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

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

May 6, 2015

#3
+22

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 = 522 - 482

(CD+14)2 = 400

CD = sqrt(400) - 14

CD = 6

May 6, 2015

#1
+5 AWESOMEEE May 6, 2015
#2
+11

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

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   May 6, 2015
#3
+22

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 = 522 - 482

(CD+14)2 = 400

CD = sqrt(400) - 14

CD = 6

civonamzuk May 6, 2015
#4
+19

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 May 6, 2015
#5
+6

points

this is for points

May 6, 2015