Question 1: 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

Answer:

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s54Question 2: In the given triangle P, Q \ and \  R are mid points of sides AB, BC \ and \  AC respectively. Prove that \triangle PQR \sim \triangle ABC .

Answer:

In \triangle ABC, PR \parallel BC

Consider \triangle ABC \ and\  \triangle PAR

\angle ABC = \angle APR  (alternate angles)

\angle ARP =\angle ACB  alternate)

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

\frac{PR}{BC} = \frac{AP}{AB}

Since P is the mid point of AB

AB = 2 AP

\frac{PR}{BC} = \frac{1}{2}

Similarly,  we can prove \frac{PQ}{AC} = \frac{1}{2} and \frac{RQ}{AB} = \frac{1}{2}

Therefore \frac{PR}{BC} = \frac{PQ}{AC} = \frac{RQ}{AB}

Therefore \triangle ABC \sim \triangle PQR  (SSS postulate)

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s53.jpgQuestion 3: In the following figure AD \ and \  CE are medians of \triangle ABC. DF \parallel CE . Prove that :

(i) EF=FB

(ii) AG : GD = 2:1

Answer:

(i) Consider \triangle BDF \ and\  \triangle BCE

DF \parallel CE

\angle BDF = \angle BCE  (alternate angles)

\angle BFD = \angle BEC (alternate angles)

Therefore \triangle BDF \sim \triangle BCE (AAA postulate)

\frac{BD}{BC} = \frac{BF}{BE}

\frac{BD}{2BD} = \frac{BF}{BE}

\Rightarrow 2 BF = BE

\Rightarrow BF = FE

(ii)  Consider \triangle AFD \ and \  \triangle AEG

FD \parallel EG

\angle DFA = \angle GEA  (alternate angles)

\angle FDA =\angle EGA (alternate angles)

Therefore \triangle AFD \sim \triangle AEG (AAA postulate)

\frac{AG}{GD} = \frac{AE}{FE} By basic proportionality theorem

Given AE = EB

AE = 2 FE \Rightarrow \frac{AE}{FE} = 2

\frac{AG}{GD} = \frac{2}{1}

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Question 4: In the given figure \triangle ABC \sim  \triangle PQR . AM \ and \ PN are altitudes  where as AX \ and \  PY are medians. Prove:

\frac{AM}{PN} = \frac{AX}{PY} 

s52

Answer:

(Since \triangle ABC \sim \triangle PQR

\frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR}

Given AX \ and \  PY are medians

Therefore 2BX = BC and 2 YR = QR

\frac{AB}{PQ} = \frac{BC}{QR} = \frac{2BX}{2QY} =  \frac{BX}{QY}    … … … … (i)

Consider \triangle ABM \ and\  \triangle PQN

\angle ABN = \angle PQN    Since \triangle ABC \sim \triangle PQR

\angle AMB = \angle PNQ = 90^o (alternate angles)

Therefore \triangle ABM \sim \triangle PQN (AAA postulate)

\frac{AB}{PQ} = \frac{AM}{PN} … … … … (ii)

From (i) and (ii) we get

\frac{AB}{PQ} = \frac{AM}{PN}

and \angle ABX = \angle PQY

Therefore \triangle ABX \sim \triangle PQY

\Rightarrow \frac{AB}{PQ} = \frac{AX}{PY}

Hence \frac{AM}{PN} = \frac{AX}{PY} 

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Question 5: Two similar triangles are equal in area. Prove that the triangles are congruent.

Answer:

Let the two triangles be ABC \ and \  PQR

Since the two triangles are similar, We know

\frac{Ar. \ \triangle ABC}{Ar. \ \triangle PQR} = \frac{AB^2}{PQ^2} = \frac{BC^2}{Qr^2} = \frac{AC^2}{PR^2} 

Since the are of the two triangles is equal

\Rightarrow AB=PQ, BC=QR \ and \ AC = PR 

Therefore \triangle ABC \cong \triangle PQR 

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Question 6: The ratio between the altitudes of two similar triangles is 3:5 . Write the ratios between their (i) medians (ii) perimeters (iii) areas.

Answer:

The ratio of the altitude of two similar triangles  is the same as the ratio of their sides. Given ratio = 3:5

(i) Ratio between their median = 3:5

(ii) Ratio between their perimeter = 3:5

(iii) Ratio between their areas = 3^2:5^2 = 9:25

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Question 7: The ratio between the altitudes of two similar triangles is 16:25 . Find the ration between their: (i) perimeters (ii) altitudes (iii) medians

Answer:

The ratio between the altitudes of two similar triangles is 16:25 .

This means that the ratio of the sides of the triangles = 4:5

(i)  Ratio between their perimeter = 4:5

(ii) Ratio between their altitude = 4:5

(iii) Ratio between their median = 4:5

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s51Question 8: The following figure shows a \triangle PQR in which XY \parallel QR . If PX:XQ=1:3 and QR =9 \ cm , find the length of XY .

Answer:

Given PX:XQ=1:3 and QR =9 \ cm

Consider \triangle PXY \ and \  \triangle PQR

\angle PXY = \angle PQR  (alternate angles)

\angle PYX =\angle PRQ (alternate angles)

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

\frac{PX}{PQ} = \frac{XY}{QR}

\frac{PX}{PX+XQ} = \frac{XY}{QR}

\frac{PX/XQ}{PX/XQ+1} = \frac{XY}{QR}

\frac{1}{4} = \frac{XY}{9}

\Rightarrow XY = \frac{9}{P4} 

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Question 9: 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 10: On a map, drawn to a scale of 1:20000 , a rectangular plot of land ABCD has  AB = 24 \ cm and BC = 32 \ cm . Calculate:

(i) the diagonal distance  of the plot in km

(ii) the area of the plot in km^2

Answer:

k = \frac{1}{20000}

Length of AB on map = k \times actual length of AB

Actual length of  AB = 24 \times 20000 \ cm = 4.8 \ km

Similarly Actual length of BC = 32 \times 20000 \ cm = 6.4 \ km

(i) Therefore the diagonal = \sqrt{4.8^2+6.4^2} = 8 \ km

(ii) Area of the plot = 4.8 \times 6.4 = 30.72 \ km

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Question 11: The dimension of a model of a multi storied building are 1  \ m by 60 \ by 1.20 \ m .  If the scale factor is 1:50 , find the actual dimensions of the building. Also find:

(i) the floor area of a room of the building, if the floor are of the corresponding room in the model is 50 \ cm^2

(ii) the space inside the room of them model if the space inside the corresponding room if the building is 90 \ m^3

Answer:

Dimension of model = 100 cm \times 60 \ cm \times 120 \ cm

Scale factor (k) = \frac{1}{50}

Actual length = 100 \times 50 = 50 m

Actual breadth = 60 \times 50 = 30 m

Actual height = 120 \times 50 = 60 m

Therefore the actual dimension of the building = 50 \ m \times 30 \ m \times 60 \ m

(i) Floor area = 50^2 \times 50 cm^2 = 125000 \ cm^2 = 12.5 \ m^2

(ii) Volume of the model = \frac{90 \ m^3}{50^3} = 720 \ cm^3

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sm4Question 12: In \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 13: In \triangle ABC, \angle ACB = 90^o and CD \perp AB . Prove that: \frac{BC^2}{AC^2} = \frac{BD}{AD} .s62.jpg

Answer:

Consider \triangle ACD \ and \  \triangle ABC

\angle DAC = \angle BAC (Common)

\angle CDA = \angle ACB = 90^o (Given)

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

\frac{AC}{AB} = \frac{AD}{AC}

AC^2=AD \times AB   … … … … (ii)

Consider \triangle BCD \ and \  \triangle ABC

\angle CBD = \angle CBA (Common)

\angle BDC = \angle ACB = 90^o (Given)

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

\frac{BC}{AB} = \frac{BD}{BC}

BC^2=BD \times AB … … … … (ii)

From (i) and (ii)

\frac{BC^2}{AC^2} = \frac{BD \times AB}{AD \times AB} 

Hence Proved.

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Question 14: A \triangle ABC with AB=3 \ cm, BC = 6 \ cm \ and \ AC = 4 \ cm is enlarged to a \triangle DEF such that the longest side of \triangle DEF = 9 \ cm . Find the scale factor and hence, the lengths of the other sides of \triangle DEF .

Answer:

Scale factor (k) = \frac{EF}{BC} = \frac{9}{6} = 1.5 

Therefore \frac{ED}{AB} = 1.5 \Rightarrow ED = 3 \times 1.5 = 4.5 \ cm 

\frac{DF}{AC} = 1.5 \Rightarrow ED = 4 \times 1.5 = 6 \ cm 

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Question 15: Two isosceles triangles have equal vertical angles. Show that the triangles are similar.  If the ratios between the areas of these two triangles is 16:25 , find the ratio between their corresponding altitudes.

Answer:

Consider \triangle ABC \ and \  \triangle DEF

\angle BAC = \angle EDF (Common)

AB=AC \ and \ DE = DF (Given)

\triangle ABC \sim \triangle EDF (SAS postulate)

\frac{Ar. \ \triangle ABC}{Ar. \ \triangle DEF} = \frac{AD^2}{PS^2}

\frac{16}{25} = ( \frac{AD}{PS} )^2 

\Rightarrow \frac{AD}{PS}= \frac{4}{5} 

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Question 16: In \triangle ABC, AP:PB=2:3. PQ \parallel BC and is extended to Q so that CQ \parallel BA .  Find:

(i) area \ of \ \triangle APO : area \ of \  \triangle ABC

(ii) area \ of  \ \triangle APO : area \ of \  \triangle CQO

Answer:

(i) Consider \triangle APO \ and \  \triangle ABC

\angle APO = \angle ABC (alternate angles)

\angle AOP = \angle ACB  (alternate angles)

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

\frac{AP}{PB} = \frac{2}{3}

or \frac{AP}{AB} = \frac{2}{5}

\frac{area \ of \ \triangle APO}{area \ of \  \triangle ABC} = \frac{2^2}{5^2} = \frac{4}{25}

(ii) Consider \triangle APO \ and \  \triangle QOC

\angle AOP = \angle QOC (vertically opposite angles)

\angle PAO = \angle OCQ (alternate angles)

\triangle APO \sim \triangle QOC (AAA postulate)

\frac{area \ of \ \triangle APO}{area \ of \  \triangle QOC} = \frac{AP^2}{CQ^2} = \frac{AP^2}{PB^2} = \frac{2^2}{3^2} = \frac{4}{9}

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s61.jpgQuestion 17:The following figure shows a \triangle ABC in which AD \perp BC and BE \perp AC . Show that:

(i) \triangle ADC \sim \triangle BEC

(ii) CA \times CE = CB \times CD

(iii) \triangle ABC \sim \triangle DEC

(iv) CD \times AB = CA \times DE

Answer:

(i) Consider \triangle ADC \ and \  \triangle BEC

\angle ADC = \angle BEC = 90^o (alternate angles)

\angle ADC = \angle BCE  (common angle)

\triangle ADC \sim \triangle BEC (AAA postulate)

(ii) Therefore \frac{CA}{CB} = \frac{CD}{CE}

CA \times CE = CB \times CD

(iii) Consider \triangle ABC \ and \  \triangle DEC

\angle ADC = \angle BCE  (common angle)

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

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

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

(iv) Therefore \frac{AC}{DC} = \frac{AB}{DE}

AC \times DE = CD \times AB

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Question 18: 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 19: 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 20: 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