UK Secondary (7-11)
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Applications of Similarity and Symmetry
Lesson

Similar triangles in the real world

Similar triangles show up everywhere in the real world, even when you can't see them. We can use what we've learnt about similar triangles to figure out distances and angles when they might be too large to just measure with a ruler or compass.

There are 5 steps to any problem involving similar triangles.

QUESTION Read the question closely
DIAGRAM Draw and label a diagram
IDENTIFY Identify the similar triangles and their corresponding sides
EQUATION Form an equation with the equivalent ratios of the corresponding sides
SOLVE Solve the equation for the unknown value

 

Examples

Question 1

A very common kind of similar triangles problem involves shadows being cast over straight objects. If two objects are in roughly the same place on the earth, the sun's rays will pass over the top of them at the same angle. Suppose we are given the following problem.

A stick of height $1.2$1.2 m casts a shadow of length $3.1$3.1 m. Nearby, a tree casts a shadow $27$27 m long. What is the height of the tree? Round your answer to one decimal place.

Solution

After we have read the question and understood the information, we can construct a diagram to help us solve the problem. We will call the unknown height of the tree $h$h.

<Image coming soon>

We can tell that these two triangles must be similar because they are both right angled and have equal corresponding angles where the sun's rays come down.

Neither triangle is a reflection or rotation of the other, so the corresponding sides are easy to see. The shadows represent corresponding sides, and the heights also correspond. 

Since the triangles are similar, these both pairs of corresponding sides will be in the same proportion. Hence, we can form the following equation relating the ratios.

$\frac{h}{27}=\frac{1.2}{3.1}$h27=1.23.1

We then solve for $h$h as follows.

$h$h $=$= $\frac{1.2}{3.1}\times27$1.23.1×27
  $=$=

$10.451613$10.451613...

  $\approx$

$10.5$10.5 to 1 decimal place

So, the tree is approximately $10.5$10.5 m tall, and we figured this out with a stick and some shadows!

QUESTION 2

Another common similar triangles problem involves figuring out horizontal distances on the earth. Suppose we are given the following problem.

Paula wants to figure out the width of a river. She stands $4$4 m from the south bank, looking across at two trees on the north bank that are known to be $38$38 m apart. Her assistant stands on the south bank, stands in between Paula and the first tree, and places a stick in the ground. The assistant then walks $8.4$8.4 m across the bank and stands in between Paula and the second tree, placing a stick in the ground again. Find the width of the river, rounding your answer to the nearest centimetre. 

Solution

Again, we construct a diagram and label the relevant information. We can call the unknown width of the river $w$w.

<Image coming soon>

We can tell that these triangles are similar because one is nestled inside the other and the river banks are parallel.

Neither triangle is a reflection or rotation of the other, so the corresponding sides are easy to see. The river banks represent corresponding sides, and Paula's distance of $4$4 m from the south bank corresponds with her total distance from the first tree directly across the river, which is $w+4$w+4 m.

Again, since the triangles are similar, these both pairs of corresponding sides will be in the same proportion. Hence, we can form the following equation relating the ratios.

$\frac{w+4}{4}=\frac{38}{8.4}$w+44=388.4

We then solve for $w$w as follows.

$w+4$w+4 $=$= $\frac{38}{8.4}\times4$388.4×4
$w$w $=$= $\frac{38}{8.4}\times4-4$388.4×44
  $=$=

$14.095238$14.095238...

  $\approx$

$14.10$14.10 to 2 decimal places

So, the river is approximately $14.10$14.10 m wide.

Symmetry

Many problems involve symmetrical patterns.

Examples

Question 3

Paper usually comes in eleven standard $A$A series sizes starting at the largest, $A0$A0, through to the smallest, $A10$A10.

Each new size is obtained by cutting the previous one in half, as shown below.

a) Serena has some gift paper and wants to cut out a rectangle of $A7$A7 size. She knows that the area of an $A0$A0 sheet of paper is always $1$1 m2.

i) What area should Serena's $A7$A7 cut-out have? Express your answer in cm2.

ii) Given that $A0$A0 paper is $841$841 mm wide, what will be the length of Serena's $A7$A7 cut-out? Round your answer to $2$2 decimal places.

b) $A$A Series paper uses this simple halving method to reduce waste but still manages to keep all sheets in the same rectangular proportion. 

What must the ratio of the sides be in order for this to happen?

c) Find the perimeter of Serena's $A7$A7 cut-out to $2$2 decimal places.

Solution

a) i) Since each new sized sheet is just half the area of the previous sheet, we just have to keep halving $1$1 m2 until we arrive at the area for an $A7$A7 sheet.

$A0$A0 $=$= $1$1 m2
$A1$A1 $=$= $1\div2$1÷​2 m2
$A2$A2 $=$= $1\div2\div2$1÷​2÷​2 m2
$A3$A3 $=$= $1\div2\div2\div2$1÷​2÷​2÷​2 m2

Notice the pattern. We have to divide by $2$2 $n$n times for the area of an $An$An sheet of paper. Hence we can calculate our area for an $A7$A7 sheet by dividing by $2$2 seven times..

$A7$A7 $=$= $1\div2\div2\div2\div2\div2\div2\div2$1÷​2÷​2÷​2÷​2÷​2÷​2÷​2 m2
  $=$= $0.0078125$0.0078125 m2

Remember that there are $10000$10000 cm2 in $1$1 m2. Hence our answer becomes:

$A7$A7 $=$= $0.0078125\times10000$0.0078125×10000 cm2
  $=$= $78.125$78.125 cm2

a) ii) Notice that the width of the $A0$A0 sheet becomes the length of the $A1$A1 sheet. Then half of that is the width of the $A2$A2 sheet, which becomes the length of the $A3$A3 sheet.

The pattern is that the width of even numbered sheets keep halving, then turn into the length of the odd sheet after it. Hence, we can figure out the width of an $A6$A6 sized sheet, which will be the length of an $A7$A7 sized sheet.

$A0$A0 $=$= $841$841 mm wide
$A2$A2 $=$= $841\div2$841÷​2 mm wide
$A4$A4 $=$= $841\div2\div2$841÷​2÷​2 mm wide
$A6$A6 $=$= $841\div2\div2\div2$841÷​2÷​2÷​2 mm wide
  $=$= $105.13$105.13 mm wide (to $2$2 d.p.)
     
$A7$A7 $=$= $105.13$105.13 mm long

b) Let's call the length and width of the $A1$A1 sheet $l_1$l1 and $w_1$w1 respectively. Similarly, let's call the length and width of the $A2$A2 sheet $l_2$l2 and $w_2$w2 respectively.

Firstly, we are halving the sheets as in the diagram, so we have the following:

$l_2=w_1$l2=w1 and $w_2=\frac{l_1}{2}$w2=l12 ... (1)

Secondly, we want the rectangles to all be similar, so we have the following:

$\frac{l_1}{w_1}=\frac{l_2}{w_2}$l1w1=l2w2 ... (2)

Substituting (1) into (2), we get:

$\frac{l_1}{w_1}=\frac{w_1}{\frac{l_1}{2}}$l1w1=w1l12

Which we can then rearrange to get:

$\frac{l_1}{w_1}$l1w1 $=$= $\frac{w_1}{\frac{l_1}{2}}$w1l12
$\frac{l_1}{w_1}$l1w1 $=$= $\frac{2w_1}{l_1}$2w1l1
$\frac{l_1^2}{w_1}$l21w1 $=$= $2w_1$2w1
$\frac{l_1^2}{w_1^2}$l21w21 $=$= $2$2
$\left(\frac{l_1}{w_1}\right)^2$(l1w1)2 $=$= $2$2
$\frac{l_1}{w_1}$l1w1 $=$= $\sqrt{2}$2

So the length and width of each sheet are in the same ratio $\sqrt{2}:1$2:1.

c) We know that the length of Serena's $A7$A7 cut-out is $105.13$105.13 mm. Given the ratio we just found, we know that:

$\frac{105.13}{w}=\frac{\sqrt{2}}{1}$105.13w=21

Which we can rearrange to get:

$w=\frac{105.13}{\sqrt{2}}$w=105.132$=$=$74.34$74.34 mm (to $2$2 d.p.)

Hence, we have the following perimeter for the $A7$A7 sheet:

$105.13+105.13+74.34+74.34=358.94$105.13+105.13+74.34+74.34=358.94 mm (to $2$2 d.p.)

Worked Examples

Question 1

A stick of height $1.1$1.1 m casts a shadow of length $2.2$2.2 m. At the same time, a tree casts a shadow of $6.2$6.2 m.

  1. If the tree has a height of $h$h metres, solve for $h$h.

Question 2

James is $1.7$1.7 m tall and casts a shadow $2$2 m long. At the same time, a tower casts a shadow $13$13 m long. If the tower is $h$h metres high, solve for $h$h correct to one decimal place.

Question 3

A school building reaching $h$h metres high casts a shadow of $30$30 m while a $3$3 m high tree casts a shadow of $6$6 m. Solve for $h$h.

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