Video Transcript
A triangle has vertices at the
points 𝐴, 𝐵, and 𝐶 with coordinates two, negative two; four, negative two; and
zero, two respectively. Work out the perimeter of the
triangle 𝐴𝐵𝐶. Give your solution to two decimal
places. Work out the area of triangle
𝐴𝐵𝐶.
So, we are asked to do two
different things: find the perimeter and find the area of triangle 𝐴𝐵𝐶. First, let’s draw a quick
sketch. We have point 𝐴 at two, negative
two; point 𝐵 at four, negative two; and point 𝐶 at zero, two. So first, we are asked to find the
perimeter. So, we need to know the length of
each side. This means we’re going to have to
use the distance formula.
The distance formula is the square
root of 𝑥 two minus 𝑥 one squared plus 𝑦 two minus 𝑦 one squared. And to find the perimeter, we need
to take the length of 𝐴𝐵 plus the length of 𝐵𝐶 plus the length of 𝐴𝐶 and add
them together. However, we can actually already
find the length of 𝐴𝐵 because it is parallel to the 𝑥-axis and it has a length of
two. So, we just need to use the
distance formula to find the length of 𝐵𝐶 and 𝐴𝐶.
So, to find the length of 𝐵𝐶,
point 𝐵 can be our 𝑥 one, 𝑦 one and point 𝐶 can be our 𝑥 two, y two. And now we plug that into our
formula, the distance formula. So, we need to take zero minus four
and square it plus two minus negative two and square that. And it’s all underneath the square
root. Zero minus four is four and two
minus negative two is really two plus two. So, that’s four.
So, we have negative four squared,
which is positive 16, and positive four squared, which is positive 16. And be aware, when you square a
negative, it is a positive. 16 plus 16 is 32. Now when we take the square root of
32, we get about 5.6568. Now we are supposed to round our
solution to two decimal places. So, let’s use two decimal
places. So, we will either keep the five a
five or round it up to a six. We’ll look at the number to the
right, which is a six. So, the five will round up to a
six. So, the length of 𝐵𝐶 is about
5.66.
Now let’s find the length of
𝐴𝐶. So, to find the length of 𝐴𝐶, 𝐴
can be 𝑥 one, 𝑦 one and 𝐶 can be 𝑥 two, 𝑦 two. So, plugging this in, we take the
square root of zero minus two squared plus two minus negative two squared. Zero minus two is negative two and
two minus negative two is positive four. Negative two squared will be
positive four and four squared will be 16. And four plus 16 is 20. And the square root of 20 is about
4.472. So, after rounding two decimal
places, we have around 4.47 for the length of 𝐴𝐶.
So, now to find the perimeter, we
simply add them all together. Two plus 5.66 plus 4.47, making the
perimeter 12.13.
Now to find the area, we need to
take one-half times the length of the base times the height of the triangle. Our base and our height need to be
perpendicular. So, we can use 𝐴𝐵 as our
base. And we know that that length is
two. And then our height needs to be
perpendicular to that.
So how tall is this triangle, if we
use 𝐴𝐵 as the base? It will be this tall, which would
be four units. And that height is perpendicular to
our base. So, we take one-half times two
times four. The twos can cancel. And we’re left with an area of
four.Therefore, again, the perimeter is 12.13 and the area is four.
