A triangle has vertices at the points 𝐴, 𝐵, and 𝐶 with the coordinates three, three; negative one, three; and seven, negative six, respectively. Work out the perimeter of triangle 𝐴𝐵𝐶. Give your answer solution to two decimal places.
We’re trying to work out the perimeter of this triangle. This means we need to add the distance from 𝐴𝐵 to the distance from 𝐵 to 𝐶 to the distance from 𝐶 to 𝐴, side one of the triangle plus side two of the triangle plus side three of the triangle. Together they’ll equal the perimeter. But to do that, we’ll need to know how to find the distance between two points. And we have a formula for calculating the distance between two points. The distance between two points is equal to the square root of 𝑥 two minus 𝑥 one squared plus 𝑦 two minus 𝑦 one squared. We know that 𝐴 is located at three, three. And 𝐵 is located at negative one, three, while 𝐶 is found at seven, negative. Six. If we want to find the distance from 𝐴 to 𝐵, we can let 𝐴 be 𝑥 one, 𝑦 one and 𝐵 be 𝑥 two, 𝑦 two. 𝑥 two minus 𝑥 one is negative one minus three. Then square that value. 𝑦 two minus 𝑦 one, in this case, would be three minus three. The 𝑦-coordinates of both 𝐴 and 𝐵 are three.
Negative one minus three equals negative four. We need to square that value. Three minus three equals zero. And zero squared equals zero. And that means we can drop that term. We need to square negative four which is 16 and the square root of 16 is four. This means the distance from point 𝐴 to point 𝐵 in this triangle is four units. We’ll follow the same procedure to find the distance from 𝐵 to 𝐶. In this case, we label 𝐵 as 𝑥 one, 𝑦 one and 𝐶 as 𝑥 two, 𝑦 two. After we plug in our values, we have the square root of seven minus negative one squared plus negative six minus three squared. Seven minus negative one is seven plus one, so we have eight squared. Negative six minus three is negative nine, so we have negative nine squared.
We’ll take the square root of eight squared plus negative nine squared. Eight squared is 64. Nine squared, or negative nine squared, is 81. Now, we’ll need to add 64 and 81 which is 145. Line 𝐵𝐶 has a length of the square root of 145. We will round our answer solution to two decimal places. But for now, we’ll keep it as the square root of 145 and round in our final step. We now need to find the distance from 𝐴 to 𝐶 or from 𝐶 to 𝐴. Either way will work as long as you’re consistent in you’re labeling. We’ll let 𝐴 be 𝑥 one, 𝑦 one and 𝐶 be 𝑥 two, 𝑦 two. The distance from 𝐴 to 𝐶 will be equal to the square root of seven minus three squared plus negative six minus three squared. Seven minus three is four, so we’ll have four squared. Negative six minus three is negative nine. We have four squared plus negative nine squared. Negative nine squared equals 81. Four squared equals 16. And so we can say that the distance from 𝐴 to 𝐶 is the square root of 97. 16 plus 81 is 97.
At this point, we could find the perimeter. The perimeter equals four plus the square root of 145 plus the square root of 97. This method, using the distance between two points’ formula does not require us to sketch a picture. However, sometimes it’s helpful to use a coordinate plane to see what’s going on. If we sketch a coordinate plane, we can plot the three points of our triangle 𝐴 located at three, three; 𝐵 located at negative one, three; and 𝐶 located at seven, negative six. From there, we need to connect the three points, 𝐴𝐵, 𝐵𝐶, and 𝐴𝐶. By sketching the graph of this triangle, we can find the distance from 𝐴 to 𝐵 visually. 𝐴 to 𝐵 is four units because they fall on the same horizontal line. Instead of doing the long calculation, we could have just inspected to see that it was four units.
Now, we can’t do that kind of inspection, even on the graph from 𝐴 to 𝐶 or from 𝐵 to 𝐶. But there’s something else we can do here. We can create a right triangle that has a high hypotenuse of 𝐵𝐶. The vertical distance from 𝐵 to 𝐶 is from negative six to three. And that’s nine units. The horizontal distance from 𝐵 to 𝐶 is from negative one to seven, which is eight units. To find the length 𝐵𝐶, we could say that 𝐵𝐶 squared equals nine squared plus eight squared. We’re using the Pythagorean theorem to find the length of 𝐵𝐶. Nine squared is 81; eight squared is 64; 𝐵𝐶 squared equals 145. We take the square root of both sides and we see that 𝐵𝐶 equals the square root of 145. And what you’re seeing is that the distance formula is actually the Pythagorean theorem rearranged.
The distance formula takes the vertical distance and squares it, the horizontal distance and squares it, and then takes the square root. We could also do this to find the distance from 𝐴 to 𝐶. The distance from 𝐴 to 𝐶 squared will be equal to the horizontal distance of four, four squared, plus the vertical distance of nine squared. Again we’ll have 16 plus 81. 𝐴𝐶 squared equals 97, and if we take the square root of both sides, we confirm that 𝐴𝐶 does in fact equal the square root of 97. Back to our original goal of finding the perimeter, if we add four plus the square root of 145 plus the square root of 97, our calculator will give us something like 28.890452 continuing. We’re rounding to two decimal places to the hundredths place we look to the digit to the right of the hundredths place to do our rounding, which in this case is a zero. So we’ll round down. And we’ll say that the perimeter of this triangle is 28.89 units, 28 and 89 hundredths.