### Video Transcript

Calculate, to two decimal places, the area of the triangle πππ
, where the coordinates of its vertices are at the point π: four, zero, two; the point π: two, one, five; and the point π
: negative one, zero, one.

In this question, weβre asked to find the area of a triangle πππ
. And to find the area of this triangle, weβre given the coordinates of its three vertices: the point π, the point π, and the point π
. And we need to give our answer to two decimal places. Usually, when weβre asked to find the area of a triangle, the first thing we will think of is a half the base multiplied by the height. However, itβs difficult to see how we would find these values for our triangle. Weβre only given the coordinates of its three vertices, so itβs going to be difficult to find the angles of our triangle. However, there is another way of calculating the area of a triangle. We need to recall Heronβs formula. Heronβs formula gives us a method of calculating the area of a triangle when all we know is the length of its three sides.

We recall if we have a triangle with side lengths π, π, and π and we let π be equal to π plus π plus π all divided by two, then Heronβs formula tells us the area of this triangle is the square root of π times π minus π times π minus π multiplied by π minus π. And this is really useful in this case because it gives us a way of finding the area of a triangle when we donβt know the angles of our triangle. However, we do need to know the lengths of our triangle. And the lengths of the triangle πππ
are going to be the distance between its vertices. And we know how to find the distance between two points in three dimensions, so we can find the length of all three sides of our triangle πππ
.

Letβs start by recalling how we find the distance between two points in three dimensions. The distance between the points π₯ sub one, π¦ sub one, π§ sub one and π₯ sub two, π¦ sub two, π§ sub two is given by the square root of π₯ sub two minus π₯ sub one all squared plus π¦ sub two minus π¦ sub one all squared plus π§ sub two minus π§ sub one all squared. Now, we can use this to find the length of the sides of our triangle πππ
. And we might want to sketch this triangle. However, this will be difficult because itβs given in three dimensions. And if we instead just wanted to sketch our triangle in two dimensions, this would also be difficult because we donβt know the angles of our triangle. And itβs not necessary anyway because weβre just applying Heronβs formula, so weβre not going to sketch this in this case.

Letβs start by finding the distance between points π and π. Weβll represent this as π sub ππ. And we can find this by using our formula for the distance between two points. We square the difference in the π₯-coordinates. Then we add onto this the square of the distance between the π¦-coordinates. Then we add onto this the square of the distance between the π§-coordinates. And we take the square root of this value. The distance between points π and π is the square root of two minus four all squared plus one minus zero all squared plus five minus two all squared. And if we calculate this expression, we can simplify it to give us the square root of 14.

Weβll now do exactly the same thing to find the distance between point π and π
. Applying our formula, we get the square root of negative one minus four all squared plus zero minus zero all squared plus one minus two all squared. And once again, we can simplify this expression. Weβll see that itβs equal to the square root of 26. Finally, we want to find the length of the last side of our triangle. Thatβs going to be the distance between the two vertices π and π
.

Once again, we just apply our formula. We get that this distance is equal to the square root of negative one minus two all squared plus zero minus one all squared plus one minus five all squared. And we can evaluate everything inside the square root symbol. We get that the distance between π and π
is root 26, and this gives us an interesting piece of information. We can see that our triangle πππ
is isosceles because two of the sides have the same length.

We can now use this to sketch the triangle πππ
. However, itβs not necessary when weβre using Heronβs formula. So instead, letβs just find the value for our area. Letβs start by finding our value of π . We know our value of π is the perimeter of our triangle divided by two. And the perimeter is the sum of the length of our triangle. And we just found these. Theyβre root 14, root 26, and root 26, so π is root 14 plus root 26 plus root 26 all divided by two. And we can simplify the expression in our numerator since root 26 plus root 26 is equal to two root 26.

Now, all we need to do is evaluate the expression for the area. Weβll start by substituting the values of π, π, and π into our formula. Thatβs the length of the sides of our triangle. We get the area is equal to the square root of π times π minus root 14 multiplied by π minus root 26 times π minus root 26. And at this point, we can simplify because we see we have π minus root 26 multiplied by π minus root 26. We can just write this as π minus root 26 all squared. This gives us the following expression for our area. However, we can simplify this even further.

We now see that we have a square inside of our square root symbol. So instead, we can just take the square root of the first two factors of this and then remove the square. And it can be useful at this point to remember weβre calculating an area, which has to be positive. And luckily, weβve already calculated the value of π and we know that itβs bigger than negative root 26. So weβve simplified our formula to the area to be the square root of π times π minus root 14 all multiplied by π minus root 26.

At this point, itβs possible to substitute our expression for π into our formula for the area and then simplify this by using our laws of radicals to find an exact value for the area. However, itβs not necessary to do this because we only need to give our answer to two decimal places. So instead, we can just write this into our calculator. However, if we were to find this area exactly, we would see that itβs three times the square root of 35 all divided by two, which to two decimal places is 8.87, which is our final answer.

In this question, we were able to find the area of a triangle in three dimensions only given the coordinates of its vertices. All we had to do was use our distance formula to find the length of all of the sides of our triangle and then apply Heronβs formula to find the area. We found that the area of triangle πππ
to two decimal places was 8.87.