Question 1: In $\triangle ABC, \angle ABC = \angle DAC, AB= 8 \ cm,$ $AC = 4 \ cm \ and \ AD = 5 \ cm$.

(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]

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

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

(i) $EF=FB$

(ii) $AG : GD = 2:1$

(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:

$\frac{AM}{PN}$ $=$ $\frac{AX}{PY}$

(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.

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.

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

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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Question 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 $XY$.

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]

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 $km^2$

$k =$ $\frac{1}{20000}$

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 $90 \ m^3$

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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Question 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]

(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: $\frac{BC^2}{AC^2}$ $=$ $\frac{BD}{AD}$.

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 $\triangle DEF$.

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.

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) $area \ of \ \triangle APO : area \ of \ \triangle CQO$

(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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Question 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) $CD \times AB = CA \times DE$

(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: In 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]

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: In 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]

(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: In 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]

(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 $Area \ of \ \triangle ABC : Area \ of \ \triangle DEC = 9:4$