Question

If A(4 -6) B(3 -2) and C (5 2) are the vertices of a triangle ABC fine the length of the median AD from A to BC. Also verify that area of triangle ABD = Area of triangle ACD

Answer 1

a) The median AD from A to BC has a length of 6.

b) Areas of triangles ABD and ACD are the same.

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((x_(B)+x_(C))/(2),(y_(B)+y_(C))/(2) \right)`

`D(x,y) = \left((3 + 5)/(2),(-2 + 2)/(2) \right)`

`D(x,y) = (4,0)`

The length of the median AD is calculated by the Pythagorean Theorem:

`AD = \sqrt{(x_(D)-x_(A))^(2)+ (y_(D)-y_(A))^(2)}`

`AD = \sqrt{(4-4)^(2)+[0-(-6)]^(2)}`

`AD = 6`

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)}`

`AB = \sqrt{(3-4)^(2)+[-2-(-6)]^(2)}`

`AB \approx 4.123`

`AC = \sqrt{(x_(C)-x_(A))^(2)+ (y_(C)-y_(A))^(2)}`

`AC = \sqrt{(5-4)^(2)+[2-(-6)]^(2)}`

`AC \approx 4.123`

`BC = \sqrt{(x_(C)-x_(B))^(2)+ (y_(C)-y_(B))^(2)}`

`BC = \sqrt{(5-3)^(2)+[2-(-2)]^(2)}`

`BC \approx 4.472`

`BD = CD = (1)/(2)\cdot BC` (by the definition of median)

`BD = CD = (1)/(2) \cdot (4.472)`

`BD = CD = 2.236`

`AD = 6`

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:

`A_(ABD) = \sqrt{s_(ABD)\cdot (s_(ABD)-AB)\cdot (s_(ABD)-BD)\cdot (s_(ABD)-AD)}`, where
`s_(ABD) = (AB+BD+AD)/(2)`

`A_(ACD) = \sqrt{s_(ACD)\cdot (s_(ACD)-AC)\cdot (s_(ACD)-CD)\cdot (s_(ACD)-AD)}`, where
`s_(ACD) = (AC+CD+AD)/(2)`

Finally,

`s_(ABD) = (4.123+2.236+6)/(2)`

`s_(ABD) = 6.180`

`A_(ABD) = √((6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6))`

`A_(ABD) \approx 3.004`

`s_(ACD) = (4.123+2.236+6)/(2)`

`s_(ACD) = 6.180`

`A_(ACD) = √((6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6))`

`A_(ACD) \approx 3.004`

Therefore, areas of triangles ABD and ACD are the same.