# Question Video: Finding the Perimeter and Area of a Triangle Mathematics

A triangle has vertices at the points 𝐴, 𝐵, and 𝐶 with coordinates (2, −2), (4, −2), and (0, 2) respectively. Work out the perimeter of the triangle 𝐴𝐵𝐶. Give your solution to two decimal places. Work out the area of triangle 𝐴𝐵𝐶.

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### 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.

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