Question

Find the perimeter of the triangle.

Answer 1

The perimeter of a triangle is the sum of its side lengths so we first need to find the three side lengths of the triangle.
Since the side from (−3,−1) to (2,−1) is a horizontal segment, we can count the number of units between the two points to find its length. Since 2−(−3)=5, then the length of this side is 5.
Since the side from (2,−1) to (2,3) is a vertical segment, we can count the number of units between the two points to find its length. Since 3−(−1)=4, then the length of this side is 4.
The side between from (-3,-1) to (2,3) is not a horizontal or vertical segment so we need to use the distance formula \(\displaystyle{d}=\sqrt{{{\left({x}{2}-{x}{1}\right)}^{{{2}}}+{\left({y}{2}-{y}{1}\right)}^{{{2}}}}}\) to find the lenght of this side:
\(\displaystyle{d}=\sqrt{{{\left(-{3}-{2}\right)}^{{{2}}}+{\left(-{1}-{3}\right)}^{{{2}}}}}\) \ Substitute.
\(\displaystyle=\sqrt{{{\left(-{5}\right)}^{{{2}}}+{\left(-{4}\right)}^{{{2}}}}}\) \ Substract.
\(\displaystyle=\sqrt{{{\left({25}+{16}\right)}}}\) \ Evaluate the powers.
\(\displaystyle=\sqrt{{{41}}}\) \ Add.
The length of the third side is then \(\displaystyle√{41}≈{6.4}.\)
The perimeter of the triangle is then about 5+4+6.4=15.4. The exact perimeter is \(\displaystyle{5}+{4}+\sqrt{{{41}}}={9}+\sqrt{{{41}}}\)