Question

A triangle has vertices A(2,1,6), B(4,7,9), and C(8,5,-6). Determine the area of the triangle. Prove that the triangle is a right angle triangle. ?

Answer 1

The Pythagorean Theorem works great for 3D points.

`|AB|^2=(4-2)^2+(7-1)^2+(9-6)^2=2^2+6^2+3^2=49`

`|BC|^2=(8-4)^2+(5-7)^2+(-6-9)^2= 245`

`|AC|^2=(8-2)^2+(5-1)^2+(-6 -6)^2=196`

Since `|BC|^2=|AB|^2+|AC|^2` we have a right triangle, right angle A, by the converse of the Pythagorean Theorem.

Answer 2

One robust way to determine the area is to use Heron's formula (http://www.mathwarehouse.com/geometry/triangles/area/herons-formula-triangle-area.php), which works for non-right-angled triangles also.

However, the hint in the second part of the question that this triangle is right-angled tells us that it will be easier to simply find the two-sides adjacent to the right-angle and then use the triangle area formula `A=1/2bh` on them - area = half base x height.

Find side lengths: show right-angled by Pythagoras' Theorem

Pythagoras' Theorem tells us that in a right-angled triangle (and only in right-angled triangles), the three side lengths `a` `b` `c` relate with `a^2+b^2=c^2`. So showing that the lengths of the sides here fit this formula will tell us that the triangle is right-angled. Use the distance formula between two points in 3D (which is itself a simple application of Pythagoras' Theorem) to work the side lengths out:
`d=sqrt((x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2)`

Length of side `ul(AB)`:
`ul(AB)=sqrt((4-2)^2+(7-1)^2+(9-6)^2)=sqrt(4+36+9)=sqrt(49)=7`
Length of side `ul(AC)`:
`ul(AC)=sqrt((8-2)^2+(5-1)^2+(-6-6)^2)=sqrt(36+16+144)=sqrt(196)=14`
Length of side `ul(BC)`:
`ul(BC)=sqrt((8-4)^2+(5-7)^2+(-6-9)^2)=sqrt(16+4+225)=sqrt(245)=7sqrt(5)`

Looking at the squares of the side lengths, we see that indeed `ul(AB)^2+ul(AC)^2=ul(BC)^2`. The long side is `ul(BC)`, so this is the hypotenuse, and the right-angle is in the corner of the triangle at point `A`.

Find triangle area

Now this is made very simple - as the triangle is right-angled, the two sides adjacent to the right-angle are the base and height (note that these are simply terms - the orientation of the actual triangle doesn't matter).

So
`A=1/2bh=1/2*ul(AB)*ul(AC)=1/2*7*14=49` `units^2`