NZ Level 6 (NZC) Level 1 (NCEA)

Pythagoras in 3D

Lesson

The familiar theorem of Pythagoras applies to right-angled triangles drawn on a plane surface. In this setting, triangles are made of lines in a two-dimensional space.

We can extend the idea by considering lines in ordinary three-dimensional space.

Any single line in 3-space can belong to many different planes but when two lines meet, there is only one plane they can both belong to. In this sense, we can say that a pair of intersecting lines defines a plane.

Furthermore, if we select pairs of points, one from each intersecting line, and connect them by another line, the third line belongs to the plane defined by the first two and an ordinary triangle is formed. This is useful because if the triangle formed in this way happens to be right-angled, then Pythagoras's theorem is applicable.

Consider a box-shaped room. Its walls are perpendicular to the floor and the walls that meet are set at right-angles to one another.

The line in a corner of the room where two walls meet is perpendicular to any line drawn on the floor. So, we can make a right-angled triangle as in the following diagram.

In this picture, there are two right-angled triangles that will be useful.

The diagonal across the floor makes a right-angled triangle with the lines formed where the floor meets two of the walls.

The body diagonal from the floor to the ceiling through the room makes a right-angled triangle with the floor diagonal and the line formed by the intersection of two of the walls.

Thus, there are two right-angled triangles and the floor diagonal is common to both. This means we should be able to calculate the length of the body diagonal in two steps if we know the wall dimensions. This is done in the working after the following diagram.

Firstly, because of the right-angled triangle on the floor of the room, we have $x^2=d^2+w^2$`x`2=`d`2+`w`2.

Next, because of the right-angled triangle cutting diagonally through the room, we have $y^2=x^2+h^2$`y`2=`x`2+`h`2.

Putting these results together, we conclude that

$y^2=d^2+w^2+h^2$`y`2=`d`2+`w`2+`h`2

How long is the body diagonal of a box with dimensions $12$12 cm, $19$19 cm and $7$7 cm?

A diagram like the one above would have dimensions $d=12$`d`=12, $w=19$`w`=19 and $h=7$`h`=7. So, the diagonal would be given by $y^2=12^2+19^2+7^2$`y`2=122+192+72. Thus, $y^2=554$`y`2=554.

The final step in finding the diagonal is to take the square root. So, $y=\sqrt{554}\approx23.5$`y`=√554≈23.5 cm.

A square prism has sides of length $11$11cm, $11$11cm and $15$15cm as shown.

If the diagonal $HF$

`H``F`has a length of $z$`z`cm, calculate the exact length of $z$`z`, leaving your answer in surd form.Now, we want to find $y$

`y`, the length of the diagonal $DF$`D``F`.Calculate $y$

`y`to two decimal places.

A triangular divider has been placed into a box, as shown in the diagram.

If the length of the base of the divider $AC$

`A``C`is $z$`z`cm long, calculate the exact value of $z$`z`.Use the exact value of $z$

`z`from part (a) to calculate the length of the diagonal $AD$`A``D`correct to two decimal places.Calculate the area of the triangular divider correct to two decimal places.

Use trigonometric ratios and Pythagoras’ theorem in two and three dimensions

Apply geometric reasoning in solving problems

Apply right-angled triangles in solving measurement problems