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. 

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