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.

Answer:

\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}  

Answer:

\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.

Answer:

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 .

Answer:

\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 .

Answer:

\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}  

Answer:

\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  

Answer:

\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]

Answer:

(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 .

Answer:

\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

Answer:

(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  

Answer:

\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]

Answer:

\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]  

Answer:

\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 .

Answer:

\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}  

Answer:

\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.

Answer:

\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. }