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Median question

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Point D is the midpoint of median $$\overline{AM}$$ of triangle ABC. Point E is the midpoint of $$\overline{AB}$$, and point T is the intersection of $$\overline{BD}$$ and $$\overline{ME}$$. Find the area of triangle DMT if [ABC] =150. [ABC] is the area of triangle ABC

Apr 20, 2020

#1
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The area of triangle DMT is 15.

Apr 20, 2020
#2
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Point D is the midpoint of median $$\overline{AM}$$ of triangle ABC.
Point E is the midpoint of $$\overline{AB}$$, and point T is the intersection of $$\overline{BD}$$ and $$\overline{ME}$$.
Find the area of triangle DMT if [ABC] =150. [ABC] is the area of triangle ABC

My attempt:

$$\begin{array}{|rcll|} \hline 2[ABC] &=& 2u*2w*\sin(M) \\ [ABC] &=& 2uw\sin(M)\quad \text{ or } \quad \mathbf{\sin(M) = \dfrac{[ABC]}{2uw}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \text{In \triangle BDM } \\ \hline \dfrac{\sin(M)}{\overline{BD}} &=& \dfrac{\sin(D)}{u} \quad | \quad \mathbf{\sin(M) = \dfrac{[ABC]}{2uw}} \\ \dfrac{[ABC]}{2uw\overline{BD}} &=& \dfrac{\sin(D)}{u} \\\\ \dfrac{[ABC]}{2w\overline{BD}} &=& \sin(D) \\\\ \mathbf{\dfrac{[ABC]}{2\overline{BD}}} &=& \mathbf{w\sin(D)} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline 2[DMT] &=& w*\overline{DT}*\sin(D) \\ 2[DMT] &=& w\sin(D)*\overline{DT} \quad | \quad \mathbf{w\sin(D) = \dfrac{[ABC]}{2\overline{BD}}} \\ 2[DMT] &=& \dfrac{[ABC]}{2\overline{BD}}*\overline{DT} \\\\ \mathbf{[DMT]} &=& \mathbf{\dfrac{[ABC]}{4} * \dfrac{\overline{DT}} {\overline{BD}}} \\ \hline \end{array}$$

The Centroid of a Triangle ABM Divides Each Median in the Ratio $$1:2$$

$$\begin{array}{|rcll|} \hline \overline{DT} &=& \dfrac{1}{3}*\overline{BD} \\\\ \overline{BT} &=& \dfrac{2}{3}*\overline{BD} \\\\ \dfrac{\overline{DT}} {\overline{BD}} &=& \dfrac{\overline{DT}} { \overline{DT}+\overline{BT}} \\\\ \dfrac{\overline{DT}} {\overline{BD}} &=& \dfrac{\dfrac{1}{3}*\overline{BD}} { \dfrac{1}{3}*\overline{BD}+\dfrac{2}{3}*\overline{BD}} \\\\ \dfrac{\overline{DT}} {\overline{BD}} &=& \dfrac{\dfrac{1}{3}*\overline{BD}} { \dfrac{3}{3}*\overline{BD}} \\\\ \dfrac{\overline{DT}} {\overline{BD}} &=& \dfrac{\dfrac{1}{3} } { \dfrac{3}{3} } \\\\ \dfrac{\overline{DT}} {\overline{BD}} &=& \dfrac{\dfrac{1}{3} } { 1 } \\\\ \mathbf{\dfrac{\overline{DT}} {\overline{BD}} } &=& \mathbf{ \dfrac{1}{3}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{[DMT]} &=& \mathbf{\dfrac{[ABC]}{4} * \dfrac{\overline{DT}} {\overline{BD}}} \quad | \quad \mathbf{\dfrac{\overline{DT}} {\overline{BD}} = \dfrac{1}{3}} \\\\ [DMT] &=& \dfrac{[ABC]}{4}* \dfrac{1}{3} \\\\ [DMT] &=& \dfrac{[ABC]}{12} \quad | \quad [ABC] =150 \\\\ [DMT] &=& \dfrac{150}{12} \\\\ \mathbf{[DMT]} &=& \mathbf{12.5} \\ \hline \end{array}$$

Apr 22, 2020