7.09 Pythagoras in 3D

We can identify right-angled triangles in 3D space the same way that we do in 2D space, by looking for a right angle.

In 3D space, right angles occur between lines that are perpendicular in the same 2D plane. In other words, if two lines meet at a right-angle in some 2D slice of 3D space, the triangle formed with these two lines is a right-angled triangle.

Consider this cube with its vertices labelled.

If we take the 2D slice through the vertices A, B, C and D we can see that any triangle determined by three of these vertices will be right-angled, since this slice gives us a square.

In fact, any of the 2D slices that corresponds to a face of the cube will give us multiple possible right-angled triangles.

Taking a 2D slice through the vertices F, \,H,\,B and D gives us a rectangle, which means that we still form a right-angled triangle from any three of these vertices.

If we return to 3D space, we can see how these right-angled triangles look in 3D.

We can perform a similar exercise with other 3D solids like cones and prisms to find other right-angled triangles in 3D space:

In most straight-edged solids, the diagonals formed by joining two non-adjacent vertices will often be the hypotenuse of some right-angled triangle in that solid.

In order to find the length of such a diagonal, we can simply use Pythagoras' theorem to calculate the hypotenuse, provided we know the lengths of the other two sides.

In the cases where we don't know both the other side lengths, we may need to use Pythagoras' theorem on a different right-angled triangle in the solid, as seen in the example below.

Examples

Example 1

A square prism has sides of length 9\,cm, 9\,cm and 16\,cm as shown.

a

If the diagonal HF has a length of z\,cm, calculate the exact value of z, leaving your answer in surd form.

Worked Solution

Create a strategy

We can use Pythagoras' theorem to find the length of HF in the triangle \triangle FGH.

Apply the idea

\displaystyle HF^2	\displaystyle =	\displaystyle HG^2+GF^2	Use the Pythagoras' theorem
\displaystyle z^2	\displaystyle =	\displaystyle 9^2+16^2	Substitute the lengths
\displaystyle z^2	\displaystyle =	\displaystyle 337	Evaluate the right side
\displaystyle z	\displaystyle =	\displaystyle \sqrt{337}	Take the square root of both sides

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b

Find y, the length of the diagonal DF to two decimal places.

Worked Solution

Create a strategy

We can use Pythagoras' theorem to find the length of DF in the triangle \triangle DHF.

Apply the idea

Consider the right-angled \triangle FDH. The hypotenuse can be found using Pythagoras' Theorem, c^2=a^2+b^2.

\displaystyle DF^2	\displaystyle =	\displaystyle DH^2+HF^2	Use the Pythagoras' theorem
\displaystyle y^2	\displaystyle =	\displaystyle 9^2+(\sqrt{337})^2	Substitute the lengths
\displaystyle y^2	\displaystyle =	\displaystyle 81+337	Evaluate the powers
\displaystyle y^2	\displaystyle =	\displaystyle 418	Evaluate the addition
\displaystyle y	\displaystyle =	\displaystyle \sqrt{418}	Take the square root of both sides
	\displaystyle =	\displaystyle 20.45	Evaluate

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Example 2

A soft drink can has a height of 13\,cm and a radius of 5\,cm. Find L, the length of the longest straw that can fit into the can (so that the straw is not bent and fits entirely inside the can).

Round your answer down to the nearest centimetre, to ensure it fits inside the can.

Worked Solution

Create a strategy

Use the Pythagoras' theorem to find the length L using the height and radius of the can.

Apply the idea

L is the hypotenuse of the right-angled triangle. The other sides are the height which is 13 and the diameter of the can which is 2 \times 5=10.

\displaystyle c^2	\displaystyle =	\displaystyle a^2+b^2	Use the Pythagoras' theorem
\displaystyle L^2	\displaystyle =	\displaystyle 13^2+(10)^2	Substitutes L, the height and diameter
\displaystyle L^2	\displaystyle =	\displaystyle 269	Evaluate the right side
\displaystyle L	\displaystyle =	\displaystyle \sqrt{269}	Take the square root of both sides
	\displaystyle \approx	\displaystyle 16.4	Evaluate

The length of the longest straw to the nearest centimetre is 16 cm.