Class 10: Similarity – Exercise 16(c) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

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

(ii) Find $\displaystyle BC \text{ and } CD .$

(iii) Find $\displaystyle \text{ Area of } \triangle ACD : \text{ Area of } \triangle ABC .$ [2014]

Answer:

$\displaystyle \text{(i) } \text{In } \triangle ACD \text{ and } \triangle BCA$

$\displaystyle \angle C = \angle C \text{ (common angle) }$

$\displaystyle \angle ABC = \angle CAD \text{ (given) }$

$\displaystyle \triangle ACD \sim \triangle BCA \text{ (AAA postulate) }$

$\displaystyle \text{ (ii) Since } \triangle ACD \sim \triangle BCA$

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

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

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

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

$\displaystyle \text{ (iii) Since } \triangle ACD \sim \triangle ABC$

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

$\displaystyle \text{Hence } \text{ Area } \triangle ACD : \text{ Area } \triangle ABC= 1:4$

$\displaystyle \\$

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

Answer:

$\displaystyle \text{In } \triangle ABC, PR \parallel BC$

$\displaystyle \text{Consider } \triangle ABC \text{ and } \triangle PAR$

$\displaystyle \angle ABC = \angle APR \text{ (alternate angles) }$

$\displaystyle \angle ARP =\angle ACB$ alternate)

$\displaystyle \text{Therefore } \triangle ABC \sim \triangle ARP \text{ (AAA postulate) }$

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

Since $\displaystyle P$ is the mid point of $\displaystyle AB$

$\displaystyle AB = 2 AP$

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

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

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

$\displaystyle \text{Therefore } \triangle ABC \sim \triangle PQR \text{ (SSS postulate) }$

$\displaystyle \\$

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

$\displaystyle \text{(i) } EF=FB$

$\displaystyle \text{(ii) } AG : GD = 2:1$

Answer:

$\displaystyle \text{(i) } \text{Consider } \triangle BDF \text{ and } \triangle BCE$

$\displaystyle DF \parallel CE$

$\displaystyle \angle BDF = \angle BCE \text{ (alternate angles) }$

$\displaystyle \angle BFD = \angle BEC \text{ (alternate angles) }$

$\displaystyle \text{Therefore } \triangle BDF \sim \triangle BCE \text{ (AAA postulate) }$

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

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

$\displaystyle \Rightarrow 2 BF = BE$

$\displaystyle \Rightarrow BF = FE$

$\displaystyle \text{(ii) } \text{Consider } \triangle AFD \text{ and } \triangle AEG$

$\displaystyle FD \parallel EG$

$\displaystyle \angle DFA = \angle GEA \text{ (alternate angles) }$

$\displaystyle \angle FDA =\angle EGA \text{ (alternate angles) }$

$\displaystyle \text{Therefore } \triangle AFD \sim \triangle AEG \text{ (AAA postulate) }$

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

$\displaystyle \text{Given } AE = EB$

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

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

$\displaystyle \\$

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

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

Answer:

(Since $\displaystyle \triangle ABC \sim \triangle PQR$

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

$\displaystyle \text{Given } AX \text{ and } PY$ are medians

$\displaystyle \text{Therefore } 2BX = BC \text{ and } 2 YR = QR$

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

$\displaystyle \text{Consider } \triangle ABM \text{ and } \triangle PQN$

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

$\displaystyle \angle AMB = \angle PNQ = 90^{\circ} \text{ (alternate angles) }$

$\displaystyle \text{Therefore } \triangle ABM \sim \triangle PQN \text{ (AAA postulate) }$

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

From (i) and (ii) we get

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

and $\displaystyle \angle ABX = \angle PQY$

$\displaystyle \text{Therefore } \triangle ABX \sim \triangle PQY$

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

$\displaystyle \text{Hence } \frac{AM}{PN} = \frac{AX}{PY}$

$\displaystyle \\$

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

Answer:

Let the two triangles be $\displaystyle ABC \text{ and } PQR$

Since the two triangles are similar, We know

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

Since the are of the two triangles is equal

$\displaystyle \Rightarrow AB=PQ, BC=QR \text{ and } AC = PR$

$\displaystyle \text{Therefore } \triangle ABC \cong \triangle PQR$

$\displaystyle \\$

Question 6: The ratio between the altitudes of two similar triangles is $\displaystyle 3:5 .$ 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 $\displaystyle = 3:5$

(i) Ratio between their median $\displaystyle = 3:5$

(ii) Ratio between their perimeter $\displaystyle = 3:5$

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

$\displaystyle \\$

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

Answer:

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

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

(i) Ratio between their perimeter $\displaystyle = 4:5$

(ii) Ratio between their altitude $\displaystyle = 4:5$

(iii) Ratio between their median $\displaystyle = 4:5$

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Given } PX:XQ=1:3 \text{ and } QR =9 \text{ cm }$

$\displaystyle \text{Consider } \triangle PXY \text{ and } \triangle PQR$

$\displaystyle \angle PXY = \angle PQR \text{ (alternate angles) }$

$\displaystyle \angle PYX =\angle PRQ \text{ (alternate angles) }$

$\displaystyle \text{Therefore } \triangle PXY \sim \triangle PQR \text{ (AAA postulate) }$

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

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

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

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

$\displaystyle \Rightarrow XY = \frac{9}{P4}$

$\displaystyle \\$

Question 9: In the following figure, $\displaystyle AB, CD \text{ and } EF$ are parallel lines. $\displaystyle AB=6 \text{ cm }, CD= y \text{ cm }, EF=10 \text{ cm }, AC = 4 \text{ cm } \text{ and } CF= x \text{ cm}.$ Calculate: $\displaystyle x \text{ and } y .$ [1985]

Answer:

$\displaystyle \text{Consider } \triangle FDC \text{ and } \triangle FBA$

$\displaystyle \angle FDC = \angle FDA \text{ (Corresponding angles) }$

$\displaystyle \angle DFC = \angle BFA \text{ (common angle) }$

$\displaystyle \triangle FDC \sim \triangle FBA \text{ (AAA postulate) }$

$\displaystyle \text{Therefore } \frac{CD}{ AB} = \frac{FC}{ FA}$

$\displaystyle \frac{y}{ 6} = \frac{x}{ x+4}$

Now $\displaystyle \text{Consider } \triangle FCE \text{ and } \triangle ACB$

$\displaystyle \angle FCE = \angle ACB \text{ (Vertically opposite angles) }$

$\displaystyle \angle CFE = \angle CAB \text{ (alternate angles) }$

$\displaystyle \triangle FCE \sim \triangle ACB \text{ (AAA postulate) }$

$\displaystyle \text{Therefore } \frac{FC}{ AC} = \frac{EF}{ AB}$

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

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

$\displaystyle \\$

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

(i) the diagonal distance of the plot in km

(ii) the area of the plot $\displaystyle \text{In } \text{ km }^2$

Answer:

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

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

Actual length of $\displaystyle AB = 24 \times 20000 \text{ cm } = 4.8 \text{ km }$

Similarly Actual length of $\displaystyle BC = 32 \times 20000 \text{ cm } = 6.4 \text{ km }$

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

(ii) Area of the plot $\displaystyle = 4.8 \times 6.4 = 30.72 \text{ km }$

$\displaystyle \\$

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

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

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

Answer:

Dimension of model $\displaystyle = 100 \text{ cm } \times 60 \text{ cm } \times 120 \text{ cm }$

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

$\displaystyle \text{Actual length } = 100 \times 50 = 50\text{ m }$

$\displaystyle \text{Actual breadth } = 60 \times 50 = 30\text{ m }$

$\displaystyle \text{Actual height } = 120 \times 50 = 60\text{ m }$

$\displaystyle \text{Therefore the actual dimension of the building } = 50\text{ m } \times 30\text{ m } \times 60\text{ m }$

$\displaystyle \text{(i) Floor area } = 50^2 \times 50 \text{ cm }^2 = 125000 \text{ cm }^2 = 12.5 \text{ m }^2$

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

$\displaystyle \\$

Question 12: $\displaystyle \text{In } \triangle PQR, L \text{ and } M$ are two points on the base $\displaystyle QR$ , such that $\displaystyle \angle LPQ = \angle QRP \text{ and } \angle RPM = \angle RQP .$ Prove that:

$\displaystyle \text{(i) } \triangle PQL \sim \triangle RPM$

$\displaystyle \text{(ii) } QL \times RM = PL \times PM$

$\displaystyle \text{(iii) } PQ^2= QR \times QL .$ [2003]

Answer:

$\displaystyle \text{(i) } \text{Consider } \triangle PQL \text{ and } \triangle RMP$

$\displaystyle \angle LPQ = \angle QRP \text{ (given) }$

$\displaystyle \angle RQP = \angle RPM \text{ (given) }$

$\displaystyle \triangle PQL \sim \triangle RMP \text{ (AAA postulate) }$

$\displaystyle \text{ (ii) Since } \triangle PQL \sim \triangle RMP$

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

$\displaystyle \Rightarrow QL \times RM = PL \times PM$

$\displaystyle \text{(iii) } \text{Consider } \triangle PQL \text{ and } \triangle RQP$

$\displaystyle \angle LPQ = \angle QRP \text{ (given) }$

$\displaystyle \angle Q \text{ (common angle) }$

$\displaystyle \triangle PQL \sim \triangle RQP \text{ (AAA postulate) }$

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

$\displaystyle \Rightarrow PQ^2 = QR \times QL$

$\displaystyle \\$

Question 13: $\displaystyle \text{In } \triangle ABC, \angle ACB = 90^{\circ} \text{ and } CD \perp AB .$

$latex \displaystyle \text{Prove that: } \frac{BC^2}{AC^2} = \frac{BD}{AD} . $

Answer:

$\displaystyle \text{Consider } \triangle ACD \text{ and } \triangle ABC$

$\displaystyle \angle DAC = \angle BAC$ (Common)

$\displaystyle \angle CDA = \angle ACB = 90^{\circ} \text{ (given) }$

$\displaystyle \triangle ACD \sim \triangle ABC \text{ (AAA postulate) }$

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

$\displaystyle AC^2=AD \times AB .$ .. … … … (ii)

$\displaystyle \text{Consider } \triangle BCD \text{ and } \triangle ABC$

$\displaystyle \angle CBD = \angle CBA$ (Common)

$\displaystyle \angle BDC = \angle ACB = 90^{\circ} \text{ (given) }$

$\displaystyle \triangle BCD \sim \triangle ABC \text{ (AAA postulate) }$

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

$\displaystyle BC^2=BD \times AB .$ .. … … … (ii)

From (i) and (ii)

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

Hence Proved.

$\displaystyle \\$

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

Answer:

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

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Consider } \triangle ABC \text{ and } \triangle DEF$

$\displaystyle \angle BAC = \angle EDF$ (Common)

$\displaystyle AB=AC \text{ and } DE = DF \text{ (given) }$

$\displaystyle \triangle ABC \sim \triangle EDF \text{ (SAS postulate) }$

$\displaystyle \frac{\text{ Area } \triangle ABC}{\text{ Area } \triangle DEF} = \frac{AD^2}{PS^2}$

$\displaystyle \frac{16}{25} = ( \frac{AD}{PS} )^2$

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

$\displaystyle \\$

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

$\displaystyle \text{(i) } \text{ Area of } \triangle APO : \text{ Area of } \triangle ABC$

$\displaystyle \text{(ii) } \text{ Area of } \triangle APO : \text{ Area of } \triangle CQO$

Answer:

$\displaystyle \text{(i) } \text{Consider } \triangle APO \text{ and } \triangle ABC$

$\displaystyle \angle APO = \angle ABC \text{ (alternate angles) }$

$\displaystyle \angle AOP = \angle ACB \text{ (alternate angles) }$

$\displaystyle \triangle APO \sim \triangle ABC \text{ (AAA postulate) }$

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

$\displaystyle \text{or } \frac{AP}{AB} = \frac{2}{5}$

$\displaystyle \frac{ \text{ Area of } \triangle APO}{ \text{ Area of } \triangle ABC} = \frac{2^2}{5^2} = \frac{4}{25}$

$\displaystyle \text{(ii) } \text{Consider } \triangle APO \text{ and } \triangle QOC$

$\displaystyle \angle AOP = \angle QOC \text{ (Vertically opposite angles) }$

$\displaystyle \angle PAO = \angle OCQ \text{ (alternate angles) }$

$\displaystyle \triangle APO \sim \triangle QOC \text{ (AAA postulate) }$

$\displaystyle \frac{ \text{ Area of } \triangle APO}{ \text{ Area of } \triangle QOC} = \frac{AP^2}{CQ^2} = \frac{AP^2}{PB^2} = \frac{2^2}{3^2} = \frac{4}{9}$

$\displaystyle \\$

Question 17:The following figure shows a $\displaystyle \triangle ABC$ in which $\displaystyle AD \perp BC \text{ and } BE \perp AC .$ Show that:

$\displaystyle \text{(i) } \triangle ADC \sim \triangle BEC \hspace{1.0cm} \text{(ii) } CA \times CE = CB \times CD \hspace{1.0cm} \\ \\ \text{(iii) } \triangle ABC \sim \triangle DEC \hspace{1.0cm} \text{(iv) } CD \times AB = CA \times DE$

Answer:

$\displaystyle \text{(i) } \text{Consider } \triangle ADC \text{ and } \triangle BEC$

$\displaystyle \angle ADC = \angle BEC = 90^{\circ} \text{ (alternate angles) }$

$\displaystyle \angle ADC = \angle BCE \text{ (common angle) }$

$\displaystyle \triangle ADC \sim \triangle BEC \text{ (AAA postulate) }$

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

$\displaystyle CA \times CE = CB \times CD$

$\displaystyle \text{(iii) } \text{Consider } \triangle ABC \text{ and } \triangle DEC$

$\displaystyle \angle ADC = \angle BCE \text{ (common angle) }$

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

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

$\displaystyle \triangle ABC \sim \triangle DEC \text{ (SAS postulate) }$

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

$\displaystyle AC \times DE = CD \times AB$

$\displaystyle \\$

Question 18: In the given figure, $\displaystyle ABC$ is a triangle with $\displaystyle \angle EDB = \angle ACB .$ Prove that $\displaystyle \triangle ABC \sim \triangle EBD .$ If $\displaystyle BE=6 \text{ cm }, EC = 4 \text{ cm }, BD = 5 \text{ cm }$ and area of $\displaystyle \triangle BED = 9 \text{ cm }^2 .$ Calculate the:

(i) length of $\displaystyle AB$ (ii) area of $\displaystyle \triangle ABC$ [2010]

Answer:

$\displaystyle \text{Consider } \triangle ABC \text{ and } \triangle EBD$

$\displaystyle \angle EDB = \angle ACB \text{ (given) }$

$\displaystyle \angle DBE = \angle ABC$ (common)

$\displaystyle \text{Therefore } \angle DEB = \angle BAC$

$\displaystyle \triangle ABC \sim \triangle EBD \text{ (AAA postulate) }$

$\displaystyle \text{(i) } \text{Given } BE=6 cm, EC=4 cm, BD=5 \text{ cm }$

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

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

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

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

$\displaystyle \\$

Question 19: In the given figure $\displaystyle \triangle ABC$ is a right angled triangle with $\displaystyle \angle BAC = 90^{\circ} .$

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

(ii) If $\displaystyle BD = 18 \text{ cm } \text{ and } CD = 8 \text{ cm }$ , find $\displaystyle AD$

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

Answer:

$\displaystyle \text{(i) Let } \angle DAB = \theta$

$\displaystyle \text{Therefore } \angle DAC = 90^{\circ} - \theta$

$\displaystyle \angle DBA = 90^{\circ} - \theta$

$\displaystyle \angle DCA = \theta$

$\displaystyle \text{Therefore } \triangle ADB \sim \triangle CDA \text{ (AAA postulate) }$

$\displaystyle \text{(ii) } \frac{CD}{AD} = \frac{AD}{BD}$

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

$\displaystyle \text{Therefore } AD = \sqrt{144} = 12$

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