Pythagoras' Theorem in Two and Three Dimensions

Pythagoras' theorem in 2 dimensions

Pythagoras' theorem is a rule that applies specifically to right-angled triangles. It describes the relationship between the lengths of the three sides. The longest side of a right-angled triangle is called the hypotenuse, and it is always located directly opposite the right angle.

The theorem states that: $a^2 + b^2 = c^2$

In the context of the ESAT, you must be able to perform several tasks using this theorem:

1. Find a missing side length when the other two sides are known.
2. Calculate the distance between two points on a coordinate grid.
3. Determine if a triangle is right-angled by checking if the side lengths satisfy the theorem.
4. Find missing lengths in composite shapes that contain right angles.

Pythagoras' theorem in 3 dimensions

Pythagoras' theorem can be extended to find lengths in three-dimensional shapes. A common application is finding the internal diagonal of a cuboid, which is the line segment connecting two opposite corners through the centre of the shape.

The extended formula for the diagonal $d$ of a cuboid with dimensions $a$, $b$, and $c$ is: $a^2 + b^2 + c^2 = d^2$

You should be prepared to find missing lengths in various 3D shapes and solve multi-step problems that may involve keeping values in surd form for accuracy.

Finding a missing side

When calculating a missing side, the method depends on whether you are looking for the hypotenuse or one of the shorter sides.

Case 1: Finding the hypotenuse In triangle ABC below, we need to find the length of the hypotenuse, AC.

$AC^2 = 8^2 + 12^2$ $AC^2 = 64 + 144 = 208$ $AC = \sqrt{208}$ $AC = 4\sqrt{13} \text{ cm}$

Case 2: Finding a shorter side In triangle DEF, we know the hypotenuse is 17 and one side is 8. We need to find DF.

$17^2 = 8^2 + DF^2$ $DF^2 = 17^2 - 8^2$ $DF^2 = 289 - 64 = 225$ $DF = \sqrt{225} = 15$

Distance between two points given as coordinates

To find the distance between two points M $(-1, -1)$ and N $(2, 3)$, you should treat the line segment MN as the hypotenuse of a right-angled triangle.

By sketching the points, you can find the horizontal and vertical distances. The horizontal distance is $2 - (-1) = 3$. The vertical distance is $3 - (-1) = 4$.

$MN^2 = 3^2 + 4^2$ $MN^2 = 9 + 16 = 25$ $MN = \sqrt{25} = 5$

Identifying whether or not a triangle is right-angled

We can use the converse of Pythagoras' theorem to check if a triangle contains a right angle. If $a^2 + b^2 = c^2$ holds true for the three sides, the triangle is right-angled.

In triangle PQR, the longest side is 17. We check if the square of 17 equals the sum of the squares of 15 and 8: $PQ^2 + QR^2 = 15^2 + 8^2 = 225 + 64 = 289$ $PR^2 = 17^2 = 289$

Because $PQ^2 + QR^2 = PR^2$, the theorem applies and triangle PQR is right-angled.

Missing lengths in other shapes containing right angles

Complex shapes can often be broken down into simpler right-angled triangles. Consider finding the perimeter of the trapezium ABCD.

To find the unknown side $x$, we drop a vertical line to form a right-angled triangle. This reveals a triangle with a hypotenuse of 7 and a base of 2 (since $5 - 3 = 2$).

$7^2 = 2^2 + x^2$ $x^2 = 49 - 4 = 45$ $x = \sqrt{45} = 3\sqrt{5}$

The total perimeter is the sum of all outer sides: $3 + 2 + 3 + 3\sqrt{5} + 7 = 15 + 3\sqrt{5} \text{ cm}$.

Missing lengths in 3-dimensional shapes

To find the internal diagonal $d$ of a cuboid with sides 3 cm, 4 cm, and 12 cm, you can use two methods.

Method 1: The 3D Formula $d^2 = a^2 + b^2 + c^2$ $d^2 = 3^2 + 4^2 + 12^2 = 9 + 16 + 144 = 169$ $d = \sqrt{169} = 13 \text{ cm}$

Method 2: Using 2D Pythagoras twice First, find the diagonal of the base (let's call it $a$): $a^2 = 3^2 + 12^2 = 9 + 144 = 153$ Then, use this base diagonal and the height to find $d$: $d^2 = a^2 + 4^2 = 153 + 16 = 169$ $d = \sqrt{169} = 13 \text{ cm}$

Complex problems in 3 dimensions

Pythagoras' theorem is also vital for finding heights in pyramids. Consider a square-based pyramid where F is the centre of the base and E is the apex directly above F.

To find the vertical height FE:

1. Find the diagonal of the square base ACD: $AC^2 = 4^2 + 4^2 = 32$. Therefore, $AC = \sqrt{32} = 4\sqrt{2}$.
2. The distance from a corner to the centre (AF) is half of AC: $AF = 2\sqrt{2}$.
3. Use triangle AFE to find FE, where AE is given as $\sqrt{17}$: $AE^2 = AF^2 + FE^2$ $(\sqrt{17})^2 = (2\sqrt{2})^2 + FE^2$ $17 = 8 + FE^2$ $FE^2 = 9$ $FE = 3 \text{ cm}$