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

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