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}  

\displaystyle \\

Question 20: In the given figure \displaystyle AB \text{ and } DE are perpendiculars to \displaystyle BC .

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

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

(iii) Find the ratio of the \displaystyle \text{ Area of } \triangle ABC : \text{ Area of } \triangle DEC . [2013]

Answer:

\displaystyle \text{(i) From } \triangle ABC \text{ and } \triangle DEC  

\displaystyle \angle ABC = \angle DEC = 90^{\circ} \text{ (given) }

\displaystyle \angle ACB = \angle DCE \text{ (common angle) }

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

\displaystyle \text{ (ii) Since }  \triangle ABC \sim \triangle DEC  

\displaystyle \text{In } \triangle ABC \text{ and } \triangle DEC ,

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

\displaystyle AB = 6 cm, DE = 4 \text{ cm } \text{ and } AC = 15 \text{ cm }  

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

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

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

Therefore the \displaystyle \text{ Area of } \triangle ABC : \text{ Area of } \triangle DEC = 9:4