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3 years ago

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You want to install a 3 yd wide walk around a circular swimming pool. The diameter of the pool is 16 yd. What is

1 answer:

elena-s [515]3 years ago

4 0

3d becauvshsbdbbdhdhdhdhd

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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 tha

Gnoma [55]

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:

$D (x,y) = \left(\frac{x_{B}+x_{C}}{2},\frac{y_{B}+y_{C}}{2} \right)$

$D(x,y) = \left(\frac{3 + 5}{2},\frac{-2 + 2}{2} \right)$

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

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:

$AB = \sqrt{(x_{B}-x_{A})^{2}+ (y_{B}-y_{A})^{2}}$

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

$AB \approx 4.123$

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

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

$AC \approx 4.123$

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

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

$BC \approx 4.472$

$BD = CD = \frac{1}{2}\cdot BC$ (by the definition of median)

$BD = CD = \frac{1}{2} \cdot (4.472)$

$BD = CD = 2.236$

$AD = 6$

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 $s_{ABD} = \frac{AB+BD+AD}{2}$

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

Finally,

$s_{ABD} = \frac{4.123+2.236+6}{2}$

$s_{ABD} = 6.180$

$A_{ABD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}$

$A_{ABD} \approx 3.004$

$s_{ACD} = \frac{4.123+2.236+6}{2}$

$s_{ACD} = 6.180$

$A_{ACD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}$

$A_{ACD} \approx 3.004$

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

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4 years ago

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