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 $km^2$

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]

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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Question 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]

(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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Question 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 $\triangle ABC \ and \ \triangle DEC$.

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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Question 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)  $\frac{Ar. \ \triangle FEB}{Ar. \ \triangle ABC}$

(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.

$15 \ cm$ 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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