We think you are located in South Africa. Is this correct?

Ratio And Proportion

8.2 Ratio and proportion (EMCJ8)

Ratio

A ratio describes the relationship between two quantities which have the same units. We can use ratios to compare weights, heights, lengths, currencies, etc. A ratio is a comparison between two quantities of the same kind and has no units.

Example: if the length of a rectangle is \(\text{20}\) \(\text{cm}\) and the width is \(\text{60}\) \(\text{cm}\), then we can express the ratio between the length and width of the rectangle as:

\begin{align*} \text{length to width } &= 20 \text{ to } 60 \\ &= 1 \text{ to } 3 \\ & \\ \text {Or } \quad \frac{\text{length}}{\text{width}} &= \frac{20}{60} \\ &= \frac{1}{3} \\ & \\ \text {Or } \quad \text{length} : \text{width } &= 20: 60 \\ &= 1: 3 \end{align*}
  • The ratio of \(\frac{1}{3}\) describes the length of the rectangle relative to its width.
  • A ratio written as a fraction is usually given in its simplest form.
  • A ratio gives no indication of actual length. For example,

    \[\frac{\text{length}}{\text{width}} = \frac{\text{50}\text{ cm}}{\text{150}\text{ cm}} \enspace \text{also gives a ratio of} \enspace \frac{1}{3}\] \[\text{And} \enspace \frac{\text{length}}{\text{width}} = \frac{\text{0,8}\text{ m}}{\text{2,4}\text{ m}} \enspace \text{also gives a ratio of} \enspace \frac{1}{3}\]
  • Do not convert a ratio to a decimal (even though \(\frac{1}{3}\) and \(\text{0,}\dot{3}\) have the same numerical value).

Proportion

Predicting heights

A record of heights is given.

If the ratio \(\frac{\text{height of a person at two years old}} {\text{height of a person as an adult}}\) is \(\text{1}\) to \(\text{2}\), complete the table below:

NameHeight at two yearsHeight as an adult
Hendrik\(\text{84}\) \(\text{cm}\)
Kagiso\(\text{162}\) \(\text{cm}\)
Linda\(\text{86}\) \(\text{cm}\)
Mandisa\(\text{0,87}\) \(\text{m}\)
Prashna\(\text{1}\) \(\text{m}\) \(\text{64}\) \(\text{cm}\)

Consider the diagram below:

69ec46c7fe1dbbb18be1bdf001b669ba.png

If two or more ratios are equal to each other, then we say that they are in the same proportion. Proportionality describes the equality of ratios.

If \(\dfrac{w}{x} = \dfrac{y}{z}\), then \(w\) and \(x\) are in the same proportion as \(y\) and \(z\).

  1. \(wz = xy\)
  2. \(\dfrac{x}{w} = \dfrac{z}{y}\)
  3. \(\dfrac{w}{y} = \dfrac{x}{z}\)
  4. \(\dfrac{y}{w} = \dfrac{z}{x}\)
51f06e56c0d570569958c9de8935fe0c.png

Given

\[\frac{AB}{BC}=\frac{x}{y}=\frac{kx}{ky}=\frac{DE}{EF}\]

The line segments \(AB\) and \(BC\) are in the same proportion as \(DE\) and \(EF\). The following statements are also true:

ProportionReciprocal proportionCross multiplication
\(\frac{AB}{BC}=\frac{DE}{FE}\)\(\frac{BC}{AB}=\frac{FE}{DE}\)\(AB \cdot FE = BC \cdot DE\)
\(\frac{AB}{AC}=\frac{DE}{DF}\)\(\frac{AC}{AB}=\frac{DF}{DE}\)\(AB \cdot DF = AC \cdot DE\)
\(\frac{BC}{AC}=\frac{EF}{DF}\)\(\frac{AC}{BC}=\frac{DF}{EF}\)\(BC \cdot DF = AC \cdot EF\)

We can also substitute \(x\), \(y\), \(kx\), and \(ky\) to show algebraically that the statements are true.

For example,

\begin{align*} BC \cdot DF &= y \times (kx + ky) \\ &= ky (x + y) \\ \text{And } AC \cdot EF &= (x+y) \times ky \\ &= ky(x+y) \\ \therefore BC \cdot DF &= AC \cdot EF \end{align*}

Test yourself now

High marks in maths are the key to your success and future plans. Test yourself and learn more on Siyavula Practice.

Sign up and test yourself

Ratio and proportion

Exercise 8.2

Solve for \(p\):

\(\frac{8}{40} = \frac{p}{25}\)
\begin{align*} \frac{8}{40} &= \frac{p}{25} \\ \frac{8 \times 25}{40} &= p \\ \frac{200}{40} &= p \\ \therefore 5 &= p \end{align*}
\(\frac{6}{9} = \frac{29 + p}{54}\)
\begin{align*} \frac{6}{9} &= \frac{29 + p}{54} \\ \frac{6 \times 54}{9} &= 29 + p \\ 36 &= 29 + p \\ \therefore 7 &= p \end{align*}
\(\frac{3}{1 + \frac{p}{4}} = \frac{4}{p + 1}\)
\begin{align*} \frac{3}{1 + \frac{p}{4}} &= \frac{4}{p + 1} \\ 3(p + 1) &= 4 \left( 1 + \frac{p}{4} \right) \\ 3p + 3 &= 4 + p \\ 2p &= 1 \\ \therefore p &= \frac{1}{2} \end{align*}
\(\frac{14}{100 - p} = \frac{49}{343}\)
\begin{align*} \frac{14}{100 - p} &= \frac{49}{343} \\ 14 \times 343 &= 49( 100 - p) \\ \frac{4802}{49} &= 100 - p \\ 98 &= 100 - p \\ \therefore p &= 2 \end{align*}

A packet of \(\text{160}\) sweets contains red, blue and yellow sweets in the ratio of \(3:2:3\) respectively. Determine how many sweets of each colour there are in the packet.

\begin{align*} 3 + 2 + 3 &= 8 \\ \text{Red } &= \frac{3}{8} \times 160 \\ &= 60 \\ \text{Blue } &= \frac{2}{8} \times 160 \\ &= 40 \\ \text{Yellow } &= \frac{3}{8} \times 160 \\ &= 60 \end{align*}

A mixture contains \(\text{2}\) parts of substance \(A\) for every \(\text{5}\) parts of substance \(B\). If the total weight of the mixture is \(\text{50}\) \(\text{kg}\), determine how much of substance \(B\) is in the mixture (correct to \(\text{2}\) decimal places).

\begin{align*} \text{Ratio substance } A \text{ to substance B } &= 2 : 5 \\ 2 + 5 &= 7 \\ \text{Substance } B &= \frac{5}{7} \times 50 \\ &= \text{35,71}\text{ kg} \end{align*}

Given the diagram below.

87fa220c43faff46ef5a412a3abcdaa5.png

Show that:

\(\frac{AB}{BC} = \frac{FE}{ED}\)
\begin{align*} \frac{AB}{BC} &= \frac{12}{18} \\ &= \frac{2}{3} \\ \frac{FE}{ED} &= \frac{36}{54} \\ &= \frac{2}{3} \\ \therefore \frac{AB}{BC} &= \frac{FE}{ED} \end{align*}
\(\frac{AC}{BC} = \frac{FD}{EF}\)
\begin{align*} \frac{AC}{BC} &= \frac{12 + 18}{18} \\ &= \frac{30}{18} \\ &= \frac{5}{3} \\ \frac{FD}{EF} &= \frac{36 + 54}{54} \\ &= \frac{90}{54} \\ &= \frac{5}{3} \\ \therefore \frac{AC}{BC} &= \frac{FD}{EF} \end{align*}
\(AB \cdot DF = AC \cdot FE\)
\begin{align*} AB \cdot DF &= 12 \times (36 + 54) \\ &= 12 \times 90 \\ &= 1080 \\ AC \cdot FE &= (12 + 18) \times 36 \\ &= 30 \times 36 \\ &= 1080 \\ \therefore AB \cdot DF &= AC \cdot FE \end{align*}

Consider the line segment shown below.

1be0ad169891344fb3d15cb3ea1129a7.png

Express the following in terms of \(a\) and \(b\):

\(PT: ST\)
\((a + b): b\)
\(\frac{PS}{TQ}\)
\(\frac{a}{2a} = \frac{1}{2}\)
\(\frac{SQ}{PQ}\)
\(\frac{2a + b}{3a + b}\)
\(QT: TS\)
\(2a: b\)

\(ABCD\) is a parallelogram with \(DC = \text{15}\text{ cm}\), \(h = \text{8}\text{ cm}\) and \(BF = \text{9}\text{ cm}\).

Calculate the ratio \(\dfrac{\text{area } ABF}{\text{area } ABCD}\).

88295721a375a24b791cff8532615f6a.png

The area of a parallelogram \(ABCD =\) base \(\times\) height:

\begin{align*} \text{Area} &= \text{15} \times \text{8} \\ &= \text{120}\text{ cm$^{2}$} \end{align*}

The perimeter of a parallelogram \(ABCD = 2DC + 2BC\).

To find the length of \(BC\), we use \(AF \perp BC\) and the theorem of Pythagoras.

\begin{align*} \text{In \(\triangle ABF\):} \quad AF^2 &= AB^2 - BF^2 \\ &= \text{15}^2 - \text{9}^2 \\ &= \text{144} \\ \therefore AF &= \text{12}\text{ cm} \\ \therefore \text{area } ABF &= \frac{1}{2} AF \cdot BF \\ &= \frac{1}{2}(12)(9)\\ &= \text{54}\text{ cm} \end{align*} \begin{align*} \therefore \frac{\text{area } ABF}{\text{area } ABCD} &= \frac{54}{150} \\ &= \frac{9}{25} \end{align*}

\(AB = \text{36}\text{ m}\) and \(C\) divides \(AB\) in the ratio \(4:5\). Determine \(AC\) and \(CB\).

a6a15db3875447fa39f75f2250f37855.png
\begin{align*} \frac{AC}{AB} &= \frac{4k}{(4k + 5k)} \\ \therefore AC &= 36 \times \frac{4}{9} \\ &= \text{16}\text{ m} \\ CB &= 36 - 16 \\ &= \text{20}\text{ m} \\ \text{Or } \frac{CB}{AC} &= \frac{5k}{(4k + 5k)} \\ \therefore CB &= 36 \times \frac{5}{9} \\ &= \text{20}\text{ m} \end{align*}

If \(PQ = \text{45}\text{ mm}\) and the ratio of \(TQ:PQ\) is \(2:3\), calculate \(PT\) and \(TQ\).

f8e54b2bd6ceee8639af76aed4308419.png
\begin{align*} \frac{TQ}{PQ} &= \frac{2}{3} \\ \therefore TQ &= PQ \times \frac{2}{3} \\ &= 45 \times \frac{2}{3} \\ &= \text{30}\text{ mm} \\ PT &= 45 - 30 \\ &= \text{15}\text{ mm} \\ \text{Or } \quad \frac{PT}{PQ} &= \frac{1}{3} \\ \therefore PT &= 45 \times \frac{1}{3} \\ &= \text{15}\text{ mm} \end{align*}

Luke's biology notebook is \(\text{30}\) \(\text{cm}\) long and \(\text{20}\) \(\text{cm}\) wide. The dimensions of his desk are in the same proportion as the dimensions of his notebook.

If the desk is \(\text{90}\text{ cm}\) wide, calculate the area of the top of the desk.

\begin{align*} \text{Ratio } &= \frac{\text{table width}}{\text{book width}} \\ &=\frac{90}{20} \\ &=\text{4,5} \\ \therefore \text{Length of table } &= 30 \times \text{4,5} \text{cm} \\ &=\text{135}\text{ cm}\\ \text{Area table} &= 135 \times 90 \\ &=\text{12 150}\text{ cm$^{2}$}\\ &=\text{1,2}\text{ m$^{2}$} \end{align*}

Luke covers each corner of his desk with an isosceles triangle of cardboard, as shown in the diagram:

1e1d2dd2e7e566e1d135cb6ac1beec24.png

Calculate the new perimeter and area of the visible part of the top of his desk.

\begin{align*} x^2&=15^2+15^2 \\ x&=\text{21,2}\text{ cm}\\ \text{New length}&=135-2(15) \\ &=\text{105}\text{ cm} \\ \text{New breadth}&=90-2(15) \\ &=\text{60}\text{ cm}\\ \text{New perimeter }&=2(105)+2(60)+4(\text{21,2})\\ &=\text{414,8}\text{ cm} \end{align*} \begin{align*} \text{Area cut off}&=2\times(15^2) \\ &=\text{450}\text{ m$^{2}$} \\ \text{New area } &= \text{12 150}\text{ cm$^{2}$} - \text{450}\text{ m$^{2}$} \\ &= \text{11 700}\text{ cm$^{2}$} \end{align*}
Use this new area to calculate the dimensions of a square desk with the same desk top area.
\begin{align*} s^2&= \text{11 700}\text{ cm$^{2}$} \\ \therefore s&= \text{108,2}\text{ cm}\\ \text{Square table of length} &\approx 108 \times 108 \text{cm$^{2}$} \end{align*}