Pythagoras

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Notes

Pythagoras' Theorem

Pythagoras was a Greek mathematician who lived over 2500 years ago.
**Pythagoras' theorem** links the lengths of the three sides of a **right-angled triangle**.
The **hypotenuse** is the longest side, opposite the right angle.
Formula: **a² $+ b^{2} =$ c²**, where c is the hypotenuse and a, b are the shorter sides.
It does not matter which shorter side is labelled a or b.

Finding the Hypotenuse

To find the hypotenuse: square the two shorter sides, add them, then take the **positive square root**.
Formula: **c $= √(a^{2} +$ b²)**.
When finding the hypotenuse, you **add** inside the square root.
Example: If $a = 3, b = 4$ , then $c = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5$ .

Finding a Shorter Side

To find a shorter side: square the hypotenuse and the other shorter side, **subtract** the smaller from the larger, then take the square root.
Formula: **a $= √(c^{2}$ – b²)**.
When finding a shorter side, you **subtract** inside the square root.
Ensure the hypotenuse is longer than any other side; otherwise, check your working.

Using Pythagoras with Other Shapes

Pythagoras can be applied to any shape that can be split into right-angled triangles.
For a rectangle or square, the **diagonal** splits the shape into two identical right-angled triangles.
The diagonal becomes the **hypotenuse** of those triangles.
Example: In a rectangle 14 cm by w cm with diagonal 23 cm, use Pythagoras to find $w: w = √(23^{2}$ – $14^{2})$ .

Multi-Step Problems

In multi-step problems, leave intermediate answers as **exact values** $(e.g., \sqrt{63})$ to avoid rounding errors.
Only round the final answer to the required degree of accuracy.
Work through each right-angled triangle step by step.
Example: Find BC in a trapezium by first finding BD using triangle ABD, then $DC = AC$ – AD, then $BC = \sqrt{BD^{2} + DC^{2}}$ .

Common Mistakes & Tips

The hypotenuse must be the longest side; if it isn't, you've made an error.
When subtracting for a shorter side, always subtract the smaller square from the larger square.
Otherwise, you'll get a negative number and a 'Math Error' when taking the square root.
Always check that your answer is reasonable.

Right-Angled Triangle with Sides Labelled

Diagonal of a Rectangle

Flagpole and Rope (Real-World Application)

Multi-Step: Two Right Triangles

Practice questions

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1.Which side of a right-angled triangle is the hypotenuse?
Easy
- AThe side opposite the right angle
- BThe side adjacent to the right angle
- CThe shortest side
- DAny side
2.Pythagoras' theorem states that for a right-angled triangle with sides a, b and hypotenuse c:
Easy
- A $a^{2} + b^{2} = c^{2}$
- B $a + b = c$
- C $a^{2} = b^{2} + c^{2}$
- D $a^{2} - b^{2} = c^{2}$
3.A right-angled triangle has legs of lengths 3 cm and 4 cm. What is the length of the hypotenuse?
Easy
- A5 cm
- B7 cm
- C12 cm
- D25 cm
4.A flagpole is 25 m tall. A rope from the top is tied to the ground 8 m from the base. What is the length of the rope?
Easy
- A26.2 m
- B24.0 m
- C33.0 m
- D17.0 m
5.In a right-angled triangle, the hypotenuse is 10 cm and one leg is 6 cm. What is the length of the other leg?
Easy
- A8 cm
- B4 cm
- C16 cm
- D11.7 cm
6.A rectangle is 14 cm wide and has a diagonal of 23 cm. What is the length of the rectangle?
Medium
- A18.2 cm
- B26.9 cm
- C9.0 cm
- D18.0 cm
7.A right-angled triangle has legs of lengths x and 8 cm, and hypotenuse 17 cm. What is x?
Medium
- A15 cm
- B18.8 cm
- C9 cm
- D25 cm
8.The length of one side of a rectangle is 12 cm and the diagonal is 13 cm. What is the area of the rectangle?
Medium
- A $60 cm^{2}$
- B $156 cm^{2}$
- C $30 cm^{2}$
- D $144 cm^{2}$

Summary

## Pythagoras Theorem - **Pythagoras** was a Greek mathematician (2500+ years ago). - **Pythagoras' theorem** links the three sides of a **right-angled triangle**. - The **hypotenuse** is the longest side, opposite the right angle. - Formula: **a² + b² = c²** where c is the hypotenuse, a and b are the shorter sides. ### Finding the Hypotenuse - Square the two shorter sides, add them, then take the positive square root. - c = √(a² + b²) ### Finding a Shorter Side - Square the hypotenuse and the other shorter side, subtract, then take the positive square root. - a = √(c² – b²) ### Tips - The hypotenuse must be the longest side. If it's not, check your working. - When finding a shorter side, subtract the smaller square from the larger to avoid a negative under the square root. - In multi-step problems, keep answers exact (e.g., √63) until the final step, then round. - Pythagoras can be used with any shape that can be split into right-angled triangles (e.g., rectangles, squares). ### Common Question Types - Find the hypotenuse given two shorter sides. - Find a shorter side given the hypotenuse and one side. - Find the diagonal of a rectangle or square. - Combine with other shapes (trapezium, multiple triangles). - Calculate perimeter or area after finding missing lengths. > **Example**: In a right triangle with legs 12 cm and 9 cm, hypotenuse = √(12²+9²) = √(144+81) = √225 = 15 cm.

Key terms

Pythagoras' theorem: a² + b² = c², where c is the length of the hypotenuse and a, b are the lengths of the two shorter sides of a right-angled triangle.
Hypotenuse: The longest side of a right-angled triangle, opposite the right angle.
Right-angled triangle: A triangle with one angle equal to 90°.

Slides

Pythagoras' Theorem

Pythagoras was a Greek mathematician who lived over 2500 years ago.
His theorem relates the sides of a right-angled triangle.
The longest side is called the hypotenuse, opposite the right angle.

What is the name of the longest side in a right-angled triangle?

Hypotenuse
Adjacent
Opposite
Base

The Formula

Pythagoras' theorem states: $a^{2} + b^{2} = c^{2}$
c is the hypotenuse, a and b are the two shorter sides.
It doesn't matter which shorter side is a or b.

In a right-angled triangle, the formula $is ____ + b^{2} = c^{2}$ .

Finding the Hypotenuse

To find the hypotenuse: square both shorter sides, add them, then take the square root.
This can be written as $c = \sqrt{a^{2} + b^{2}}$ .
Remember to add inside the square root.

If $a = 3$ and $b = 4$ , what is the hypotenuse?

5
7
25
$\sqrt{7}$

Finding a Shorter Side

To find a shorter side: square the hypotenuse and the known shorter side, subtract, then take the square root.
This can be written as $a = \sqrt{c^{2} - b^{2}}$ .
Subtract the smaller square from the larger one.

To find a shorter side, use $a = \sqrt{c^{2} ____ b^{2}}$ .

Using Pythagoras with Other Shapes

Pythagoras can be used in any shape that can be split into right-angled triangles.
For example, the diagonal of a rectangle splits it into two right-angled triangles.
The diagonal becomes the hypotenuse.

Which of these are true?

The diagonal of a rectangle is the hypotenuse of a right-angled triangle.
Pythagoras' theorem only works for squares.
You can use Pythagoras to find the side of a trapezium.

Examiner Tips

If the hypotenuse is shorter than another side, you made a mistake.
When finding a shorter side, subtract the smaller value from the bigger value.
Leave answers as exact $(e.g., \sqrt{63})$ until the final step, then round.

What should you do if your calculated hypotenuse is shorter than a shorter side?

Check your working
Use the answer anyway
Add 1 to the hypotenuse
Ignore it

Worked Example: Step 1

In triangle $ABD, AB = 12, AD = 9$ , find BD.
BD is a shorter side: $BD = \sqrt{12^{2} - 9^{2}} = \sqrt{144-81} = \sqrt{63}$ ≈ 7.937.
Keep $as \sqrt{63}$ for accuracy.

In triangle $ABD, BD = \sqrt{12^{2} - 9^{2}} = √____$ .

Worked Example: Step 2

$AC = 22, AD = 9$ , so $DC = 22 - 9 = 13$ .
Now in triangle $BCD, BD = \sqrt{63}, DC = 13$ , find BC (hypotenuse).
$BC = √((\sqrt{63})^{2} + 13^{2}) = \sqrt{63 + 169} = \sqrt{232}$ ≈ 15.2 cm.

In the worked example, what is BC to 1 decimal place?

15.2 cm
15.1 cm
15.3 cm
15.0 cm

Practice: Flagpole Problem

A flagpole is 25 m tall. A rope from the top is tied 8 m from the base.
The rope is the hypotenuse of a right-angled triangle.
Length $= \sqrt{25^{2} + 8^{2}} = \sqrt{625+64} = \sqrt{689}$ ≈ 26.2 m.

A rope from a 25 m flagpole is tied 8 m from the base. How long is the rope?

26.2 m
24.0 m
33.0 m
17.0 m

Summary

Pythagoras' theorem: $a^{2} + b^{2} = c^{2}$ for right-angled triangles.
c is the hypotenuse, a and b are the shorter sides.
Use addition for the hypotenuse, subtraction for a shorter side.
Apply to any shape with right-angled triangles.

Which statements are correct?

The hypotenuse is the longest side.
To find a shorter side, add the squares.
Pythagoras can be used in rectangles.

Flashcards

What is Pythagoras' theorem?

For a right-angled triangle, $a^{2} + b^{2} = c^{2}$ , where c is the hypotenuse (longest side) and a, b are the two shorter sides.

How do you find the length of the hypotenuse using Pythagoras?

Square the two shorter sides, add them, then take the positive square root: $c = \sqrt{a^{2} + b^{2}}$ .

How do you find the length of a shorter side using Pythagoras?

Square the hypotenuse and the other shorter side, subtract the smaller square from the larger, then take the positive square root: $a = \sqrt{c^{2} - b^{2}}$ .

What is the hypotenuse?

The longest side of a right-angled triangle, opposite the right angle.

In a right triangle with legs 3 cm and 4 cm, find the hypotenuse.

$c = \sqrt{3^{2} + 4^{2}} = \sqrt{9+16} = \sqrt{25} = 5$ cm.

In a right triangle with hypotenuse 13 cm and one leg 5 cm, find the other leg.

$a = \sqrt{13^{2} - 5^{2}} = \sqrt{169-25} = \sqrt{144} = 12$ cm.

A flagpole is 25 m tall. A rope from its top is tied to the ground 8 m from its base. Find the rope length.

Rope length $= \sqrt{25^{2} + 8^{2}} = \sqrt{625+64} = \sqrt{689}$ ≈ 26.2 m.

A rectangle has sides 14 cm and w cm, diagonal 23 cm. Find w.

$w = \sqrt{23^{2} - 14^{2}} = \sqrt{529-196} = \sqrt{333}$ ≈ 18.2 cm.

Find the diagonal of a rectangle with sides 12 cm and 5 cm.

Diagonal $= \sqrt{12^{2} + 5^{2}} = \sqrt{144+25} = \sqrt{169} = 13$ cm.

A rectangle has one side 12 cm and diagonal 13 cm. Find its area.

Other side $= \sqrt{13^{2} - 12^{2}} = \sqrt{169-144} = \sqrt{25} = 5$ cm. Area $= 12 \times 5 = 60 cm^{2}$ .

A guy rope of length 6.9 m is attached to a flagpole top and fixed 3.2 m from its base. Find the flagpole height.

Height $= \sqrt{6.9^{2} - 3.2^{2}} = \sqrt{47.61 - 10.24} = \sqrt{37.37}$ ≈ 6.11 m.

In triangle $ABC, AB = 12$ cm, $AD = 9$ cm, $AC = 22$ cm, and BD is perpendicular to AC. Find BC (x).

$BD = \sqrt{12^{2} - 9^{2}} = \sqrt{63}$ ≈ 7.937 cm. $DC = 22 - 9 = 13$ cm. $BC = \sqrt{BD^{2} + DC^{2}} = \sqrt{63+169} = \sqrt{232}$ ≈ 15.2 cm.

A trapezium has $AB=150 m, BC=90 m, CD=120 m$ , with right angles at B and C. Show AD ≈ 95 m.

Drop perpendicular from D to AB, forming a right triangle with legs $(150-120)=30 m$ and 90 m. $AD = \sqrt{30^{2} + 90^{2}} = \sqrt{900+8100} = \sqrt{9000}$ ≈ 94.9 m, correct to nearest metre 95 m.

For the trapezium above, find the cost of a fence at $48 per 5-metre section.

Perimeter $= AB+BC+CD+AD = 150+90+120+95 = 455 m$ . Number of sections $= \frac{455}{5} = 91$ . Cost $= 91 \times$ $48 = $4368.

A right triangle has sides 5 cm, 12 cm, and 13 cm. Find its perimeter.

Perimeter $= 5 + 12 + 13 = 30$ cm.

A rectangular pond ABCD has $AB = 12 m$ , diagonal $AC = 15 m$ . Find BC.

$BC = \sqrt{15^{2} - 12^{2}} = \sqrt{225-144} = \sqrt{81} = 9 m$ .

For the pond above, find the length of fence around its edge.

Perimeter $= 2(12+9) = 42 m$ .

What common mistake should you avoid when using Pythagoras?

Ensure the hypotenuse is the longest side; if it ends up shorter than another side, you made an error. When finding a shorter side, subtract the smaller square from the larger to avoid a negative under the square root.

How can Pythagoras be used with rectangles?

The diagonal splits the rectangle into two right-angled triangles, with the diagonal as the hypotenuse. So $diagonal^{2} = length^{2} + width^{2}$ .

In a right triangle, if $a = 6$ and $b = 8$ , find c.

$c = \sqrt{6^{2} + 8^{2}} = \sqrt{36+64} = \sqrt{100} = 10$ .

In a right triangle, if $c = 17$ and $a = 15$ , find b.

$b = \sqrt{17^{2} - 15^{2}} = \sqrt{289-225} = \sqrt{64} = 8$ .

A 25 m flagpole has a rope from top to ground 8 m from base. Find rope length (exact form).

Rope $= \sqrt{25^{2} + 8^{2}} = \sqrt{625+64} = \sqrt{689} m$ .

A rectangle 14 cm by w cm has diagonal 23 cm. Find w (exact form).

$w = \sqrt{23^{2} - 14^{2}} = \sqrt{529-196} = \sqrt{333}$ cm.

A right triangle has legs 9 cm and 12 cm. Find the hypotenuse.

$c = \sqrt{9^{2} + 12^{2}} = \sqrt{81+144} = \sqrt{225} = 15$ cm.

A right triangle has hypotenuse 10 cm and one leg 6 cm. Find the other leg.

Other $leg = \sqrt{10^{2} - 6^{2}} = \sqrt{100-36} = \sqrt{64} = 8$ cm.

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