  (ii) DC:AD = BC:AB  s13.jpg

Answer:

Given \triangle ABC, \angle B = 2 \angle C  

\angle ABD = \angle DBC 

Therefore \angle ABD = \angle DBC = \angle ACB 

Consider \triangle ABC  \ and \  \triangle ABD

\angle BAC = \angle DAB  (Given)

\angle ACB = \angle ABD 

Therefore \triangle ABC \sim \triangle ABD    (AAA postulate)

\frac{BC}{BD} \frac{AB}{AD} 

\frac{BC}{AB} = \frac{BD}{AD} 

\frac{BC}{AB} = \frac{DC}{AD}  ( \angle DBC = \angle DCB \Rightarrow DC = BD )

DC:AD = BC:AB 

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Question 11: In \triangle PQR, \angle Q=90^o and QM \perp  PR . Prove that:

(i) PQ^2 = PM \times PR s15

(ii) QR^2 = PR \times MR

(iii) PQ^2 + QR^2 = PR^2

Answer:

(i)  Given  \triangle PQR, \angle Q=90^o and QM \perp  PR

Consider \triangle PQM \ and\  \triangle PRQ

\angle PMQ = \angle PQR = 90^o (given)

\angle QPM =\angle RPQ  (common angle)

Therefore \triangle PQM \sim \triangle PRQ    (AAA postulate)

Hence \frac{PQ}{PR} = \frac{PM}{PQ} 

 Therefore  PQ^2 = PM \times PR

(ii) Consider \triangle QMR \ and\  \triangle PQR

\angle QMR = \angle PQR = 90^o (given)

\angle QRM =\angle QRP  (common angle)

Therefore \triangle QMR \sim \triangle PQR    (AAA postulate)

Hence \frac{QR}{PR} = \frac{MR}{QR} 

 Therefore  QR^2 = PR \times MR

(iii) Adding (i) and (ii)

PQ^2 + QR^2 = PM \times PR + PR \times MR

\Rightarrow PQ^2 + QR^2 = PR \times (PM + MR)

\Rightarrow PQ^2 + QR^2 = PR \times PR = PR^2

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Question 12: In \triangle ABC, \angle C= 90^o, CD \perp AB . Prove that CD^2 = AD \times DB .

Answer: s16

\triangle ABC, \angle C= 90^o, CD \perp AB .

Consider \triangle PQM \ and\  \triangle PRQ

\angle PMQ = \angle PQR = 90^o (given)

\angle QPM =\angle RPQ  (common angle)

Therefore \triangle PQM \sim \triangle PRQ    (AAA postulate)

Hence \frac{PQ}{PR} = \frac{PM}{PQ} 

Therefore  PQ^2 = PM \times PR

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Question 13: In \triangle ABC, \angle B= 90^o, BD \perp AC .

(i) If CD=10 \ cm \ and \ BD = 8 \ cm, \ find \ AD

(ii) If AC=18 \ cm \ and \ AD = 6 \ cm, \ find \ BD

(iii) If AC=9 \ cm \ and \ AB = 7 \ cm, \ find \ AD

Answer:s16

\triangle ABC, \angle B= 90^o, BD \perp AC

Consider \triangle CDB \ and\  \triangle BCD

\angle ADB = \angle BDC = 90^o (given)

\angle ABD =\angle BCD

Therefore \triangle CDB \sim \triangle BDA    (AAA postulate)

(i)   \frac{CD}{BD} = \frac{BD}{AD} 

Therefore AD = \frac{BD^2}{CD} = \frac{8^2}{10} = 6.4\ cm

(ii)  \frac{BD}{DA} = \frac{CD}{BD} 

Therefore BD = \sqrt{CD \times DA} = \sqrt{72}= 8.49 \ cm

(iii)  \frac{AD}{AB} = \frac{AB}{AC} 

Therefore AD = \frac{AB \times AB}{AC} = \frac{7 \times 7}{9} = 5.44 \ cm

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Question 14: sm11In the figure, PQRS is a parallelogram with PQ=16 \ cm and QR=10 \ cm . L is a point on PR such that RL:LP=2:3. QL produced meets RS at M and PS produced at N .  Find the lengths of PN and RM  [1997]

Answer:

In \triangle RLQ and \triangle PLN

\angle RLQ = \angle PLN (vertically opposite angles)

\angle LRQ = \angle LPN (alternate angles)

Therefore \triangle RLQ \sim \triangle PLN (AAA postulate)

Therefore \frac{RL}{LP} = \frac{RQ}{PN}

\Rightarrow PN = \frac{10}{2} \times 3 = 15 \ cm

In \triangle RLM \ and \  \triangle PLQ

\angle RLM = \angle PLQ (vertically opposite angles)

\angle LRM = \angle LPQ (alternate angles)

Therefore \triangle RLM \sim \triangle PLQ (AAA postulate)

Therefore \frac{RM}{PQ} = \frac{RL}{LP}

\Rightarrow RM = \frac{2}{3} \times 16 = 10.667 \ cm

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Question 15:  In quadrilateral ABCD , Diagonal AC \ and \ BD intersect at point E such that: AE:EC=BE:ED . Show that ABCD is a parallelogram.s17

Answer:

Given \frac{AE}{EC} = \frac{BE}{ED}  … … … … (i)

In \triangle ADB , because EF \parallel AB , by proportionality theorem

\frac{DF}{FA} = \frac{ED}{BE} … … … … (i)

Therefore  from (i) and (ii)

\frac{DF}{FA} = \frac{EC}{AE}

Similarly in \triangle DCA

\frac{DF}{FA} = \frac{CE}{AE} \Rightarrow FE \parallel DC

Also it is given that FE \parallel AB

Therefore DC \parallel AB . Therefore ABCD is a parallelogram.

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s5Question 16: Given AB \parallel DE and BC \parallel EF . Prove that:

(i) \frac{AD}{DG} = \frac{CF}{FG}

(ii) \triangle DFG \sim \triangle ACG

Answer:

(i) Given AB \parallel DE and BC \parallel EF

In \triangle AGB, \frac{AD}{DG} = \frac{EB}{GE} … … … … (i)

In \triangle BGC, \frac{EB}{GE} = \frac{CF}{FG} … … … … (ii)

From (i) and (ii)  you get \frac{AD}{DG} = \frac{CF}{FG}

(ii)  Consider \triangle DFG \ and\  \triangle ACG

We have \frac{AD}{DG} = \frac{CF}{FG}

and \angle DGF = \angle AGC (common angle)

Therefore  \triangle DFG \sim \triangle ACG

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Question 17: In \triangle ABC, AD \perp BC and AD^2 = BD \times DC . Show that \angle BAC = 90^o .s18

Answer:

Given AD^2 = BD \times DC

\Rightarrow \frac{AD}{DC} = \frac{BD}{AD}

Also given \angle ADB = \angle ADC = 90^o

Therefore \triangle DBA \sim \triangle DAC (SAS postulate)

Therefore \angle C = \angle  BAD

\angle B = \angle  DAC

Adding \angle C + \angle  B = \angle  BAD +  \angle  DAC

\Rightarrow \angle A =\angle C + \angle  B  … … … (i)

We know \angle A + \angle B + \angle C = 180^o  … … … … (ii)

From (i) and (ii) 2 \angle A = 180^o

\Rightarrow \angle A = 90^o

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s6Question 18:  In the given figure AB \parallel EF \parallel DC ; AB=67.5 \ cm, DC = 40.5 \ cm and AE = 52.5 \ cm .

(i) Name the three pairs of similar triangles

(ii) Find the lengths of EC \ and \ EF

Answer:

Given AB \parallel EF \parallel DC ,  The three pairs of similar triangles are

\triangle ABC \sim \triangle CEF

\triangle BCD \sim \triangle BEF

\triangle ABE \sim \triangle CDE

Since \triangle ABE \sim \triangle CDE

\frac{AB}{CD} = \frac{AE}{CE} \Rightarrow CE = \frac{52.5 \times 40.5}{67.5} = 31.5 \ cm

Since \triangle ABC \sim \triangle CEF

\frac{CE}{CA} = \frac{EF}{AB} \Rightarrow EF =  \frac{67.5 \times 31.5}{52.5+31.5} = 25.3125 \ cm

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s7Question 19: In the given figure QR \parallel AB , and DR \parallel QB .  Prove that PQ^2 = PD \times PA .

Answer:

Given QR \parallel AB , and DR \parallel QB

Using basic proportionality theorem

In \triangle PQR \ and \ \triangle PAB

\frac{PQ}{PA} = \frac{PR}{PB}    … … … … (i)

In \triangle PDR \ and \  \triangle PQB

\frac{PD}{PQ} = \frac{PR}{PB}    … … … … (ii)

From (i) and (ii)  we get

\frac{PQ}{PA} = \frac{PD}{PQ}

\Rightarrow PQ^2 = PD \times PA

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Question 20: Through the mid point M of the side CD of a parallelogram ABCD , the line BM is drawn intersecting diagonal AC \ in \ L and AD produced in E . Prove that EL=2BL .s19

Answer:

Consider \triangle DEM \ and \  \triangle CBM

\angle AEB = \angle EBC (alternate angles)

\angle EMD = \angle CMB (opposite angles)

DM = MC (as M is the midpoint of CD)

Therefore \triangle DEM \cong \triangle CBM

Therefore DE = BC (corresponding angles)

AD = BC (opposite sides of a parallelogram)

Therefore AE = AD + DE = 2BC

Now consider \triangle ELA \ and \  \triangle LBC

\angle AEB = \angle EBC

\angle ELA = \angle EBC

Therefore \triangle ELA \sim \triangle LBC

Hence \frac{EL}{BL} = \frac{EA}{BC}

\frac{EL}{BL} = \frac{2BC}{BC}  = 2

Hence EL = 2 BL

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Question 21: sm12In the figure given below,  P is a point on AB such that AP:PB=4:3 . PQ \parallel AC .

(i) Calculate the ratio PQ:AC , giving reasons for your answer.

(ii) In \triangle ARC, \angle ARC = 90^o and in \triangle PQS, \angle PSQ=90^o . Given QS=6 \ cm , calculate length of AR  [1999]

Answer:

(i)   Given AP:PB=4:3

Also PQ \parallel AC , applying basic proportionality theorem

\frac{AP}{PB} = \frac{CQ}{QB}

\Rightarrow \frac{CQ}{QB} = \frac{4}{3}

\Rightarrow \frac{BQ}{BC} = \frac{3}{7}

\angle QBP = \angle ACB (corresponding angles)

\angle QPB = \angle CAB (corresponding angles)

Therefore \triangle PBQ \sim \triangle ABC (AAA postulate)

\frac{PQ}{ AC} = \frac{BQ}{ BC} =  \frac{3}{ 7}

(ii) Given \angle ARC = \angle QSP = 90^o

\angle ACR = \angle SPQ (alternate angles)

Therefore \triangle ARC \sim \triangle QSP (AAA postulate)

\frac{AR}{ SQ} = \frac{AC}{ PQ} = \frac{7}{ 3}

\Rightarrow AR = \frac{7 \times 6}{3} = 14 \ cm

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Question 22: sm131.jpgIn the right angled \triangle QPR, PM  is the altitude. Given that QR= 8 cm and MQ = 3.5 cm , calculate the value of PR . [2000]

Answer:

\angle QPR = \angle PMR = 90^o

\angle PRQ = \angle PRM (common)

\triangle PQR \sim \triangle MPR (AAA postulate)

\frac{QR}{ PR} = \frac{ PR}{ MR}

PR^2 = QR \times MR = 8 \times  4.5 = 36 \Rightarrow PR = 6 \ cm

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Question 23: In the figure given below, median BD \ and \ CE of the \triangle ABC meet at G . Prove that:s8.jpg

(i) \triangle EGD \sim \triangle CGB

(ii) BG = 2 GD from (i)

Answer:

Given BD \ and  \ CE  are medians

Therefore AD = DC and AE = BE 

Applying converse of proportionality theorem ED \parallel BC 

In \triangle EGD and \triangle CGB

\angle DEG = \angle GCB (alternate angles)

\angle EGD = \angle BGC (vertically opposite angles)

Therefore \triangle EGD \sim \triangle CGB (AAA postulate)

Therefore \frac{GD}{GB} = \frac{ED}{BC}

In \triangle EAD and \triangle BAC

\angle AED = \angle ABC (corresponding angles)

\angle EAD = \angle BAC (common angle)

Therefore \triangle EAD \sim \triangle BAC (AAA postulate)

Therefore \frac{ED}{BC} = \frac{AE}{AB} = \frac{1}{2}

\frac{ED}{BC} = \frac{1}{2} 

From 1,  \frac{GD}{GB} = \frac{1}{2} 

GB = 2GD