Question

select the correct answer from each drop - down menu. △abc has the points a(1, 7), b(-2, 2), and c(4, 2) as its vertices. the measure of the longest side of △abc is units. △abc is triangle. if △abd is formed with the point d(1, 2) as its third vertex, then △abd is triangle. the length of side ad is units.

Explanation:

Step1: Calculate side - lengths of $\triangle ABC$ using distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$

For side $AB$ with $A(1,7)$ and $B(-2,2)$:
$AB=\sqrt{(-2 - 1)^2+(2 - 7)^2}=\sqrt{(-3)^2+(-5)^2}=\sqrt{9 + 25}=\sqrt{34}$
For side $BC$ with $B(-2,2)$ and $C(4,2)$:
$BC=\sqrt{(4+2)^2+(2 - 2)^2}=\sqrt{6^2+0^2}=6$
For side $AC$ with $A(1,7)$ and $C(4,2)$:
$AC=\sqrt{(4 - 1)^2+(2 - 7)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9 + 25}=\sqrt{34}$
The longest side of $\triangle ABC$ is $\sqrt{34}$ units. Since $AB = AC=\sqrt{34}$ and $BC = 6$, $\triangle ABC$ is an isosceles triangle.

Step2: Calculate side - lengths of $\triangle ABD$

For side $AD$ with $A(1,7)$ and $D(1,2)$:
$AD=\sqrt{(1 - 1)^2+(2 - 7)^2}=\sqrt{0+(-5)^2}=5$
For side $BD$ with $B(-2,2)$ and $D(1,2)$:
$BD=\sqrt{(1 + 2)^2+(2 - 2)^2}=\sqrt{3^2+0^2}=3$
For side $AB=\sqrt{34}$ (calculated above)
Since $AB^2=34$, $AD^2 = 25$, $BD^2=9$ and $AD^2+BD^2=25 + 9=34=AB^2$, $\triangle ABD$ is a right - triangle.

Answer:

The measure of the longest side of $\triangle ABC$ is $\sqrt{34}$ units. $\triangle ABC$ is an isosceles triangle. $\triangle ABD$ is a right - triangle. The length of side $AD$ is $5$ units.