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

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

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


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

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

Answer:

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

(ii)  \frac{Ar.  \ of \ \triangle 1}{Ar.  \ of  \ \triangle 2} = \frac{side1 ^2}{side2^2} 

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

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

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

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

Answer:s36

Given AP = \frac{1}{2} PB

Consider \triangle APQ \ and\  \triangle ABC

\angle APQ = \angle ABC  (alternate angles)

\angle AQP =\angle ACB  (alternate angles)

Therefore \triangle APQ \sim \triangle ABC    (AAA postulate)

Hence \frac{Ar. \triangle ABC}{Ar. \triangle APQ} = \frac{AB^2}{AP^2} = \frac{(AP+PB)^2}{AP^2} = ( \frac{1+\frac{PB}{AP}}{1} )^2 = \frac{16}{1}

Also \frac{Ar. \ \triangle APQ}{Ar. \ trapezium PBCQ} =  \frac{Ar. \ \triangle APQ}{Ar. \ \triangle ABC - Ar. \ \triangle APQ} = \frac{1}{16-1} = \frac{1}{15} 

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

Answer:

Since the two given triangles are similar, we have

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

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

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

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s35Question 4: In the given figure, AX:XB=3:5 . Find:

(i) the length of BC , if the length of XY is 18 \ cm .

(ii) the ratio between the areas of trapezium XBCY and \triangle ABC .

Answer:

Given AX:XB=3:5

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

Consider \triangle ABC \ and\  \triangle AXY

\angle ABC = \angle AXY  (alternate angles)

\angle BAC =\angle XAY  (common angle)

Therefore \triangle ABC \sim \triangle AXY

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

\frac{Ar. \ \triangle AXY}{Ar. \ \triangle ABC} = \frac{AX^2}{AB^2} = \frac{9}{64} 

Also \frac{Ar. \ \triangle ABC - Ar. \ \triangle AXY}{Ar. \ \triangle ABC} = 1- \frac{9}{64} = \frac{55}{64} 

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

Answer:s36.jpg

Given Ar. \ \triangle APQ = \frac{1}{2} Ar.  \ \triangle ABC

Consider \triangle APQ \ and\  \triangle ABC

\angle APB = \angle ABC  (corresponding angles)

\angle BAC =\angle PAQ  (common angle)

Therefore \triangle APQ \sim \triangle ABC

Therefore \frac{Ar. \ \triangle APQ}{Ar. \ \triangle ABC} = \frac{AP^2}{AB^2}   

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

(1+ \frac{PB}{AB} )^2 = 2 

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

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s34Question 6: In the given \triangle PQR, LM \parallel QR and PM:MR=3:4 . Calculate the value of the ratio: 

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

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

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

Answer:

(i) Given \triangle PQR, LM \parallel QR and PM:MR=3:4 .

\Rightarrow PM:PR=3:7

Consider \triangle PLM \ and\  \triangle PQR

\angle PLM = \angle PQR  (alternate angles)

\angle LPM =\angle QPR  (common angle)

Therefore \triangle PLM \sim \triangle PQR

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

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

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

\frac{Ar. \ \triangle LMN}{Ar. \ \triangle MNR} = \frac{LN}{NR}  

Consider \triangle LMN \ and\  \triangle QRN

\angle NLM = \angle NRQ  (alternate angles)

\angle LMN =\angle NQR  (common angle)

Therefore \triangle LMN \sim \triangle QRN

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

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

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

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

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s33Question 7: The given diagram shows two isosceles triangles which are similar. PQ \ and \ BC are not parallel. PC=4, AQ=3, QB-=12, BC=15 and AP=PQ . Calculate:

(i) the length of AP

(ii) the ratio of the areas of \triangle APQ \ and \  \triangle ABC

Answer:

Given AB = AC = 15

AP = PQ = x

Since \triangle APQ \sim \triangle ABC

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

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

Solving x^2+4x-45 = 0 \Rightarrow x = 5 \ cm

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

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Question 8:  In the figure given below, ABCD is a parallelogram. P is a point on BC such that BP:PC = 1:2 . DP produced meets AB produced at Q . Given the area of \triangle CPQ=20 \ cm^2 . Calculate:sm81

(i) area of \triangle CDP

(ii) area of parallelogram ABCD  [1996]

Answer:

(i)   In \triangle BPQ \ and \ \triangle CPD

\angle BPQ = \angle CPD (vertically opposite angles)

\angle BQP = \angle PDC (alternate angles)

\triangle BPQ \sim \triangle CPD

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

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

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

(ii) In \triangle BAP  \ and \ \triangle AQD

BP \parallel AD (Given)

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

\angle BQP = \angle AQD (common angle)

\triangle BQP \sim \triangle AQD

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

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

Area \ \triangle AQD = 3^2 \times 10 = 90 \ cm^2

Area (ABCD) = Area \ \triangle AQD -Area \ \triangle BQP + Area \ \triangle CDP 

= 90-10+40 = 120 \ cm^2

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s32Question 9: In the given figure, BC \parallel DE . Area of \triangle ABC = 25 \ cm^2 , area of trapezium  BCED = 24 \ cm^2 and DE = 14 \ cm . Calculate the length of BC . Also find the area of \triangle BCD .

Answer:

Given  BC \parallel DE

Consider \triangle ABC \ and\  \triangle ADE

\angle ABC = \angle ADE  (alternate angles)

\angle ACB =\angle AED  (common angle)

Therefore \triangle ABC \sim \triangle ADE

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

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

\frac{25}{25+24} = \frac{BC^2}{14^2} 

\Rightarrow BC = 10 \ cm 

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

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

\Rightarrow height = \frac{24 \times 2}{24} = 2 \ cm 

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

= \frac{1}{2} \times 10 \times 2 = 10 \ cm^2 

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s31Question 10: The given figure shows a trapezium in which AB \parallel DC and diagonals AC \ and \  BD intersect at point P . If AP:CP=3:5 . Find:

(i) \triangle APB : \triangle CPB

(ii) \triangle DPC : \triangle APB

(iii)  \triangle ADP : \triangle APB

(iv) \triangle APB : \triangle ADB

Answer:

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

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

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

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

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

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

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

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

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

(i) the actual length of QR \ and \ PR in kilometers

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

Answer:

Scale factor k =\frac{1}{2500000}

(i) Length of side PQ   in the map = k \times  the actual length of the side PQ in the land

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

\Rightarrow Actual \ length \ of \ PQ = 3 \times 2500000 = 7500000 \ m = 7.5 \ km

Length of side QR   in the map = k \times the actual length of the side QR in the land

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

\Rightarrow \ Actual \ length \ of \ QR = 4 \times 2500000 = 10000000 \ m = 10 \ km

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

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

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

(i) The length of the model is 4 \ m ; calculate the length of the ship.

(ii) The area of the deck of the ship is 160000 \ m^2 ; find the area of the deck of the model.

(iii) The volume of the model is 200 \ liters ; calculate the volume of the ship in m^3  [1995]

Answer:

Scale factor = \frac{1}{k}

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

\Rightarrow Actual length of the ship = 4 \times 200 = 800 \ m 

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

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

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

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

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sm7Question 13: In the figure given below ABC is a triangle. DE is parallel to BC and \frac{AD}{DB}=\frac{3}{2} .

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

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

Answer:

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

\triangle ADE \ and\  \triangle ABC 

\angle BAC = \angle DAC   (common angle)

\angle ADE = \angle ABC

\triangle ADE \sim \triangle ABC

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

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

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

(ii) In \triangle DEF and \triangle CBF

\angle FDE  =  \angle FCB (alternate angles)

\angle DFE  =  \angle BFC (vertically opposite angles)

\triangle DEF \sim \triangle CBF

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

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

(iii) We know

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

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s42Question 14: In the given figure \angle B = \angle E , \angle ACD = \angle BCE . AB = 10.4 \ cm \ and \ DE=7.8 \ cm . Find the ratio between the area of the \triangle ABC \ and \  \triangle DEC .

Answer:

Consider \triangle ABC and \triangle CDE

\angle CBA = \angle CED (given)

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

Therefore \triangle ABC \sim \triangle CDE

We know that for similar triangles

\frac{Ar. \ \triangle ABC}{Ar. \  \triangle DEC} = \frac{AB^2}{DE^2} = \frac{10.4^2}{7.8^2} = \frac{16}{9}    = 1.778

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s41Question 15: \triangle ABC  is an isosceles triangle in which AB=AC=13 \ cm and BC=10 \ cm. AD \perp BC . If CE = 8 \ cm \ and \ EF \perp AB , find:

(i) \frac{Ar. \ \triangle ADC}{Ar. \  \triangle FEB} 

(ii)  \frac{Ar. \ \triangle FEB}{Ar. \  \triangle ABC} 

Answer:

(i) Consider \triangle ADC and \triangle FEB

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

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

Therefore \triangle ADC \sim \triangle FEB

We know that for similar triangles

\frac{Ar. \ \triangle ADC}{Ar. \  \triangle FEB} = \frac{AC^2}{BE^2} = \frac{13^2}{18^2} = \frac{169}{324}

(ii)  Now Consider \triangle ABD and \triangle ACD

\angle ABC = \angle ACB  (given)

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

Therefore \triangle ABD \cong \triangle ACD

Therefore \frac{Ar. \ \triangle FEB}{Ar. \  \triangle ABC} = \frac{Ar. \ \triangle FEB}{2 \times Ar. \  \triangle ADC} =  = \frac{324}{2 \times 169} = \frac{162}{169}

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

Answer:

15  \ cm of the model represent 30 \ m of actual airplane

Therefore 1 \ cm wold represent 2 \ m of actual airplane

Hence 1 \ cm^2 would represent 4 \ m^2 of the surface are of the actual airplane

Given that the surface area of the model = 150 \ cm^2 .

Therefore the actual surface are of the actual airplane = 150 \times 4 = 600 \ m^2

Area to be painted = 600-50 = 550 \ m^2

Cost of painting = 120 Rs. /m^2

Therefore the total cost of painting = 120 \times 550 = 66000 \ Rs.