Pythagoras’ Theorem Revision Notes for VCE SSCE General Mathematics

Pythagoras' Theorem

Understanding right-angled triangles

Pythagoras' theorem shows how the three sides of a right-angled triangle are mathematically linked together. Before we explore this relationship, you need to understand two important terms.

In any right-angled triangle, the longest side is called the hypotenuse. This side always sits directly opposite the right angle.

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The right angle itself is marked with a small square in the corner of the triangle. Being able to identify these features is essential for using Pythagoras' theorem correctly.

The theorem explained

Pythagoras' theorem states that in any right-angled triangle, if you square the lengths of the two shorter sides and add them together, you get the square of the hypotenuse.

We can write this as a formula:

$a^2 + b^2 = c^2$

where c represents the length of the hypotenuse, and a and b represent the lengths of the other two sides.

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The area of the square on the hypotenuse equals the combined areas of the squares on the other two sides. This visual relationship demonstrates why the theorem works.

This theorem is incredibly useful because if you know the lengths of any two sides of a right-angled triangle, you can always calculate the length of the third side.

Calculating the hypotenuse

When you need to find the hypotenuse (the longest side), you can use Pythagoras' theorem directly. Here's a systematic approach:

Step 1: Write down the formula $a^2 + b^2 = c^2$

Step 2: Substitute the known values for the two shorter sides

Step 3: Take the square root of both sides to find $c$

Step 4: Calculate the answer and round to the required number of decimal places

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Worked Example: Finding the Hypotenuse

Calculate the length of the hypotenuse in the triangle below to two decimal places.

Solution:

We start with Pythagoras' theorem:

$a^2 + b^2 = c^2$

Substituting the known values ( $a = 4$ cm and $b = 10$ cm):

$4^2 + 10^2 = c^2$

Rearranging to make $c$ the subject:

$c = \sqrt{4^2 + 10^2}$

$c = \sqrt{16 + 100}$

$c = \sqrt{116}$

$c = 10.770...$

Therefore, the length of the hypotenuse is 10.77 cm (to two decimal places).

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Calculator Tip: Always ensure your calculator is in decimal mode when performing these calculations. This will give you the decimal answer directly rather than leaving it in surd form.

Calculating other sides

Pythagoras' theorem can be rearranged to find any side of a right-angled triangle, not just the hypotenuse. When the unknown side is one of the shorter sides, you need to rearrange the formula.

If the hypotenuse is known along with one other side, the formula becomes:

$a = \sqrt{c^2 - b^2}$

The key steps are:

Step 1: Write down Pythagoras' theorem $a^2 + b^2 = c^2$

Step 2: Substitute the known values and the variable for the unknown side

Step 3: Rearrange to make the unknown the subject

Step 4: Calculate and round to the required accuracy

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Worked Example: Finding a Shorter Side

Calculate the length of the unknown side, $x$ , in the triangle below to one decimal place.

Solution:

Starting with Pythagoras' theorem:

$a^2 + b^2 = c^2$

Here, $x$ is one of the shorter sides, $4.7$ mm is the other shorter side, and $11$ mm is the hypotenuse. Substituting:

$x^2 + 4.7^2 = 11^2$

Rearranging to solve for $x$ :

$x^2 = 11^2 - 4.7^2$

$x = \sqrt{11^2 - 4.7^2}$

$x = \sqrt{121 - 22.09}$

$x = \sqrt{98.91}$

$x = 9.945...$

Therefore, the length of $x$ is 9.9 mm to one decimal place.

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Key Difference: Notice that when finding a shorter side, we subtract the squares before taking the square root. When finding the hypotenuse, we add the squares.

Solving practical problems

Pythagoras' theorem is extremely useful for solving real-world problems involving distances, heights, and lengths. The key is to identify or draw a right-angled triangle that represents the situation.

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Worked Example: Helicopter Distance Problem

A helicopter hovers at a height of $150$ m above the ground and is a horizontal distance of $220$ m from a landing pad. Find the direct distance of the helicopter from the landing pad to two decimal places.

Solution:

First, we sketch a diagram showing the situation. The helicopter's height, the horizontal distance to the landing pad, and the direct distance form a right-angled triangle.

The vertical distance is $150$ m, the horizontal distance is $220$ m, and we need to find the direct distance, $c$ .

Using Pythagoras' theorem:

$c^2 = a^2 + b^2$

$c^2 = 150^2 + 220^2$

$c^2 = 22500 + 48400$

$c^2 = 70900$

$c = \sqrt{70900}$

$c = 266.270...$

Therefore, the helicopter is 266.27 m from the landing pad (to two decimal places).

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Problem-Solving Strategy: Always draw a clear diagram for practical problems. This helps you identify which measurements correspond to which sides of the triangle.

Working in three dimensions

Pythagoras' theorem can also be applied to three-dimensional shapes like cubes, boxes, and prisms. The technique involves identifying right-angled triangles within the 3D shape and working with them one at a time.

When working with three-dimensional problems, remember that a plane is a flat surface. Right-angled triangles exist within these flat planes, even though the overall shape is three-dimensional.

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Strategy for 3D Problems:

Carefully examine the 3D shape to identify the right-angled triangle containing the unknown side
Draw this triangle separately as a flat, two-dimensional diagram
Mark all known measurements on your triangle
Apply Pythagoras' theorem
If you need to find another distance, repeat the process with a different triangle

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Worked Example: Finding Diagonals in a Cube

The cube in the diagram has side lengths of $5$ cm. Find the length AC and AD, both to one decimal place.

Part a) Finding AC:

First, we locate the right-angled triangle in the base of the cube that contains the diagonal AC. This is triangle ABC.

Drawing this triangle separately:

The triangle ABC has two sides of $5$ cm each (the edges of the cube). We can now apply Pythagoras' theorem:

$(AC)^2 = (AB)^2 + (BC)^2$

$(AC)^2 = 5^2 + 5^2$

$(AC)^2 = 25 + 25$

$(AC)^2 = 50$

$AC = \sqrt{50}$

$AC = 7.071...$

Therefore, the length AC is 7.1 cm to one decimal place.

Part b) Finding AD:

Now we need to find the space diagonal AD. This requires us to use triangle ACD, which is vertical and contains the edge we just calculated.

Drawing triangle ACD separately, we know:

AC = $7.07$ cm (using one extra decimal place for accuracy)
CD = $5$ cm (the vertical edge of the cube)
AD is unknown

Applying Pythagoras' theorem:

$(AD)^2 = (AC)^2 + (CD)^2$

$(AD)^2 = 7.07^2 + 5^2$

$(AD)^2 = 49.9849 + 25$

$(AD)^2 = 74.9849$

$AD = \sqrt{74.9849}$

$AD = 8.659...$

Therefore, the length AD is 8.7 cm to one decimal place.

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Maintaining Accuracy in Multi-Step Problems: When solving multi-step 3D problems, keep extra decimal places in intermediate calculations to maintain accuracy. Only round your final answer to the required number of decimal places.

Quick Check Formula: For a rectangular box with dimensions $a$ , $b$ , and $c$ , the space diagonal (corner to opposite corner) is $\sqrt{a^2 + b^2 + c^2}$ . This can be used to verify your answer.

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Key Points to Remember:

Pythagoras' theorem links the three sides of a right-angled triangle: $a^2 + b^2 = c^2$
The hypotenuse is always the longest side and sits opposite the right angle
To find the hypotenuse: $c = \sqrt{a^2 + b^2}$ (add the squares, then take the square root)
To find a shorter side: $a = \sqrt{c^2 - b^2}$ (subtract the squares, then take the square root)
For 3D problems: identify the right-angled triangles within the shape, draw them separately, then apply Pythagoras' theorem to each triangle in turn
Always draw clear diagrams to help identify the correct triangle and measurements

Pythagoras’ Theorem (VCE SSCE General Mathematics): Revision Notes