Question 1: sm1In the given figure, AB and DE are perpendicular to BC . If AB = 9 \ cm, DE=3 \ cm \ and \ AC=24 cm , calculate AD  [2005]

Answer:

In \triangle ABC and \triangle DEC :

\angle ABC = \angle DEC (perpendiculars)

\angle C (common angle)

\triangle ABC \sim \triangle DEC (by AAA postulate)

\Rightarrow \frac{AC}{DC}=\frac{AB}{DE}

\Rightarrow DC = \frac{3}{9} \times 24 = 8 \  cm

Therefore AD = AC - DC = 24-8 = 16 \  cm

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Question 2:  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 (right angles)

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

AM = \sqrt{AP^2-PM^2} = \sqrt{15^2-12^2} = \sqrt{121} = 11

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 3: 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 4: 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 5: 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 6: sm14In the given figure DE \parallel BC .

(i) Prove that \triangle ADE and \triangle ABC are similar.

(ii) Given that AD=\frac{1}{2}BD , calculate DE , if BC=4.5 \ cm . Also find:

\frac{Ar. (\triangle ADE)}{Ar. (\triangle ABC)}

and \frac{Ar. (\triangle ADE)}{Ar. (trapezium \ BCED)} . [2004]

Answer:

(i) DE \parallel BC

\Rightarrow \angle ADE = \angle ABC (corresponding angles)

and \angle AED = \angle ACB (corresponding angles)

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

(ii) Since \triangle ADE \sim \triangle ABC

\frac{DE}{ BC} = \frac{AD}{ AB}

\Rightarrow DE = \frac{1}{ 3} \times 4.5 = 1.5 \ cm

Given AD = \frac{1}{ 2} \times BD

\Rightarrow \frac{AD}{ BD} = {1}{ 2}

\Rightarrow \frac{AD} {AD+BD}= \frac{1} {1+2}

\Rightarrow \frac{AD}{ AB} = \frac{1}{ 3}

(iii) Since \triangle ADE \sim \triangle ABC

\frac{Area(\triangle ADE)}{Area (\triangle ABC)} = \frac{DE^2}{BC^2} = \frac{1.5^2}{4.5^2} = \frac{1}{9}

\Rightarrow \frac{Area (\triangle ADE)}{Area (\triangle ABC)- Area(\triangle ADE)}=\frac{1}{9-1}

\Rightarrow \frac{Area(\triangle ADE)}{Area (trapezium BCED)}=\frac{1}{8}

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Question 7: sm9In the figure given below, PB  and QA  are perpendiculars to the line segment AB . If PO=6 \ cm, QO=9 \ cm  and area of \triangle POB = 120\  cm^2 , find the area of \triangle QOA  [2006]

Answer:

In \triangle POB \ and \  \triangle QOA

\angle PBO = \angle QAO = 90^o

\angle POB = \angle QOA (vertically opposite angles)

Therefore \triangle POB  \sim \triangle QOA  (By AAA postulate)

\Rightarrow \frac{Ar.(\triangle POB)}{Ar.(\triangle QOA)} = \frac{PO^2}{QO^2}

\Rightarrow Ar.(\triangle QOA) = \frac{9^2}{6^2} \times 120 = 270 \ cm^2

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Question 8: sm81In the figure given below, ABCD is a parallelogram. P is a point on BC such that BP:PC = 1:2 . DP produced meets AB produced at Q . Given the area of \triangle CPQ=20 \ cm^2 . Calculate:

(i) area of \triangle CDP

(ii) area of parallelogram ABCD  [1996]

Answer:

(i)   In \triangle BPQ \ and \ \triangle CPD

\angle BPQ = \angle CPD (vertically opposite angles)

\angle BQP = \angle PDC (alternate angles)

\triangle BPQ \sim \triangle CPD

Therefore \frac{BP}{ PC}= \frac{ PQ}{ PD}= \frac{ BQ}{ CD}= \frac{ 1}{ 2}

Also \frac{Area \triangle BPQ}{Area \triangle CPD} = \frac{BP^2}{PC^2}

Area \triangle CPD = \frac{4}{1} \times 10 = 40 \  cm^2

(ii) In \triangle BAP  \ and \ \triangle AQD

BP \parallel AD (Given)

\angle QBP = \angle QAD (corresponding angles are equal)

\angle BQP = \angle AQD (common angle)

\triangle BQP \sim \triangle AQD

Therefore \frac{AQ}{ BQ} =\frac{QD}{ QP} =\frac{AD}{ BP} = 3

Also \frac{Area \triangle AQD}{Area \triangle BQP} = \frac{AQ^2}{BQ^2}

Area \triangle AQD = 3^2 \times 10 = 90 \ cm^2

Area (ABCD) = Area \triangle AQD -Area \triangle BQP + Area \triangle CDP = 90-10+40 = 120 \ cm^2

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Question 9: A model of a ship is made to scale of 1:200 .

(i) The length of the model is 4 \ m ; calculate the length of the ship.

(ii) The area of the deck of the ship is 160000 \ m^2 ; find the area of the deck of the model.

(iii) The volume of the model is 200 \ liters ; calculate the volume of the ship in m^3 .   [1995]

Answer:

Scale factor = \frac{1}{k}

(i) Length of the model = k \times Actual length of the ship

\Rightarrow Actual length of the ship = 4 \times 200 = 800 \ m 

(ii) Area of the deck of the model = k^2 \times  area of the deck of the actual ship

= (\frac{1}{200})^2 \times 160000 \ m^2 = 4 \ m^2

(iii) Volume of the model = k^3 \times  Volume of the actual ship

= (\frac{1}{k})^3 \times 200 = (200)^3 \times 200 = 1600000000 \ liters = 16000000 \ m^3 

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Question 10: sm7In the figure given below ABC is a triangle. DE is parallel to BC and \frac{AD}{DB}=\frac{3}{2} .

(i) Determine the ratios \frac{AD}{AB} \ and \ \frac{DE}{BC}

(ii) Prove that \triangle DEF is similar to \triangle CBF . Hence , find \frac{EF}{FB}  [2007]

Answer:

(i)   Given DE \parallel BC \ and \  \frac{AD}{DB}=\frac{3}{2}

\triangle ADE \ and\  \triangle ABC 

\angle BAC = \angle DAC   (common angle)

\angle ADE = \angle ABC

\triangle ADE \sim \triangle ABC

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

\frac{AD }{AB} = \frac{AD }{AD+DB}  = \frac{3}{ 5}

\frac{AD}{ AE} = \frac{DE }{BC} = \frac{3}{ 5}

(ii) In \triangle DEF and \triangle CBF

\angle FDE  =  \angle FCB (alternate angles)

\angle DFE  =  \angle BFC (vertically opposite angles)

\triangle DEF \sim \triangle CBF

\frac{EF}{ FB} = \frac{DE}{ BC} = \frac{3}{ 5}

\frac{EF}{ FB }= \frac{3 }{5}

(iii) We know

\frac{Area \ of \ \triangle DEF}{Area \ of \  \triangle CBF} = \frac{EF^2}{ FB^2} = \frac{3^2 }{5^2 }= \frac{9 }{25}

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Question 11: sm6In \triangle ABC, \angle ABC = \angle DAC, AB= 8 \ cm, AC = 4 \ cm \ and \ AD = 5 \ cm .

(i) Prove that \triangle ACD is similar to \triangle BCA .

(ii) Find BC \ and \ CD .

(iii) Find area \ of \  \triangle ACD : area \ of \ \triangle ABC .    [2014]

Answer:

(i) In \triangle ACD \ and\  \triangle BCA

\angle C = \angle C (common angle)

\angle ABC = \angle CAD   (given)

\triangle ACD \sim \triangle BCA (AAA postulate)

(ii) Since \triangle ACD \sim \triangle BCA

\frac{AC}{ BC} = \frac{CD}{ CA} = \frac{AD}{ BA}

\frac{4 }{ BC}= \frac{CD}{ 4} = \frac{5}{ 8}

\Rightarrow BC = \frac{8}{ 5} \times 4 = 6.4 \ cm

\Rightarrow CD = \frac{5}{ 8} \times 4 = 2.5 \ cm

(iii) Since \triangle ACD \sim \triangle ABC

Therefore \frac{Area \triangle ACD}{Area \triangle ABC} = \frac{AC^2}{AB^2} = \frac{4^2}{8^2} = \frac{1}{4}

Hence Area \triangle ACD : Area \triangle ABC= 1:4

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Question 12: In the following figure, AB, CD \ and \ EF are parallel lines. AB=6 \ cm, \ CD= y \ cm, \ EF=10 \ cm,  AC = 4 \ cm \ and   \ CF= x \ cm . Calculate: x \ and \ y  [1985]sm5

Answer:

Consider \triangle FDC \ and\  \triangle FBA

\angle FDC = \angle FDA (Corresponding angles)

\angle DFC = \angle BFA (common angle)

\triangle FDC \sim \triangle FBA (AAA Postulate)

Therefore  \frac{CD}{ AB} = \frac{FC}{ FA}

 \frac{y}{ 6} = \frac{x}{ x+4}

Now consider \triangle FCE \ and \  \triangle ACB

\angle FCE = \angle ACB  (Vertically opposite angles)

\angle CFE = \angle CAB (Alternate angles)

\triangle FCE \sim \triangle ACB (AAA postulate)

Therefore  \frac{FC}{ AC} = \frac{EF}{ AB}

x = \frac{10}{ 6} \times 4 = 6.67 \ cm

Also y= \frac{6.67 }{ 6.67+4} \times 6 = 3.75 \ cm

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Question 13: sm4In \triangle PQR, L and M are two points on the base QR , such that \angle LPQ = \angle QRP and \angle RPM = \angle RQP . Prove that:

(i) \triangle PQL \sim \triangle RPM

(ii) QL \times RM = PL \times PM

(iii) PQ^2= QR \times QL .   [2003]

Answer:

(i)   Consider \triangle PQL \ and \  \triangle RMP

\angle LPQ = \angle QRP (Given)

\angle RQP = \angle RPM (Given)

\triangle PQL \sim \triangle RMP (AAA postulate)

(ii) Since \triangle PQL \sim \triangle RMP 

 \frac{PQ}{ RP} =  \frac{QL}{ PM }=  \frac{PL}{ RM}

\Rightarrow QL \times RM = PL \times PM

(iii) Consider  \triangle PQL \ and \  \triangle RQP

\angle LPQ = \angle QRP (Given)

\angle Q (common angle)

\triangle PQL \sim \triangle RQP (AAA postulate)

Therefore  \frac{PQ}{ RQ} =  \frac{QL}{ QP} =  \frac{PL}{ PR}

\Rightarrow PQ^2 = QR \times QL 

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Question 14: sm3In the given figure, ABC is a triangle with \angle EDB = \angle ACB . Prove that \triangle ABC \sim \triangle EBD . If BE=6 \ cm, EC = 4 \ cm, BD = 5 \ cm and area of \triangle BED = 9 cm^2 . Calculate the:

(i) length of AB

(ii) area of \triangle ABC    [2010]

Answer:

Consider \triangle ABC \ and \  \triangle EBD

\angle EDB = \angle ACB (given)

\angle DBE = \angle ABC (common)

Therefore \angle DEB = \angle BAC

\triangle ABC \sim \triangle EBD    (AAA postulate)

(i) Given BE=6\ cm, EC=4\ cm, BD=5\ cm

\frac{AB}{ EB} = \frac{BC}{ BD} = \frac{AC }{ED}

AB = \frac{BE+EC }{5} \times 6 = 2 \ cm

(ii)  \frac{Area \ of \ \triangle ABC}{Area \ of \ \triangle EBD}=\frac{AB^2}{EB^2}=\frac{144}{36}

Area of  \triangle ABC = \frac{144}{ 36} \times 9 = 36 \ cm^2

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Question 15: sm2In the given figure \triangle ABC is a right angled triangle with \angle BAC = 90^o .

(i) Prove \triangle ADB \sim \triangle CDA

(ii) If BD = 18 \ cm \ and \ CD = 8 \ cm , find AD

(iii) Find the ratio of the area of \triangle ADB is to area of \triangle CDA  [2011]

Answer:

(i)   Let \angle DAB = \theta

Therefore \angle DAC = 90^o - \theta

\angle DBA = 90^o - \theta

\angle DCA = \theta

Therefore \triangle ADB \sim \triangle CDA (AAA postulate)

(ii) \frac{CD}{AD}=\frac{AD}{BD}

\Rightarrow AD^2 = CD \times BD = 8 \times 18 = 144

Therefore AD = \sqrt{144} = 12

(iii) \frac{Area \ of \ \triangle ADB}{ Area \ of \ \triangle CDA} = \frac{\frac{1}{2} AD \times BD} {\frac{1}{2} AD \times CD} = \frac{BD}{CD} = \frac{18}{8}= \frac{9}{4}

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Question 16: sm1In the given figure AB  and DE  are perpendiculars to BC .

(i) Prove that \triangle ABC \sim \triangle DEC 

(ii) If AB =  6 \ cm, DE = 4 \ cm  and AC = 15 \ cm , calculate CD 

(iii) Find the ratio of the area \ of  \ \triangle ABC : area \ of \ \triangle DEC .   [2013]

Answer:

(i) From  \triangle ABC \ and \   \triangle DEC

\angle ABC = \angle DEC = 90^o (given)

\angle ACB = \angle DCE (common angle)

 \triangle ABC \sim  \triangle DEC (AAA postulate)

(ii) Since  \triangle ABC \sim \triangle DEC

In \triangle ABC \ and \   \triangle DEC ,

\frac{AB}{ DE} = \frac{AC}{ CD}

AB =  6 \ cm, DE = 4 \ cm  and AC = 15 \ cm

Therefore CD = \frac{15}{6} \times 4 = 10  \ cm

(iii) Since \triangle ABC \sim \triangle DEC

\frac{Area \ of \   \triangle ABC}{Area \ of \  \triangle DEC} = \frac{AB^2}{ DE^2} = \frac{6^2}{4^2}= \frac{9}{4}

Therefore the Area \ of \  \triangle ABC : Area \ of \  \triangle DEC = 9:4

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