Similarity - Maps and Models

The dimensions of a model of multi-storied building are $1 \; m$ by $60 \; cm$ by $1.20 \; m$. If the scale factor is $1 : 50$, find:

1. the floor area of a room of the building in $m^2$, if the floor area of the corresponding room in the model is $50 \; cm^2$;
2. the space inside a room of the model in $cm^3$, if the space inside the corresponding room of the building is $90 \; m^3$.

Scale factor $= k = 1 : 50 = \dfrac{1}{50}$

1. Floor area of room in the model $= 50 \; cm^2 = 50 \times 10^{-4} \; m^2$
   
   Floor area of room in model $= k^2 \times$ Floor area of room of building
   
   i.e. $\;$ $50 \times 10^{-4} = \left(\dfrac{1}{50}\right)^2 \times $ Floor area of room of building
   
   i.e. $\;$ Floor area of room of building $= 50 \times 10^{-4} \times 2500 = 7.5 \; m^2$



2. Volume of a room of the model $= k^3 \times $ Volume of a room of the building
   
   $\begin{aligned} i.e. \; \text{Volume of a room of the model} & = \left(\dfrac{1}{50}\right)^3 \times 90 \\\\ & = \dfrac{90}{125 \times 10^3} \\\\ & = 0.72 \times 10^{-3} \; m^3 \\\\ & = 0.72 \times 10^{-3} \times 10^6 \; cm^3 \\\\ & = 720 \; cm^3 \end{aligned}$

Similarity - Maps and Models

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

1. The length of the model is $4 \; m$. Calculate the length of the ship.
2. The area of the deck of the ship is $160000 \; m^2$. Find the area of the deck of the model.
3. The volume of the model is $200$ liters. Calculate the volume of the ship in meter cube.

Scale of model ship is $\;$ $1 : 200$

i.e. $\;$ scale factor $= k = \dfrac{1}{200}$

1. Length of $1 \; m$ in the model $= k \times$ the actual length of ship
   
   $\implies$ $4 \; m = \dfrac{1}{200} \times$ the actual length of ship
   
   $\implies$ the actual length of ship $= 200 \times 4 = 800 \; m$



2. Area of $1 \; m^2$ in the model $= k^2 \times$ the actual area of ship
   
   $\implies$ Area of the deck of the model $= \left(\dfrac{1}{200}\right)^2 \times 160000$
   
   $\implies$ Area of the deck of the model $= 4 \; m^2$



3. $1$ liter $= 0.001 \; m^3$
   
   $\therefore \;$ $200$ liter $= 200 \times 0.001 = 0.2 \; m^3$
   
   i.e. $\;$ Volume of model $= 0.2 \; m^3$
   
   Volume of $1 \; m^3$ in the model $= k^3 \times$ the actual volume of ship
   
   $\implies$ $0.2 \; m^3 = \left(\dfrac{1}{200}\right)^3 \times$ actual volume of ship
   
   $\implies$ actual volume of ship $= 0.2 \times 8000000 = 1600000 \; m^3$

Similarity - Maps and Models

A map is drawn to a scale of $1 : 20000$

1. The distance between the two towns on the map is $9 \; cm $. Calculate the actual distance between the towns in km.
2. Calculate the actual area in $m^2$ which represents $8 \; cm^2 $ on the map.

1. Scale of map is $\;$ $1 : 20000$
   
   i.e. $\;$ $1 \; cm$ on the map corresponds to an actual distance of $\;$ $20000 \; cm$
   
   $\therefore \;$ $9 \; cm$ on the map corresponds to an actual distance of $\;$ $\dfrac{9 \times 20000}{1} = 180000 = 1.8 \times 10^5 \; cm$
   
   Now, $\;$ $10^5 \; cm = 1 \; km$
   
   $\therefore \;$ $1.8 \times 10^5 \; cm = 1.8 \; km$
   
   $\therefore \;$ Actual distance between the towns $= 1.8 \; km$



2. Scale of map is $\;$ $1 : 20000$
   
   i.e. $\;$ $1 \; cm$ on the map corresponds to an actual distance of $\;$ $20000 = 2 \times 10^4 \; cm$
   
   i.e. $\;$ $1 \; cm^2$ on the map corresponds to an actual area of $\;$ $\left(2 \times 10^4\right)^2 = 4 \times 10^8 \; cm^2$
   
   $\therefore \;$ $8 \; cm^2$ on the map corresponds to an actual area of $\;$ $\dfrac{8 \times 4 \times 10^8}{1} = 3.2 \times 10^9 \; cm^2$
   
   Now, $\;$ $10^4 \; cm^2 = 1 \; m^2$
   
   $\therefore \;$ $3.2 \times 10^9 \; cm^2 = \dfrac{3.2 \times 10^9 \times 1}{10^4} = 3.2 \times 10^5 \; m^2$
   
   $\therefore \; $ Actual area $= 3.2 \times 10^5 \; m^2$

Similarity - Maps and Models

A map is drawn to a scale of $1:4000$.

1. Find the length on map (in cm) that represents $5 \; km$ of ground length.
2. Find the area on ground (in sq.m) that is represented by $4 \; cm^2$ on this map.

Scale of map $= 1 : 4000$

1. i.e. $\;$ $1 \; cm$ on map $\equiv$ $4000 \; cm$ on ground
   
   i.e. $\;$ $1 \; cm$ on map $\equiv$ $4000 \times 10^{-5} \; km$ on ground $\;\;$ $\left[\because \; 1 \; cm = 10^{-5} \; km\right]$
   
   i.e. $\;$ $1 \; cm$ on map $\equiv$ $4 \times 10^{-2} \; km$ on ground
   
   $\therefore \;$ $x \; cm$ on map $\equiv$ $5 \; km$ on ground
   
   $\implies$ $x = \dfrac{5}{4 \times 10^{-2}} \; cm = 125 \; cm$
   
   $\therefore \;$ $125 \; cm$ on map $\equiv$ $5 \; km$ on ground


2. $1 \; cm$ on map $\equiv$ $4000 = 4 \times 10^3 \; cm$ on ground
   
   $\therefore \;$ $1 \; cm^2$ on map $\equiv$ $16 \times 10^6 \; cm^2$ on ground
   
   $\therefore \;$ $4 \; cm^2$ on map $\equiv$ $p \; cm^2$ on ground
   
   i.e. $\;$ $p = 4 \times 16 \times 10^6 = 64 \times 10^6 \; cm^2$ on ground
   
   Now, $100 \; cm = 1 \; m$
   
   i.e. $\;$ $10^4 \; cm^2 = 1 \; m^2$
   
   $\therefore \;$ $64 \times 10^6 \; cm^2 = \dfrac{64 \times 10^6}{10^4} = 6400 \; m^2$
   
   $\therefore \;$ $4$ sq.cm on map $\equiv$ $6400$ sq.m on ground

Similarity - Maps and Models

A map of a square plot of land is drawn to a scale of $1 : 2500$. If the area of the plot on the map is $72 \; cm^2$, find

1. the actual area of the land in square kilometer;
2. the length of the diagonal in the actual plot of land in meters.

1. Scale of square plot of land $= 1 : 2500$
   
   

   On Map	Actual
   $1 \; cm$	$2500 \; cm$
   $1 \; cm^2$	$625 \times 10^4 \; cm^2$
   $72 \; cm^2$	$x \; cm^2$

   
   
   $\therefore \;$ $x = 72 \times 625 \times 10^4 = 45000 \times 10^4 = 45 \times 10^7 \; cm^2$
   
   i.e. $\;$ Actual area of the land $= 45 \times 10^7 \; cm^2$
   
   Now, $\;$ $1000 \; m \equiv 1 \; km$
   
   $\therefore \;$ $10^6 \; m^2 \equiv 1 \; km^2$
   
   $\left[\text{Note: } 1 \; m \equiv 100 \; cm \implies 1 \; m^2 = 10^4 \; cm^2 \right]$
   
   i.e. $\;$ $10^6 \times 10^4 \; cm^2 \equiv 1 \; km^2$
   
   i.e. $\;$ $10^{10} \; cm^2 \equiv 1 \; km^2$
   
   $\therefore \;$ $45 \times 10^7 \; cm^2 \equiv \dfrac{45 \times 10^7}{10^{10}} = 45 \times 10^{-3} = 0.045 \; km^2$
   
   $\therefore \;$ Actual area of the land in square kilometer $= 0.045 \; km^2$



2. Let the length of the side of the plot on the map be $= s \; cm$
   
   Then, area of the square plot on the map $= s^2 \; cm^2$
   
   Given: Area of the plot on the map $= 72 \; cm^2$
   
   i.e. $\;$ $s^2 = 72$ $\implies$ $s = \sqrt{72} \; cm$
   
   $\therefore \;$ Length of the diagonal of the square plot on the map
   
   $= \ell = \sqrt{\left(\sqrt{72}\right)^2 + \left(\sqrt{72}\right)^2} = \sqrt{72 + 72} = \sqrt{144} = 12 \; cm$
   
   

   On Map	Actual
   $1 \; cm$	$2500 \; cm$
   $12 \; cm$	$x \; cm$

   
   
   $\therefore \;$ $x = 12 \times 2500 = 30000 = 3 \times 10^4 \; cm$
   
   i.e. $\;$ length of the diagonal in the actual plot of land $= 3 \times 10^4 \; cm$
   
   Now, $\;$ $100 \; cm \equiv 1 \; m$
   
   $\therefore \;$ $3 \times 10^4 \; cm \equiv \dfrac{3 \times 10^4}{10^2} = 3 \times 10^2 = 300 \; m$
   
   $\therefore \;$ length of the diagonal (in meter) in the actual plot of land $= 300 \; m$