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:
![D (x,y) = \left(\frac{x_{B}+x_{C}}{2},\frac{y_{B}+y_{C}}{2} \right)](https://tex.z-dn.net/?f=D%20%28x%2Cy%29%20%3D%20%5Cleft%28%5Cfrac%7Bx_%7BB%7D%2Bx_%7BC%7D%7D%7B2%7D%2C%5Cfrac%7By_%7BB%7D%2By_%7BC%7D%7D%7B2%7D%20%20%5Cright%29)
![D(x,y) = \left(\frac{3 + 5}{2},\frac{-2 + 2}{2} \right)](https://tex.z-dn.net/?f=D%28x%2Cy%29%20%3D%20%5Cleft%28%5Cfrac%7B3%20%2B%205%7D%7B2%7D%2C%5Cfrac%7B-2%20%2B%202%7D%7B2%7D%20%20%5Cright%29)
![D(x,y) = (4,0)](https://tex.z-dn.net/?f=D%28x%2Cy%29%20%3D%20%284%2C0%29)
The length of the median AD is calculated by the Pythagorean Theorem:
![AD = \sqrt{(x_{D}-x_{A})^{2}+ (y_{D}-y_{A})^{2}}](https://tex.z-dn.net/?f=AD%20%3D%20%5Csqrt%7B%28x_%7BD%7D-x_%7BA%7D%29%5E%7B2%7D%2B%20%28y_%7BD%7D-y_%7BA%7D%29%5E%7B2%7D%7D)
![AD = \sqrt{(4-4)^{2}+[0-(-6)]^{2}}](https://tex.z-dn.net/?f=AD%20%3D%20%5Csqrt%7B%284-4%29%5E%7B2%7D%2B%5B0-%28-6%29%5D%5E%7B2%7D%7D)
![AD = 6](https://tex.z-dn.net/?f=AD%20%3D%206)
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:
![AB = \sqrt{(x_{B}-x_{A})^{2}+ (y_{B}-y_{A})^{2}}](https://tex.z-dn.net/?f=AB%20%3D%20%5Csqrt%7B%28x_%7BB%7D-x_%7BA%7D%29%5E%7B2%7D%2B%20%28y_%7BB%7D-y_%7BA%7D%29%5E%7B2%7D%7D)
![AB = \sqrt{(3-4)^{2}+[-2-(-6)]^{2}}](https://tex.z-dn.net/?f=AB%20%3D%20%5Csqrt%7B%283-4%29%5E%7B2%7D%2B%5B-2-%28-6%29%5D%5E%7B2%7D%7D)
![AB \approx 4.123](https://tex.z-dn.net/?f=AB%20%5Capprox%204.123)
![AC = \sqrt{(x_{C}-x_{A})^{2}+ (y_{C}-y_{A})^{2}}](https://tex.z-dn.net/?f=AC%20%3D%20%5Csqrt%7B%28x_%7BC%7D-x_%7BA%7D%29%5E%7B2%7D%2B%20%28y_%7BC%7D-y_%7BA%7D%29%5E%7B2%7D%7D)
![AC = \sqrt{(5-4)^{2}+[2-(-6)]^{2}}](https://tex.z-dn.net/?f=AC%20%3D%20%5Csqrt%7B%285-4%29%5E%7B2%7D%2B%5B2-%28-6%29%5D%5E%7B2%7D%7D)
![AC \approx 4.123](https://tex.z-dn.net/?f=AC%20%5Capprox%204.123)
![BC = \sqrt{(x_{C}-x_{B})^{2}+ (y_{C}-y_{B})^{2}}](https://tex.z-dn.net/?f=BC%20%3D%20%5Csqrt%7B%28x_%7BC%7D-x_%7BB%7D%29%5E%7B2%7D%2B%20%28y_%7BC%7D-y_%7BB%7D%29%5E%7B2%7D%7D)
![BC = \sqrt{(5-3)^{2}+[2-(-2)]^{2}}](https://tex.z-dn.net/?f=BC%20%3D%20%5Csqrt%7B%285-3%29%5E%7B2%7D%2B%5B2-%28-2%29%5D%5E%7B2%7D%7D)
![BC \approx 4.472](https://tex.z-dn.net/?f=BC%20%5Capprox%204.472)
(by the definition of median)
![BD = CD = \frac{1}{2} \cdot (4.472)](https://tex.z-dn.net/?f=BD%20%3D%20CD%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%20%284.472%29)
![BD = CD = 2.236](https://tex.z-dn.net/?f=BD%20%3D%20CD%20%3D%202.236)
![AD = 6](https://tex.z-dn.net/?f=AD%20%3D%206)
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:
, where ![s_{ABD} = \frac{AB+BD+AD}{2}](https://tex.z-dn.net/?f=s_%7BABD%7D%20%3D%20%5Cfrac%7BAB%2BBD%2BAD%7D%7B2%7D)
, where ![s_{ACD} = \frac{AC+CD+AD}{2}](https://tex.z-dn.net/?f=s_%7BACD%7D%20%3D%20%5Cfrac%7BAC%2BCD%2BAD%7D%7B2%7D)
Finally,
![s_{ABD} = \frac{4.123+2.236+6}{2}](https://tex.z-dn.net/?f=s_%7BABD%7D%20%3D%20%5Cfrac%7B4.123%2B2.236%2B6%7D%7B2%7D)
![s_{ABD} = 6.180](https://tex.z-dn.net/?f=s_%7BABD%7D%20%3D%206.180)
![A_{ABD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}](https://tex.z-dn.net/?f=A_%7BABD%7D%20%3D%20%5Csqrt%7B%286.180%29%5Ccdot%20%286.180-4.123%29%5Ccdot%20%286.180-2.236%29%5Ccdot%20%286.180-6%29%7D)
![A_{ABD} \approx 3.004](https://tex.z-dn.net/?f=A_%7BABD%7D%20%5Capprox%203.004)
![s_{ACD} = \frac{4.123+2.236+6}{2}](https://tex.z-dn.net/?f=s_%7BACD%7D%20%3D%20%5Cfrac%7B4.123%2B2.236%2B6%7D%7B2%7D)
![s_{ACD} = 6.180](https://tex.z-dn.net/?f=s_%7BACD%7D%20%3D%206.180)
![A_{ACD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}](https://tex.z-dn.net/?f=A_%7BACD%7D%20%3D%20%5Csqrt%7B%286.180%29%5Ccdot%20%286.180-4.123%29%5Ccdot%20%286.180-2.236%29%5Ccdot%20%286.180-6%29%7D)
![A_{ACD} \approx 3.004](https://tex.z-dn.net/?f=A_%7BACD%7D%20%5Capprox%203.004)
Therefore, areas of triangles ABD and ACD are the same.
The volume of a box is:
![\begin{gathered} V=W\cdot T\cdot L \\ \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20V%3DW%5Ccdot%20T%5Ccdot%20L%20%5C%5C%20%20%5Cend%7Bgathered%7D)
So for our problem will be:
Answer: 336 ways
Step-by-step explanation:
Feon the question, there are eight teams in a league and we are told to find the number of ways the teams can finish first, second, and third assuming that there are no ties.
First is 1 of 8, then 1 of 7 and finally 1 of 6.
= 8 × 7 × 6
= 336 ways