+0

triangle

0
101
1

Can anyone help with this problem?

Jul 9, 2020

1+0 Answers

#1
+25569
+2

triangle

$$\begin{array}{|rcll|} \hline \mathbf{\vec{BK}} &=& \mathbf{\dfrac{ \vec{BC} }{2} + \dfrac{ \vec{DA} }{2}} \\\\ && \boxed{ \vec{DA} = \vec{BA} - \dfrac{ \vec{BC} }{2} \\ \dfrac{\vec{DA}}{2} = \dfrac{\vec{BA}}{2} - \dfrac{ \vec{BC} }{4} } \\\\ \vec{BK} &=& \dfrac{ \vec{BC} }{2} + \dfrac{\vec{BA}}{2} - \dfrac{ \vec{BC} }{4} \\\\ \mathbf{\vec{BK}} &=& \mathbf{ \dfrac{ \vec{BC} }{4} + \dfrac{\vec{BA}}{2} } \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\vec{BL}} &=& \mathbf{ \vec{BC} + \dfrac{ \vec{CE} }{2}} \\\\ && \boxed{ \vec{CE} = \dfrac{\vec{BA}}{2} - \vec{BC} \\ \dfrac{\vec{CE}}{2} = \dfrac{\vec{BA}}{4} - \dfrac{ \vec{BC} }{2} } \\\\ \vec{BL} &=& \vec{BC} + \dfrac{\vec{BA}}{4} - \dfrac{ \vec{BC} }{2} \\\\ \mathbf{\vec{BL}} &=& \mathbf{ \dfrac{ \vec{BC} }{2} + \dfrac{\vec{BA}}{4} } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline 2A_{yellow} = 18 &=& \vec{BL} \times \vec{BK} \\ 18 &=& \left( \dfrac{ \vec{BC} }{2} + \dfrac{\vec{BA}}{4} \right) \times \left( \dfrac{ \vec{BC} }{4} + \dfrac{\vec{BA}}{2} \right) \\\\ 18 &=& \underbrace{ \left( \dfrac{ \vec{BC} }{2} \times \dfrac{ \vec{BC} }{4}\right) }_{=0} +\left( \dfrac{ \vec{BC} }{2} \times \dfrac{\vec{BA}}{2} \right) +\left( \dfrac{\vec{BA}}{4} \times \dfrac{ \vec{BC} }{4} \right) + \underbrace{\left( \dfrac{\vec{BA}}{4} \times \dfrac{\vec{BA}}{2} \right) }_{=0} \\\\ 18 &=& \left( \dfrac{ \vec{BC} }{2} \times \dfrac{\vec{BA}}{2} \right) +\left( \dfrac{\vec{BA}}{4} \times \dfrac{ \vec{BC} }{4} \right) \\\\ 18 &=& \left( \dfrac{ \vec{BC} }{2} \times \dfrac{\vec{BA}}{2} \right)-\left( \dfrac{\vec{BC}}{4} \times \dfrac{ \vec{BA} }{4} \right) \\\\ 18 &=& \dfrac{1}{4}*\left( \vec{BC} \times \vec{BA} \right)-\dfrac{1}{16}*\left( \vec{BC} \times \vec{BA} \right) \\\\ 18 &=& \dfrac{4}{16}*\left( \vec{BC} \times \vec{BA} \right)-\dfrac{1}{16}*\left( \vec{BC} \times \vec{BA} \right) \\\\ 18 &=& \dfrac{3}{16}*\left( \vec{BC} \times \vec{BA} \right) \\\\ \dfrac{16*18}{3} &=& \vec{BC} \times \vec{BA} \\\\ 16*6 &=& \vec{BC} \times \vec{BA} \\\\ 96&=& \vec{BC} \times \vec{BA} \\\\ \mathbf{ \vec{BC} \times \vec{BA} } &=& \mathbf{96} \quad | \quad 2\times \text{area } \triangle \text{ABC} = \vec{BC} \times \vec{BA} \\\\ 2\times \text{area } \triangle \text{ABC} &=& 96 \quad | \quad :2 \\\\ \mathbf{ \text{area } \triangle \text{ABC} } &=& \mathbf{48} \\ \hline \end{array}$$

The area of the triangle ABC is $$\mathbf{48\ \text{cm}^2}$$

Jul 10, 2020