Note: Relation between the areas of two similar triangles: If $\displaystyle \triangle ABC \sim \triangle DEF$ then

$\displaystyle \frac{\text{ Area of } \triangle ABC}{\text{ Area of } \triangle DEF} = \frac{AB^2}{DE^2} =\frac{BC^2}{EF^2} =\frac{AC^2}{DF^2}$

$\displaystyle \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DE} = \frac{\text{ Perimeter } \triangle ABC}{\text{ Perimeter } \triangle DEF}$

Question 1: (i) The ratio between the corresponding sides of two similar triangles is $\displaystyle 2 :5 .$ Find the ratios between the areas of these triangles.

(ii) Areas of two similar triangles is $\displaystyle 98 \text{ cm}^2 \text{ and } 128 \text{ cm}^2 .$ Find the ratios between the length of their corresponding sides.

$\displaystyle \text{(i) Required ratio of their areas } = \frac{2^2}{5^2} = \frac{4}{25}$

$\displaystyle \text{(ii) } \frac{ \text{ Area of } \triangle 1}{ \text{ Area of } \triangle 2} = \frac{\text{ side1}^2}{\text{ side2}^2}$

$\displaystyle \text{Therefore Required ratio } = \sqrt{\frac{98}{128}} = \frac{7}{8}$

$\displaystyle \$

Question 2: A line $\displaystyle PQ$ is drawn parallel to the base $\displaystyle BC \text{ of } \triangle ABC$ which meets sides $\displaystyle AB\text{ and }AC$ at points $\displaystyle P\text{ and }Q$ respectively. If $\displaystyle AP = \frac{1}{2} PB$ ; find the value of

$\displaystyle \text{(i) } \frac{ \text{ Area of } \triangle ABC}{ \text{ Area of } \triangle APQ}$

$\displaystyle \text{(ii) } \frac{ \text{ Area of } \triangle APQ}{ \text{ Area of trapezium } PBCQ}$

$\displaystyle \text{Given } AP = \frac{1}{2} PB$

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

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

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

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

$\displaystyle \text{Hence } \frac{\text{ Area} \triangle ABC}{\text{ Area} \triangle APQ} = \frac{AB^2}{AP^2} = \frac{(AP+PB)^2}{AP^2} = ( \frac{1+\frac{PB}{AP}}{1} )^2 = \frac{16}{1}$

$\displaystyle \text{Also } \frac{\text{ Area} \triangle APQ}{\text{ Area trapezium } PBCQ} \\ \\ = \frac{\text{ Area} \triangle APQ}{\text{ Area} \triangle ABC - \text{ Area} \triangle APQ} = \frac{1}{16-1} = \frac{1}{15}$

$\displaystyle \$

Question 3: The perimeter of two similar triangles are $\displaystyle 30 \text{ cm } \text{ and } 24 \text{ cm}.$ If one side of the first triangle is $\displaystyle 12 \text{ cm }$ , determine the corresponding side of the second triangle.

Since the two given triangles are similar, we have

$\displaystyle \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DE} =\frac{\text{ Perimeter } \triangle ABC}{\text{ Perimeter } \triangle DEF}$

$\displaystyle \frac{12}{DE} = \frac{30}{24}$

$\displaystyle \Rightarrow DE = \frac{12 \times 24}{30} = 9.6 \text{ cm }$

$\displaystyle \$

Question 4: In the given figure, $\displaystyle AX:XB=3:5 .$ Find:

(i) the length of $\displaystyle BC$ , if the length of $\displaystyle XY$ is $\displaystyle 18 \text{ cm}.$

(ii) the ratio between the areas of trapezium $\displaystyle XBCY \text{ and } \triangle ABC .$

$\displaystyle \text{Given } AX:XB=3:5$

$\displaystyle \Rightarrow \frac{AX}{AB} = \frac{3}{8}$

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

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

$\displaystyle \angle BAC =\angle XAY \text{ (common angle) }$

$\displaystyle \text{Therefore } \triangle ABC \sim \triangle AXY$

$\displaystyle \text{Therefore } \frac{AX}{AB} = \frac{XY}{BC} \Rightarrow BC = \frac{8 \times 18}{3} = 48 \text{ cm }$

$\displaystyle \frac{\text{ Area} \triangle AXY}{\text{ Area} \triangle ABC} = \frac{AX^2}{AB^2} = \frac{9}{64}$

$\displaystyle \text{Also } \frac{\text{ Area} \triangle ABC - \text{ Area} \triangle AXY}{\text{ Area} \triangle ABC} = 1- \frac{9}{64} = \frac{55}{64}$

$\displaystyle \$

Question 5: $\displaystyle ABC$ is a triangle. $\displaystyle PQ$ is a line segment intersecting $\displaystyle AB \text{ in } P \text{ and } AC \text{ in } Q$ such that $\displaystyle PQ \parallel BC$ and divides $\displaystyle \triangle ABC$ into two parts equal in area. Find the value of ratio $\displaystyle BP:AB .$

$\displaystyle \text{Given } \text{ Area} \triangle APQ = \frac{1}{2} \text{ Area} \triangle ABC$

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

$\displaystyle \angle APB = \angle ABC$ (corresponding angles)

$\displaystyle \angle BAC =\angle PAQ \text{ (common angle) }$

$\displaystyle \text{Therefore } \triangle APQ \sim \triangle ABC$

$\displaystyle \text{Therefore } \frac{\text{ Area} \triangle APQ}{\text{ Area} \triangle ABC} = \frac{AP^2}{AB^2}$

$\displaystyle \frac{AP^2}{AB^2} = \frac{1}{2}$

$\displaystyle (1+ \frac{PB}{AB} )^2 = 2$

$\displaystyle \Rightarrow \frac{PB}{AB} = (\sqrt{2}-1)$

$\displaystyle \$

Question 6: In the $\displaystyle \text{Given } \triangle PQR, LM \parallel QR \text{ and } PM:MR=3:4 .$ Calculate the value of the ratio:

$\displaystyle \text{(i) } \frac{PL}{PQ} \text{ and then } \frac{LM}{QR}$

$\displaystyle \text{(ii) } \frac{ \text{ Area of } \triangle LMN}{ \text{ Area of } \triangle MNR}$

$\displaystyle \text{(iii) } \frac{ \text{ Area of } \triangle LQM}{ \text{ Area of } \triangle LQN}$

$\displaystyle \text{(i) } \text{Given } \triangle PQR, LM \parallel QR \text{ and } PM:MR=3:4 .$

$\displaystyle \Rightarrow PM:PR=3:7$

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

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

$\displaystyle \angle LPM =\angle QPR \text{ (common angle) }$

$\displaystyle \text{Therefore } \triangle PLM \sim \triangle PQR$

$\displaystyle \text{Therefore } \frac{LM}{QR} = \frac{PL}{PQ} = \frac{PM}{PR}$

$\displaystyle \Rightarrow \frac{PL}{PQ} = \frac{3}{7}$

(ii) Since $\displaystyle \triangle LMN\text{ and }\triangle MNR$ have common vertex $\displaystyle L$ and their bases $\displaystyle LN\text{ and }LR$ are along the same straight line

$\displaystyle \frac{\text{ Area} \triangle LMN}{\text{ Area} \triangle MNR} = \frac{LN}{NR}$

$\displaystyle \text{Consider } \triangle LMN\text{ and }\triangle QRN$

$\displaystyle \angle NLM = \angle NRQ \text{ (alternate angles) }$

$\displaystyle \angle LMN =\angle NQR \text{ (common angle) }$

$\displaystyle \text{Therefore } \triangle LMN \sim \triangle QRN$

$\displaystyle \text{Therefore } \frac{LM}{NR} = \frac{LM}{QR} = \frac{MN}{QN} = \frac{3}{7}$

$\displaystyle \frac{ \text{ Area of } \triangle LMN}{ \text{ Area of } \triangle MNR} = \frac{3}{7}$

(iii) Since $\displaystyle \triangle LQN\text{ and }\triangle LQM$ have common vertex $\displaystyle L$ and their bases $\displaystyle MN\text{ and }MQ$ are along the same straight line

$\displaystyle \text{Therefore } \frac{ \text{ Area of } \triangle LQM}{ \text{ Area of } \triangle LQN} = \frac{10}{7}$

$\displaystyle \$

Question 7: The given diagram shows two isosceles triangles which are similar. $\displaystyle PQ\text{ and }BC$ are not parallel. $\displaystyle PC=4, AQ=3, QB-=12, BC=15 \text{ and } AP=PQ .$ Calculate:

(i) the length of $\displaystyle AP$

(ii) the ratio of the areas of $\displaystyle \triangle APQ\text{ and }\triangle ABC$

$\displaystyle \text{Given } AB = AC = 15$

$\displaystyle AP = PQ = x$

Since $\displaystyle \triangle APQ \sim \triangle ABC$

$\displaystyle \frac{AP}{AB} = \frac{PQ}{BC} = \frac{AQ}{AC}$

$\displaystyle \frac{AP}{15} = \frac{x}{15} = \frac{3}{x+4}$

Solving $\displaystyle x^2+4x-45 = 0 \Rightarrow x = 5 \text{ cm }$

$\displaystyle \text{(ii) } \frac{ \text{ Area of } \triangle APQ}{ \text{ Area of } \triangle ABC} = \frac{AQ^2}{AC^2} = \frac{3^2}{9^2} = \frac{1}{9}$

$\displaystyle \$

Question 8: In the figure given below, $\displaystyle ABCD$ is a parallelogram. $\displaystyle P$ is a point on $\displaystyle BC$ such that $\displaystyle BP:PC = 1:2 .$ $\displaystyle DP$ produced meets $\displaystyle AB$ produced at $\displaystyle Q .$ Given the area of $\displaystyle \triangle CPQ=20 \text{ cm}^2 .$ Calculate:

(i) area of $\displaystyle \triangle CDP$

(ii) area of parallelogram $\displaystyle ABCD .$ [1996]

(i) In $\displaystyle \triangle BPQ\text{ and }\triangle CPD$

$\displaystyle \angle BPQ = \angle CPD \text{ (vertically opposite angles) }$

$\displaystyle \angle BQP = \angle PDC \text{ (alternate angles) }$

$\displaystyle \triangle BPQ \sim \triangle CPD$

$\displaystyle \text{Therefore } \frac{BP}{ PC} = \frac{ PQ}{ PD} = \frac{ BQ}{ CD} = \frac{ 1}{ 2}$

$\displaystyle \text{Also } \frac{ \text{ Area } \triangle BPQ}{\text{ Area } \triangle CPD} = \frac{BP^2}{PC^2}$

$\displaystyle \text{ Area } \triangle CPD = \frac{4}{1} \times 10 = 40 \text{ cm}^2$

(ii) In $\displaystyle \triangle BAP\text{ and }\triangle AQD$

$\displaystyle BP \parallel AD \text{ (Given) }$

$\displaystyle \angle QBP = \angle QAD$ (corresponding angles are equal)

$\displaystyle \angle BQP = \angle AQD \text{ (common angle) }$

$\displaystyle \triangle BQP \sim \triangle AQD$

$\displaystyle \text{Therefore } \frac{AQ}{ BQ} = \frac{QD}{ QP} = \frac{AD}{ BP} = 3$

$\displaystyle \text{Also } \frac{\text{ Area } \triangle AQD}{\text{ Area }\triangle BQP} = \frac{AQ^2}{BQ^2}$

$\displaystyle \text{ Area } \triangle AQD = 3^2 \times 10 = 90 \text{ cm}^2$

$\displaystyle \text{ Area } (ABCD) = \text{ Area } \triangle AQD -\text{ Area } \triangle BQP + \text{ Area } \triangle CDP$

$\displaystyle = 90-10+40 = 120 \text{ cm}^2$

$\displaystyle \$

Question 9: In the given figure, $\displaystyle BC \parallel DE .$ Area of $\displaystyle \triangle ABC = 25 \text{ cm}^2$ , area of trapezium $\displaystyle BCED = 24 \text{ cm}^2 \text{ and } DE = 14 \text{ cm}.$ Calculate the length of $\displaystyle BC .$ Also find the area of $\displaystyle \triangle BCD .$

$\displaystyle \text{Given } BC \parallel DE$

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

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

$\displaystyle \angle ACB =\angle AED \text{ (common angle) }$

$\displaystyle \text{Therefore } \triangle ABC \sim \triangle ADE$

$\displaystyle \frac{\text{ Area } \triangle ABC}{\text{ Area } \triangle ADE} \frac{BC^2}{DE^2}$

$\displaystyle \frac{\text{ Area } \triangle ABC}{ \text{ Area of } \triangle ABC - \text{ Area of trapezium } BCDE} = \frac{BC^2}{DE^2}$

$\displaystyle \frac{25}{25+24} = \frac{BC^2}{14^2}$

$\displaystyle \Rightarrow BC = 10 \text{ cm }$

$\displaystyle \text{Now Area of trapezium } BCED = \frac{1}{2} ( \text{ sum of parallel sides }) \times \text{ height }$

$\displaystyle \Rightarrow 24 = \frac{1}{2} (10+14) \times \text{ height }$

$\displaystyle \Rightarrow \text{ height } = \frac{24 \times 2}{24} = 2 \text{ cm }$

$\displaystyle \text{Area of } \triangle BCD = \frac{1}{2} \times \text{ base } \times \text{ height }$

$\displaystyle = \frac{1}{2} \times 10 \times 2 = 10 \text{ cm}^2$

$\displaystyle \$

Question 10: The given figure shows a trapezium in which $\displaystyle AB \parallel DC$ and diagonals $\displaystyle AC\text{ and }BD$ intersect at point $\displaystyle P .$ If $\displaystyle AP:CP=3:5 .$ Find:

$\displaystyle \text{(i) } \triangle APB : \triangle CPB \hspace{1.0cm} \text{(ii) } \triangle DPC : \triangle APB \\ \\ \text{(iii) } \triangle ADP : \triangle APB \hspace{1.0cm} \text{(iv) } \triangle APB : \triangle ADB$

(i) Since $\displaystyle \triangle APB\text{ and }\triangle CPB$ have common vertex $\displaystyle B$ and their bases $\displaystyle AP\text{ and }PC$ are along the same straight line

$\displaystyle \text{Therefore } \frac{ \text{ Area of } \triangle APB}{ \text{ Area of } \triangle CPB} = \frac{AP}{PC} = \frac{3}{5}$

(ii) Since $\displaystyle \triangle DPC \sim \triangle BAP$

$\displaystyle \text{Therefore } \frac{ \text{ Area of } \triangle DPC}{ \text{ Area of } \triangle BPA} = \frac{PC^2}{AP^2} = \frac{25}{9}$

(iii) Since $\displaystyle \triangle ADP\text{ and }\triangle APB$ have common vertex $\displaystyle A$ and their bases $\displaystyle DP\text{ and }PB$ are along the same straight line

$\displaystyle \text{Therefore } \frac{ \text{ Area of } \triangle ADP}{ \text{ Area of } \triangle APB} = \frac{DP}{PB} = \frac{5}{3}$

(iv) Since $\displaystyle \triangle APB\text{ and }\triangle ADB$ have common vertex $\displaystyle A$ and their bases $\displaystyle BP\text{ and }BD$ are along the same straight line

$\displaystyle \text{Therefore } \frac{ \text{ Area of } \triangle APB}{ \text{ Area of } \triangle ADB} = \frac{PB}{BD} = \frac{3}{8}$

$\displaystyle \$

Question 11: On a map drawn to a scale of $\displaystyle 1:2500000$ a triangular plot of $\displaystyle PQR$ of land has the following measurements: $\displaystyle PQ = 3 \text{ cm } , QR=4 \text{ cm }\text{ and }\angle PQR=90^o .$ Calculate:

(i) the actual length of $\displaystyle QR\text{ and }PR$ in kilometers

(ii) the actual area of the plot in $\displaystyle km^2$

$\displaystyle \text{Scale factor } k =\frac{1}{2500000}$

$\displaystyle \text{(i) Length of side } PQ \text{ in the map } = k \times \text{ the actual length of the side } PQ \text{ in the land }$

$\displaystyle \Rightarrow 3 \text{ cm } = \frac{1}{2500000} \times \text{ actual length of }PQ$

$\displaystyle \Rightarrow \text{ Actual length of } PQ = 3 \times 2500000 = 7500000 m = 7.5 km$

$\displaystyle \text{Length of side } QR \text{ in the map } = k \times \text{ the actual length of the side } QR \text{ in the land }$

$\displaystyle \Rightarrow 4 \text{ cm } = \frac{1}{2500000} \times \text{ actual length of } QR$

$\displaystyle \Rightarrow \text{Actual length of } QR = 4 \times 2500000 = 10000000 \text{ m } = 10 \text{ km }$

$\displaystyle \text{Hence } PR = \sqrt{7.5^2+10^2} = \sqrt{156.5} = 12.5 \text{ km }$

$\displaystyle \text{(ii) Area of the plot } = \frac{1}{2} \times {base} \times {height} = \frac{1}{2} \times 7.5 \times 10 = 37.5 km^2$

$\displaystyle \$

Question 12: A model of a ship is made to scale of $\displaystyle 1:200 .$

(i) The length of the model is $\displaystyle 4 \text{ m}$; calculate the length of the ship.

(ii) The area of the deck of the ship is $\displaystyle 160000 \text{ m}^2$; find the area of the deck of the model.

(iii) The volume of the model is $\displaystyle 200 \text{ liters }$ ; calculate the volume of the ship in $\displaystyle \text{ m}^3 .$ [1995]

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

(i) Length of the model $\displaystyle = k \times$ Actual length of the ship

$\displaystyle \Rightarrow$ Actual length of the ship $\displaystyle = 4 \times 200 = 800 \text{ m}$

(ii) Area of the deck of the model $\displaystyle = k^2 \times$ area of the deck of the actual ship

$\displaystyle = (\frac{1}{200})^2 \times 160000 m^2 = 4 \text{ m}^2$

(iii) Volume of the model $\displaystyle = k^3 \times$ Volume of the actual ship

$\displaystyle = (\frac{1}{k})^3 \times 200 = (200)^3 \times 200 = 1600000000 \text{ liters } = 16000000 \text{ m}^3$

$\displaystyle \$

Question 13: In the figure given below $\displaystyle ABC$ is a triangle. $\displaystyle DE$ is parallel to $\displaystyle BC \text{ and } \frac{AD}{DB}=\frac{3}{2} .$

$\displaystyle \text{(i) Determine the ratios } \frac{AD}{AB}\text{ and }\frac{DE}{BC}$

$\displaystyle \text{(ii) Prove that } \triangle DEF \text{ is similar to } \triangle CBF . \text{ Hence , find } \frac{EF}{FB} . \hspace{1.0cm} [2007]$

$\displaystyle \text{(i) } \text{Given } DE \parallel BC\text{ and }\frac{AD}{DB}=\frac{3}{2}$

$\displaystyle \triangle ADE\text{ and }\triangle ABC$

$\displaystyle \angle BAC = \angle DAC \text{ (common angle) }$

$\displaystyle \angle ADE = \angle ABC$

$\displaystyle \triangle ADE \sim \triangle ABC$

$\displaystyle \frac{AD}{ AB} = \frac{AE }{AC} = \frac{DE}{ BC}$

$\displaystyle \frac{AD }{AB} = \frac{AD }{AD+DB} = \frac{3}{ 5}$

$\displaystyle \frac{AD}{ AE} = \frac{DE }{BC} = \frac{3}{ 5}$

(ii) In $\displaystyle \triangle DEF\text{ and }\triangle CBF$

$\displaystyle \angle FDE = \angle FCB \text{ (alternate angles) }$

$\displaystyle \angle DFE = \angle BFC \text{ (vertically opposite angles) }$

$\displaystyle \triangle DEF \sim \triangle CBF$

$\displaystyle \frac{EF}{ FB} = \frac{DE}{ BC} = \frac{3}{ 5}$

$\displaystyle \frac{EF}{ FB }= \frac{3 }{5}$

(iii) We know

$\displaystyle \frac{ \text{ Area of } \triangle DEF}{ \text{ Area of } \triangle CBF} = \frac{EF^2}{ FB^2} = \frac{3^2 }{5^2 }= \frac{9 }{25}$

$\displaystyle \$

Question 14: In the given figure $\displaystyle \angle B = \angle E$ , $\displaystyle \angle ACD = \angle BCE .$ $\displaystyle AB = 10.4 \text{ cm }\text{ and }DE=7.8 \text{ cm}.$ Find the ratio between the area of the $\displaystyle \triangle ABC\text{ and }\triangle DEC .$

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

$\displaystyle \angle CBA = \angle CED (given)$

$\displaystyle \angle ECD = \angle BCA$ (Since $\displaystyle \angle BCD$ is common and $\displaystyle \angle ACD = \angle BCE$ is given)

$\displaystyle \text{Therefore } \triangle ABC \sim \triangle CDE$

We know that for similar triangles

$\displaystyle \frac{\text{ Area} \triangle ABC}{\text{ Area} \triangle DEC} = \frac{AB^2}{DE^2} = \frac{10.4^2}{7.8^2} = \frac{16}{9} = 1.778$

$\displaystyle \$

Question 15: $\displaystyle \triangle ABC$ is an isosceles triangle in which $\displaystyle AB=AC=13 \text{ cm } \text{ and } BC=10 cm. AD \perp BC .$ If $\displaystyle CE = 8 \text{ cm }\text{ and }EF \perp AB$ , find:

$\displaystyle \text{(i) } \frac{\text{ Area} \triangle ADC}{\text{ Area} \triangle FEB}$

$\displaystyle \text{(ii) } \frac{\text{ Area} \triangle FEB}{\text{ Area} \triangle ABC}$

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

$\displaystyle \angle BFE = \angle ADC = 90^o (given)$

$\displaystyle \angle ACD = \angle FBE$ (Since $\displaystyle AB=AC \Rightarrow$ angles opposite equal sides are also equal)

$\displaystyle \text{Therefore } \triangle ADC \sim \triangle FEB$

We know that for similar triangles

$\displaystyle \frac{\text{ Area} \triangle ADC}{\text{ Area} \triangle FEB} = \frac{AC^2}{BE^2} = \frac{13^2}{18^2} = \frac{169}{324}$

(ii) Now $\displaystyle \text{Consider } \triangle ABD\text{ and }\triangle ACD$

$\displaystyle \angle ABC = \angle ACB (given)$

$\displaystyle \angle ADB = \angle ADC = 90^o$ (given)

$\displaystyle \text{Therefore } \triangle ABD \cong \triangle ACD$

$\displaystyle \text{Therefore } \frac{\text{ Area} \triangle FEB}{\text{ Area} \triangle ABC} = \frac{\text{ Area} \triangle FEB}{2 \times \text{ Area} \triangle ADC} = = \frac{324}{2 \times 169} = \frac{162}{169}$

$\displaystyle \$

Question 16: An airplane is $\displaystyle 30 \text{ m}$ long and its model is $\displaystyle 15 \text{ cm }$ long. If the total outer surface area of the model is $\displaystyle 150 \text{ cm}^2$ , find the cost of painting the outer surface of the airplane at the rate of $\displaystyle Rs. 120 /\text{ m}^2 .$ Given that $\displaystyle 50 \text{ m}^2$ of the surface of the airplane sis left for windows.

$\displaystyle 15 \text{ cm }$ of the model represent $\displaystyle 30 \text{ m}$ of actual airplane

$\displaystyle \text{Therefore } 1 \text{ cm }$ wold represent $\displaystyle 2 \text{ m}$ of actual airplane

$\displaystyle \text{Hence } 1 \text{ cm}^2$ would represent $\displaystyle 4 \text{ m}^2$ of the surface are of the actual airplane

Given that the surface area of the model $\displaystyle = 150 \text{ cm}^2 .$

Therefore the actual surface are of the actual airplane $\displaystyle = 150 \times 4 = 600 \text{ m}^2$

Area to be painted $\displaystyle = 600-50 = 550 \text{ m}^2$

Cost of painting $\displaystyle = 120 \text{ Rs. } /\text{ m}^2$

Therefore the total cost of painting $\displaystyle = 120 \times 550 = 66000 \text{ Rs. }$