Answer:
a) The median AD from A to BC has a length of 6.
b) Areas of triangles ABD and ACD are the same.
Step-by-step explanation:
a) A median is a line that begin in a vertix and end at a midpoint of a side opposite to vertix. As first step the location of the point is determined:
[tex]D (x,y) = \left(\frac{x_{B}+x_{C}}{2},\frac{y_{B}+y_{C}}{2} \right)[/tex]
[tex]D(x,y) = \left(\frac{3 + 5}{2},\frac{-2 + 2}{2} \right)[/tex]
[tex]D(x,y) = (4,0)[/tex]
The length of the median AD is calculated by the Pythagorean Theorem:
[tex]AD = \sqrt{(x_{D}-x_{A})^{2}+ (y_{D}-y_{A})^{2}}[/tex]
[tex]AD = \sqrt{(4-4)^{2}+[0-(-6)]^{2}}[/tex]
[tex]AD = 6[/tex]
The median AD from A to BC has a length of 6.
b) In order to compare both areas, all lengths must be found with the help of Pythagorean Theorem:
[tex]AB = \sqrt{(x_{B}-x_{A})^{2}+ (y_{B}-y_{A})^{2}}[/tex]
[tex]AB = \sqrt{(3-4)^{2}+[-2-(-6)]^{2}}[/tex]
[tex]AB \approx 4.123[/tex]
[tex]AC = \sqrt{(x_{C}-x_{A})^{2}+ (y_{C}-y_{A})^{2}}[/tex]
[tex]AC = \sqrt{(5-4)^{2}+[2-(-6)]^{2}}[/tex]
[tex]AC \approx 4.123[/tex]
[tex]BC = \sqrt{(x_{C}-x_{B})^{2}+ (y_{C}-y_{B})^{2}}[/tex]
[tex]BC = \sqrt{(5-3)^{2}+[2-(-2)]^{2}}[/tex]
[tex]BC \approx 4.472[/tex]
[tex]BD = CD = \frac{1}{2}\cdot BC[/tex] (by the definition of median)
[tex]BD = CD = \frac{1}{2} \cdot (4.472)[/tex]
[tex]BD = CD = 2.236[/tex]
[tex]AD = 6[/tex]
The area of any triangle can be calculated in terms of their side length. Now, equations to determine the areas of triangles ABD and ACD are described below:
[tex]A_{ABD} = \sqrt{s_{ABD}\cdot (s_{ABD}-AB)\cdot (s_{ABD}-BD)\cdot (s_{ABD}-AD)}[/tex], where [tex]s_{ABD} = \frac{AB+BD+AD}{2}[/tex]
[tex]A_{ACD} = \sqrt{s_{ACD}\cdot (s_{ACD}-AC)\cdot (s_{ACD}-CD)\cdot (s_{ACD}-AD)}[/tex], where [tex]s_{ACD} = \frac{AC+CD+AD}{2}[/tex]
Finally,
[tex]s_{ABD} = \frac{4.123+2.236+6}{2}[/tex]
[tex]s_{ABD} = 6.180[/tex]
[tex]A_{ABD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}[/tex]
[tex]A_{ABD} \approx 3.004[/tex]
[tex]s_{ACD} = \frac{4.123+2.236+6}{2}[/tex]
[tex]s_{ACD} = 6.180[/tex]
[tex]A_{ACD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}[/tex]
[tex]A_{ACD} \approx 3.004[/tex]
Therefore, areas of triangles ABD and ACD are the same.
