Question

the vertices of a triangle are d(-4,5), e(-7,-1), and f(-1,-1). a) determine the lengths of the sides of the triangle. b) classify △def. explain your reasoning. c) determine the perimeter of △def. round your answer to the nearest tenth of a unit.

Explanation:

Step1: Use distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
For side $DE$ with $D(-4,5)$ and $E(-7,-1)$:
$d_{DE}=\sqrt{(-7 + 4)^2+(-1 - 5)^2}=\sqrt{(-3)^2+(-6)^2}=\sqrt{9 + 36}=\sqrt{45}\approx6.7$

Step2: Calculate length of $EF$

For side $EF$ with $E(-7,-1)$ and $F(-1,-1)$:
$d_{EF}=\sqrt{(-1+7)^2+(-1 + 1)^2}=\sqrt{(6)^2+0^2}=6$

Step3: Calculate length of $FD$

For side $FD$ with $F(-1,-1)$ and $D(-4,5)$:
$d_{FD}=\sqrt{(-4 + 1)^2+(5 + 1)^2}=\sqrt{(-3)^2+(6)^2}=\sqrt{9+36}=\sqrt{45}\approx6.7$

Step4: Classify the triangle

Since $d_{DE}\approx6.7$, $d_{EF}=6$, $d_{FD}\approx6.7$, two sides are approximately equal, so $\triangle DEF$ is an isosceles triangle.

Step5: Calculate the perimeter

$P=d_{DE}+d_{EF}+d_{FD}\approx6.7 + 6+6.7=19.4$

Answer:

a) $DE\approx6.7$, $EF = 6$, $FD\approx6.7$
b) Isosceles triangle because two sides have approximately equal lengths.
c) $19.4$