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geometry problem

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There are three squares inside the triangle . Find the area of the third triangle with steps , please

Aug 9, 2020

#1
+1475
+2

When I realized that the 2 triangles between purple squares have sides 3-4-5, it was a piece of cake to solve it.

The yellow square area is  54.76 cm2

Aug 10, 2020
edited by Dragan  Aug 10, 2020
#2
+26221
+1

There are three squares inside the triangle .

Find the area of the third triangle with steps , please

$$\begin{array}{|rcl|} \hline \tan(\varphi) &=& \dfrac{4}{p_1} \qquad \tan(\varphi) = \dfrac{q_1}{4} \\ \hline p_1+q_1 &=& 4\left(\dfrac{1}{\tan(\varphi)}+\tan(\varphi) \right) \\ \hline \end{array} \begin{array}{|rclcrcl|} \hline \tan(\varphi) &=& \dfrac{3}{p_2} \qquad \tan(\varphi) = \dfrac{q_2}{3} \\ \hline p_2+q_2 &=& 3\left(\dfrac{1}{\tan(\varphi)}+\tan(\varphi) \right) \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \sin(\varphi)&=& \dfrac{x}{p_1+q_1+4} \\ \sin(\varphi)&=& \dfrac{x}{4\left(\dfrac{1}{\tan(\varphi)}+\tan(\varphi) \right)+4} \\ 4\sin(\varphi)&=& \dfrac{x}{ \dfrac{1}{\tan(\varphi)}+\tan(\varphi) +1} \\ \dfrac{1}{\tan(\varphi)}+\tan(\varphi)+1 &=& \dfrac{x}{4\sin(\varphi)}\\ \hline \end{array} \begin{array}{|rcll|} \hline \cos(\varphi)&=& \dfrac{x}{p_2+q_2+3} \\ \cos(\varphi)&=& \dfrac{x}{3\left(\dfrac{1}{\tan(\varphi)}+\tan(\varphi) \right)+3} \\ 3\cos(\varphi)&=& \dfrac{x}{\dfrac{1}{\tan(\varphi)}+\tan(\varphi) +1} \\ \dfrac{1}{\tan(\varphi)}+\tan(\varphi) +1 &=& \dfrac{x}{3\cos(\varphi)}\\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \dfrac{x}{4\sin(\varphi)} &=& \dfrac{x}{3\cos(\varphi)}\\ \mathbf{\tan(\varphi)} &=& \mathbf{\dfrac{3}{4}} \\ && \Rightarrow \cos(\varphi) = \dfrac{4}{\sqrt{3^2+4^2}} \\ && \Rightarrow \mathbf{\cos(\varphi) = \dfrac{4}{5} } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline x &=& 3\cos(\varphi) \left(\dfrac{1}{\tan(\varphi)}+\tan(\varphi) +1\right) \\\\ x &=& \dfrac{3*4}{5} \left(\dfrac{1}{ \dfrac{3}{4} }+\dfrac{3}{4} +1\right) \\\\ x &=& \dfrac{3*4}{5} \left(\dfrac{4}{3}+\dfrac{3}{4} +1\right) \\\\ x &=& \dfrac{4}{5} \left(4+\dfrac{9}{4} +3\right) \\\\ x &=& \dfrac{4}{5} \left(7+\dfrac{9}{4} \right) \\\\ x &=& \dfrac{4}{5} * \dfrac{37}{4} \\\\ x &=& \dfrac{37}{5} \\\\ \mathbf{x} &=& \mathbf{7.4} \\ \mathbf{x^2} &=& \mathbf{54.76\ \text{cm}^2} \\ \hline \end{array}$$

The area of the third triangle is $$\mathbf{54.76\ \text{cm}^2}$$

Aug 10, 2020
edited by heureka  Aug 10, 2020