In the figure below, AB = AC = 50, AD = 52, and bc = 28. Determine cd.

In the figure below, AB = AC = 50, AD = 52, and  BC = 28. Determine CD.

                                                                                                                                                                              "h" is the height of the triangle  ABC.                                                                                                                                                                                                                                                                                                           

h² = (AC)² - (BC/2)²

h² = 50² - 14²

h = sqrt(50² - 14²)

h = 48

(CD+14)² = (AD)² - h²

(CD+14)² = 52² - 48²

(CD+14)² = 400

CD = sqrt(400) - 14

CD = 6

Using the Law of Cosines, let us determine  angle BAC, first

So we have

28^2 = 50^2  + 50^2 - 2(2500)cos BAC

cos^-1 [ (28^2  - 5000) / (-5000)]  = BAC =  about 32.52°

And since AC = AB  then ACB = [180 - 32.52]/ 2 = about 73.74°

And ACD is supplemental to this = about 106.26°

And using the Law of Sines

sin ADC / 50  = sin 106.26 / 52   . so...using the sine inverse, we have

sin^-1 (50sin106.26/ 52)  =  ADC = about 67.38°

So angle CAD = 180 - 106.26 - 67.38 = about 6.36°

Now...we can find CD with the Law of Sines, again

CD / sin 6.36  = 52 / sin 106.26

CD  = 52  sin 6.36 / sin 106.26 =  6 units

In the figure below, AB = AC = 50, AD = 52, and bc = 28. Determine cd.

I. Herons Formula: Area A (ABC):

$$\small{\text{$
\begin{array}{l|l}
A=\sqrt{s\cdot (s-a)\cdot(s-b)\cdot (s-c)} & \qquad \overline{AB} = a=50, ~ \overline{AC} = b=50,~ \overline{AD} =c=28\\\\
& \qquad s = \frac{a+b+c}{2}=\frac{50+50+28}{2}=\frac{128}{2}=64 \\
A=\sqrt{s\cdot (s-50)\cdot(s-50)\cdot (s-28)} \\
A= \sqrt{64\cdot (64-50)\cdot(64-50)\cdot (64-28)} \\
A= \sqrt{64\cdot 14 \cdot 14 \cdot 36} \\
A= \sqrt{8^2\cdot 14^2 \cdot 6^2} \\
A= 8\cdot 14 \cdot 6 \\
A=672
\end{array}
$}}$$$$$$

II: Height h of (ABC):

$$\small{\text{$
\begin{array}{rcl}
c\cdot h &=& 2\cdot A\\\\
h&=& \dfrac{2\cdot A}{c} \\\\
h&=& \dfrac{2\cdot 672}{28}\\\\
h &= &48
\end{array}
$}}$$

III. Pythagoras:

$$\small{\text{$
x = \overline{CD}
$}}\\\\
\small{\text{$
\begin{array}{rcl}
h^2 + \left( \dfrac{c}{2}+x \right) ^2 &=& 52^2 \\\\
48^2 + \left( 14 +x \right) ^2 &=& 52^2 \\\\
\left( 14 +x \right) ^2 &=& 52^2 - 48^2\\\\
\left( 14 +x \right) ^2 &=& 20^2\\\\
14 +x &=& 20\\\\
x &=& 20-14\\\\
x &=& 6
\end{array}
$}}$$

CD = 6

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