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11. Applications of Trigonometry
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VCE 11 General 2023

11.01 Pythagoras' theorem

Lesson
Worksheet
Practice
Lesson

 

Pythagoras' theorem - a review

Pythagoras' theorem states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

$a^2+b^2=c^2$a2+b2=c2

where $c$c represents the length of the hypotenuse and $a$a, $b$b are the two shorter sides. 

So if two sides of a right-angled triangle are known and one side is unknown, this relationship can be used to find the length of the unknown side.

The equation can be rearranged to make any unknown side length of the triangle the subject of the formula:

Pythagoras' theorem

$c=\sqrt{a^2+b^2}$c=√a2+b2

$b=\sqrt{c^2-a^2}$b=√c2−a2

$a=\sqrt{c^2-b^2}$a=√c2−b2

 

Worked examples

Example 1

Find the length of the hypotenuse of a right-angled triangle whose two other sides measure $10$10 cm and $12$12 cm.

Think: Here we want to find $c$c, and are given $a$a and $b$b.

Do:

$c^2$c2 $=$= $10^2+12^2$102+122
$c$c $=$= $\sqrt{10^2+12^2}$√102+122
$c$c $=$= $15.62$15.62 cm ($2$2 d.p.)

 

Example 2

Find the length of unknown side $b$b of a right-angled triangle whose hypotenuse is $6$6 mm and one other side is $4$4 mm.

Think: Here we want to find $b$b, the length of a shorter side.

Do:

$6^2$62 $=$= $b^2+4^2$b2+42
$b^2$b2 $=$= $6^2-4^2$62−42
$b$b $=$= $\sqrt{6^2-4^2}$√62−42
$b$b $=$= $4.47$4.47 mm ($2$2 d.p.)

 

Practice questions

Question 1

Calculate the value of $c$c in the triangle below.

A right-angled triangle with a right angle shown at the bottom right corner. The base is labeled as 14 cm, the height on the right side is labeled as 48 cm, and the hypotenuse is labeled with c cm. There is a small square at the right angle indicating the 90-degree angle.

Question 2

Calculate the value of $b$b in the triangle below.

Give your answer correct to two decimal places.

A right-angled triangle is presented with the right angle at the top left. The side adjacent the right angle is labeled with the variable "b." The triangle's height, on the left side, is marked with the number 10, and the hypotenuse, slanting down from the left to the right, is labeled with the number 24. 

 

Question 3

Find the length of the unknown side in this right-angled triangle, expressing your answer as a decimal approximation to two decimal places.

 

Applications of Pythagoras' theorem

How to apply pythagoras' theorem to real-life problems
  1. Look for right-angled triangles
  2. Choose which side, hypotenuse or a shorter side, you are trying to find
  3. Find the lengths of the other two sides
  4. Apply the relevant formula and substitute the lengths of the other two sides

 

Practice questions

Question 4

The screen on a handheld device has dimensions $9$9 cm by $5$5 cm, and a diagonal of length $x$x cm. What is the value of $x$x?

Round your answer to two decimal places.

The rectangular screen of a handheld device is depicted in a landscape orientation. It is positioned with its longest side, measuring 9 cm, horizontal. Its shortest side, measuring 5 cm, is vertical. A diagonal line runs from the top right corner of the screen to its bottom left corner, and is labeled x cm, indicating its length. A right-angled triangle is formed by the horizontal side, vertical side and the diagonal line. The diagonal line is also the hypotenuse of the triangle and is opposite the right angle indicated by a small square is located at the bottom right corner.

Question 5

$VUTR$VUTR is a rhombus with perimeter $112$112 cm. The length of diagonal $RU$RU is $46$46 cm.

A rhombus with vertices labeled U, V, W, and T is illustrated. Dotted lines are drawn from each vertex to the opposite vertex, forming two intersecting diagonals inside the quadrilateral. The point of intersection of the diagonals is marked with a circle and labeled W. Additionally, a small solid blue square is shown, indicating a right angle at vertex W, suggesting that the diagonals are perpendicular to each other. The sides UV and WT are marked with double tick marks, indicating they are of equal length, as are sides VU and RT.

  1. First find the length of $VR$VR.

  2. Then find the length of $RW$RW.

  3. If the length of $VW$VW is $x$x cm, find $x$x correct to two decimal places.

  4. Hence, what is the length of the other diagonal $VT$VT correct to two decimal places.

Outcomes

U2.AoS4.2

Pythagoras’ theorem and the trigonometric ratios (sine, cosine and tangent) and their application including angles of elevation and depression and three figure bearings

U2.AoS4.9

solve practical problems involving right-angled triangles in the dimensions including the use of angles of elevation and depression, Pythagoras’ theorem trigonometric ratios sine, cosine and tangent and the use of three-figure (true) bearings in navigation

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