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In the figure triangle ABC is a right triangle with legs AB = 6 and AC = 8. A square is drawn as shown, with a side along BC and corners on AB and AC. Find the length of the side of the square.

 

 May 29, 2020
 #1
avatar+1326 
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Side AC = 8

 

Let  x be a side of the square

 

Triangles  JCG and AJI are similar.

 

The sine of an angle JCG = 0.8

 

The cosine of an angle AJI = 0.6

 

0.8 * x + x / 0.6 = 8                                                                                                                                                                                                                               

0.8x + 1.666667x = 8

 

2.466667x = 8   

 

Divide both sides by 2.466667

 

2.466667x / 2.466667 = 8 / 2.466667

 

x = 3.243243   indecision

 May 29, 2020
edited by Dragan  May 29, 2020
edited by Dragan  May 29, 2020
edited by Dragan  May 29, 2020
 #2
avatar+25543 
+1

In the figure triangle ABC is a right triangle with legs AB = 6 and AC = 8.

A square is drawn as shown, with a side along BC and corners on AB and AC.

Find the length of the side of the square.

 

\(\begin{array}{|rclrclrcl|} \hline \tan(C) &=& \dfrac{x}{p} & && & \tan(C) &=& \dfrac{6}{8} \\ &&& \dfrac{x}{p} &=& \dfrac{6}{8}& \\ && &\dfrac{p}{x} &=& \dfrac{8}{6}& \\ && &p &=& \dfrac{8x}{6}& \\ && &\mathbf{p} &=& \mathbf{\dfrac{4x}{3}}& \\ \hline \end{array} \begin{array}{|rclrclrcl|} \hline \tan(B) &=& \dfrac{x}{q} & && & \tan(B) &=& \dfrac{8}{6} \\ &&& \dfrac{x}{q} &=& \dfrac{8}{6}& \\ && &\dfrac{q}{x} &=& \dfrac{6}{8}& \\ && &p &=& \dfrac{6x}{8}& \\ && &\mathbf{q} &=& \mathbf{\dfrac{3x}{4}}& \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline p+x+q &=& 10 \\\\ \dfrac{3x}{4}+x+\dfrac{4x}{3} &=& 10 \\\\ x*\left( \dfrac{3}{4}+1+\dfrac{4}{3} \right) &=& 10 \\\\ x* \dfrac{37}{12} &=& 10 \\\\ x &=& 10*\dfrac{12}{37} \\\\ x &=& \dfrac{120}{37} \\\\ \mathbf{x} &=& \mathbf{3.\overline{243}} \\ \hline \end{array}\)

 

laugh

 May 29, 2020

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