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# Sorry to re-post this but I'm afraid it will get lost in the sea of questions

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+937

Thanks so much!

AnonymousConfusedGuy  Mar 20, 2018
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#1
+19207
+2

The diagonals of a trapezoid are perpendicular and have lengths 8 and 10.

Find the length of the median of the trapezoid.

Area A = ?

$$\begin{array}{|rcll|} \hline 2A &=& (8-y)x + xy + (10-x)y+(10-x)(8-y) \\ 2A &=& 8x-yx+xy+10y-xy+80-10y-8x+xy \\ 2A &=& 80 \\ \mathbf{A} & \mathbf{=} & \mathbf{40} \\ \hline \end{array}$$

h = ?

$$\begin{array}{|rclrcl|} \hline \sin(B) &=& \dfrac{h}{8} & \sin(A) &=& \dfrac{h}{10} \quad & | \quad B = 90^{\circ}-A \\ \sin(90^{\circ}-A)^2+ \sin(A)^2 &=& \dfrac{h^2}{8^2} +\dfrac{h^2}{10^2} \\ \cos(A)^2+ \sin(A)^2 &=& \dfrac{h^2}{8^2} +\dfrac{h^2}{10^2} \\ 1 &=& \dfrac{h^2}{8^2} +\dfrac{h^2}{10^2} \\ \dfrac{h^2}{8^2} +\dfrac{h^2}{10^2} &=& 1 \\ h^2\left( \dfrac{1}{8^2} + \dfrac{1}{10^2} \right) &=& 1 \\ h^2\left( \dfrac{164}{8^210^2} \right) &=& 1 \\ h &=& \dfrac{8\cdot 10}{\sqrt{164}} \\ h &=& \dfrac{8\cdot 10}{\sqrt{4\cdot 41}} \\ h &=& \dfrac{8\cdot 10}{2\sqrt{41}} \\ \mathbf{ h }&\mathbf{=}& \mathbf{\dfrac{40}{\sqrt{41}} } \\ \hline \end{array}$$

median m = ?

$$\begin{array}{|rcll|} \hline A &=& \dfrac{a+c}{2} \cdot h \quad & | \quad m = \dfrac{a+c}{2}\\\\ A &=& m \cdot h \\\\ m &=& \dfrac{A}{h} \\\\ m &=& \dfrac{40}{ \dfrac{40}{\sqrt{41} } } \\\\ m &=& \dfrac{40}{40}\cdot \sqrt{41} \\\\ m &=& \sqrt{41} \\\\ \mathbf{ m} &\mathbf{ =} & \mathbf{ 6.40312423743} \\ \hline \end{array}$$

heureka  Mar 20, 2018
edited by heureka  Mar 20, 2018
#2
+937
+3

Thanks so much for your answer Heaureka, that's some impressive Latex!

AnonymousConfusedGuy  Mar 20, 2018
#3
+92225
+2

Sorry guys, I had failed to seet he word 'perpendicular' in the question.

I better get some new specs.

Thanks Heureka :)

Melody  Mar 21, 2018
edited by Melody  Mar 21, 2018

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