## Scale Drawings

4.1  Scale Drawings

 Definition The drawing of an object with all measurements in the drawing proportional to the measurements of the object.

Interpret the scale of a scale drawing:

\begin{aligned}&\space\text{Scale}\\\\&=\dfrac{\text{Measurement of scale drawing}}{\text{Measurement of object}}\end{aligned}

The ratio is,

• Measurement of scale drawing : Measurement of object

Scale drawings in the form of ratio is,

$$1:n$$, where $$n$$ is the positive integer or fraction.

$$1:n$$ means one unit on the scale drawing will represent $$n$$ units on the object.

• If $$n \lt 1$$, then the size of the scale drawing is bigger than the size of the object.

• If $$n \gt 1$$, then the size of the scale drawing is smaller than the size of the object.

• If $$n = 1$$, then the size of the scale drawing is the same as the size of the object.

 Example The diagram below shows object $$PQRS$$ and scale drawing $$P'Q'R'S'$$ drawn on a grid of equal squares. State the scale used in the form of $$1:n$$. $$\text{Scale}=\dfrac{P'Q'}{PQ}=\dfrac{2}{4}=\dfrac{1}{2}$$ or, $$\text{Scale}= \dfrac{P'S'}{PS}=\dfrac{3}{6}=\dfrac{1}{2}$$. Thus, $$\text{scale}=1:2$$.

 Example A map is drawn to a scale of $$1 : 400 \space000$$. Calculate the actual length, in $$\text{km}$$, of a river that is $$4\text { cm}$$ long on the map. \begin{aligned} \dfrac {1 \text{ cm}}{400 \space 000 \text{ cm}}& = \dfrac{4 \text{ cm}}{ \text{Actual distance}} \end{aligned} \begin{aligned} \\&\space\text{Actual distance}\\\\&=\dfrac{4 \times 400 \space000 \text{ cm}}{1\text{ cm}} \\\\&=1 \space 600 \space 000 \text{ cm} \\\\&=16 \text{ km}. \end{aligned} Thus, the actual length of the river is $$16\text{ km}$$.

Drawing the scale drawing of an object:

Three ways to draw the scale drawing of an object are,

1. Use grid paper of the same size for different scales.
2. Use grid paper of different sizes.
3. Draw on a blank paper according to the given scale.

 Example Draw the scale drawing of shape $$PQRS$$ on a grid of equal squares using a scale of $$1 :\dfrac{1 }{2}$$. The scale given is $$1 :\dfrac{1 }{2}$$. Therefore, every side of the scale drawing is two times longer than the length of the sides of object $$PQRS$$.

Problem solving:

 Example The distance on a map between Kuantan and Gombak is $$4 \text { cm}$$. a) If the scale used to draw map is $$1\text { cm}: 50 \text{ km}$$, calculate the actual distance, in $$\text{km}$$, between Kuantan and Gombak. b) Mr. Danish wants to visit Kuantan. If he plans to drive to Kuantan at a speed of $$100\text{ km h}^{-1}$$, calculate the time taken to drive from Gombak to Kuantan in hours. a) Solution: \begin{aligned} \text{Scale}&=\dfrac{\text{Drawing distance}}{\text{Actual distance}} \\\\\dfrac{1}{50 \text{ km}}&=\dfrac{4\text{ cm}}{\text{Actual distance}} \\\\\text{Actual distance}&= \dfrac{4\text { cm}(50 \text{ km})}{1\text{ cm}} \\\\&= 200 \text{ km}. \end{aligned} b) Solution: \begin{aligned} \text{Time}&=\dfrac{\text{Distance}}{\text{Speed}}\\\\&= \dfrac {200 \text { km }} {100 \text{ km h}^{-1}} \\\\&= 2{\text{ hours}}. \end{aligned}

## Scale Drawings

4.1  Scale Drawings

 Definition The drawing of an object with all measurements in the drawing proportional to the measurements of the object.

Interpret the scale of a scale drawing:

\begin{aligned}&\space\text{Scale}\\\\&=\dfrac{\text{Measurement of scale drawing}}{\text{Measurement of object}}\end{aligned}

The ratio is,

• Measurement of scale drawing : Measurement of object

Scale drawings in the form of ratio is,

$$1:n$$, where $$n$$ is the positive integer or fraction.

$$1:n$$ means one unit on the scale drawing will represent $$n$$ units on the object.

• If $$n \lt 1$$, then the size of the scale drawing is bigger than the size of the object.

• If $$n \gt 1$$, then the size of the scale drawing is smaller than the size of the object.

• If $$n = 1$$, then the size of the scale drawing is the same as the size of the object.

 Example The diagram below shows object $$PQRS$$ and scale drawing $$P'Q'R'S'$$ drawn on a grid of equal squares. State the scale used in the form of $$1:n$$. $$\text{Scale}=\dfrac{P'Q'}{PQ}=\dfrac{2}{4}=\dfrac{1}{2}$$ or, $$\text{Scale}= \dfrac{P'S'}{PS}=\dfrac{3}{6}=\dfrac{1}{2}$$. Thus, $$\text{scale}=1:2$$.

 Example A map is drawn to a scale of $$1 : 400 \space000$$. Calculate the actual length, in $$\text{km}$$, of a river that is $$4\text { cm}$$ long on the map. \begin{aligned} \dfrac {1 \text{ cm}}{400 \space 000 \text{ cm}}& = \dfrac{4 \text{ cm}}{ \text{Actual distance}} \end{aligned} \begin{aligned} \\&\space\text{Actual distance}\\\\&=\dfrac{4 \times 400 \space000 \text{ cm}}{1\text{ cm}} \\\\&=1 \space 600 \space 000 \text{ cm} \\\\&=16 \text{ km}. \end{aligned} Thus, the actual length of the river is $$16\text{ km}$$.

Drawing the scale drawing of an object:

Three ways to draw the scale drawing of an object are,

1. Use grid paper of the same size for different scales.
2. Use grid paper of different sizes.
3. Draw on a blank paper according to the given scale.

 Example Draw the scale drawing of shape $$PQRS$$ on a grid of equal squares using a scale of $$1 :\dfrac{1 }{2}$$. The scale given is $$1 :\dfrac{1 }{2}$$. Therefore, every side of the scale drawing is two times longer than the length of the sides of object $$PQRS$$.

Problem solving:

 Example The distance on a map between Kuantan and Gombak is $$4 \text { cm}$$. a) If the scale used to draw map is $$1\text { cm}: 50 \text{ km}$$, calculate the actual distance, in $$\text{km}$$, between Kuantan and Gombak. b) Mr. Danish wants to visit Kuantan. If he plans to drive to Kuantan at a speed of $$100\text{ km h}^{-1}$$, calculate the time taken to drive from Gombak to Kuantan in hours. a) Solution: \begin{aligned} \text{Scale}&=\dfrac{\text{Drawing distance}}{\text{Actual distance}} \\\\\dfrac{1}{50 \text{ km}}&=\dfrac{4\text{ cm}}{\text{Actual distance}} \\\\\text{Actual distance}&= \dfrac{4\text { cm}(50 \text{ km})}{1\text{ cm}} \\\\&= 200 \text{ km}. \end{aligned} b) Solution: \begin{aligned} \text{Time}&=\dfrac{\text{Distance}}{\text{Speed}}\\\\&= \dfrac {200 \text { km }} {100 \text{ km h}^{-1}} \\\\&= 2{\text{ hours}}. \end{aligned}