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Six squares are inscribed in an 11 by 13 rectangle.  Find the shaded area.

 

 Jul 29, 2020
 #1
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-1

11 * 13 = 143 u2

 

112 / 9 = 13.44444444 u2

 

80.667 u2

 

143 - 80.667 = 62.33333333 u2        area of  shaded region

 Jul 29, 2020
 #2
avatar+25541 
+2

Six squares are inscribed in an 11 by 13 rectangle.  Find the shaded area.

\(\text{Let $P_2=\dbinom{x_2}{y_2} $ } \\ \text{Let $P_4=\dbinom{x_4}{y_4} $ }\)

 

\(\begin{array}{|rcll|} \hline P_1 &=& a\dbinom{\sin(\varphi)}{\cos(\varphi)} \\ \hline \end{array} \begin{array}{|rcll|} \hline P_2 &=& P_1 + 3a\dbinom{\cos(\varphi)}{-\sin(\varphi)} \\\\ P_2 &=& a\dbinom{\sin(\varphi)}{\cos(\varphi)} + 3a\dbinom{\cos(\varphi)}{-\sin(\varphi)} \\\\ P_2 &=& a\dbinom{\sin(\varphi)+3\cos(\varphi)} {\cos(\varphi)-3\sin(\varphi)} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{x_2} &=& \mathbf{a\Big( \sin(\varphi)+3\cos(\varphi) \Big)} \quad | \quad x_2 = 11 \\ a\Big( \sin(\varphi)+3\cos(\varphi) \Big) &=& 11 \\ a &=& \dfrac{11}{ \Big( \sin(\varphi)+3\cos(\varphi) \Big) } \qquad (1) \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline P_3 &=& 2a\dbinom{\cos(\varphi)}{-\sin(\varphi)} \\ \hline \end{array} \begin{array}{|rcll|} \hline P_4 &=& P_3 - 3a\dbinom{\sin(\varphi)}{\cos(\varphi)} \\\\ P_4 &=& 2a\dbinom{\cos(\varphi)}{-\sin(\varphi)} - 3a\dbinom{\sin(\varphi)}{\cos(\varphi)} \\\\ P_4 &=& a\dbinom{2\cos(\varphi)-3\sin(\varphi)} {-2\sin(\varphi)-3\cos(\varphi)} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{|y_4|} &=& \mathbf{a\Big( 2\sin(\varphi)+3\cos(\varphi)\Big)} \quad | \quad y_4 = 13 \\ a\Big( 2\sin(\varphi)+3\cos(\varphi) \Big) &=& 13 \\ a &=& \dfrac{13} { \Big( 2\sin(\varphi)+3\cos(\varphi) \Big) } \qquad (2) \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1): & a &=& \dfrac{11}{ \Big( \sin(\varphi)+3\cos(\varphi) \Big) } \\ (2): & a &=& \dfrac{13} { \Big( 2\sin(\varphi)+3\cos(\varphi) \Big) } \\ \hline & a = \dfrac{11}{ \Big( \sin(\varphi)+3\cos(\varphi) \Big) } &=& \dfrac{13} { \Big( 2\sin(\varphi)+3\cos(\varphi) \Big) } \\\\ & \dfrac{11}{ \Big( \sin(\varphi)+3\cos(\varphi) \Big) } &=& \dfrac{13} { \Big( 2\sin(\varphi)+3\cos(\varphi) \Big) } \\\\ & 11 \Big( 2\sin(\varphi)+3\cos(\varphi) \Big) &=& 13 \Big( \sin(\varphi)+3\cos(\varphi) \Big) \\\\ & 22\sin(\varphi)+33\cos(\varphi) &=& 13 \sin(\varphi)+39\cos(\varphi) \\\\ & 22\sin(\varphi)- 13 \sin(\varphi) &=& 39\cos(\varphi)-33\cos(\varphi) \\\\ & 9\sin(\varphi) &=& 6\cos(\varphi) \\\\ & \mathbf{\tan(\varphi)} &=& \mathbf{\dfrac{2}{3}} \\ \hline \end{array}\)


\(\mathbf{a=\ ?}\)

\(\begin{array}{|rclrcl|} \hline a &=& \dfrac{11}{ \Big( \sin(\varphi)+3\cos(\varphi) \Big) } \\ &&& \sin(\varphi) &=& \dfrac{\tan(\varphi) }{\sqrt{1+\tan^2(\varphi)}} \quad | \quad \mathbf{\tan(\varphi)=\dfrac{2}{3}} \\ &&& \sin(\varphi) &=& \dfrac{\dfrac{2}{3}}{\sqrt{1+\dfrac{4}{9}}} \\ &&& \mathbf{\sin(\varphi)} &=& \mathbf{\dfrac{2\sqrt{13}}{13}} \\\\ &&& \cos(\varphi) &=& \dfrac{1}{\sqrt{1+\tan^2(\varphi)}} \quad | \quad \mathbf{\tan(\varphi)=\dfrac{2}{3}} \\ &&& \cos(\varphi) &=& \dfrac{1}{\sqrt{1+\dfrac{4}{9}}} \\ &&& \mathbf{\cos(\varphi)} &=& \mathbf{\dfrac{3\sqrt{13}}{13}} \\\\ a &=& \dfrac{11}{ \dfrac{2\sqrt{13}}{13}+3*\dfrac{3\sqrt{13}}{13} } \\\\ a &=& \dfrac{11}{ \dfrac{2\sqrt{13}}{13}+\dfrac{9\sqrt{13}}{13} } \\\\ a &=& \dfrac{11}{ \dfrac{11\sqrt{13}}{13} } \\\\ a &=& \dfrac{13}{\sqrt{13}} \\\\ \mathbf{a} &=& \mathbf{\sqrt{13}} \\ \hline \end{array}\)

 

The shaded area:

\(\begin{array}{|rcll|} \hline \text{The shaded area} &=& 11*13 - 6a^2 \quad | \quad \mathbf{a=\sqrt{13}} \\ \text{The shaded area} &=& 11*13 - 6*13 \\ \text{The shaded area} &=& 5*13 \\ \mathbf{\text{The shaded area}} &=& \mathbf{65} \\ \hline \end{array}\)

 

laugh

 Jul 29, 2020
 #3
avatar+1042 
+7

Wow! Wonderfully presented answer, as usual. Just wondering, how did you get the boxes? With the \boxed{} command? For me, that only worked for one line, so how did you put all those lines of math into the box? 

ilorty  Jul 29, 2020
 #4
avatar+25541 
+2

Hallo ilorty,

 

i use:

 

\(\begin{array}{|rcll|} \hline Example~line~1 \ldots \\ Example~line~2 \ldots \\ \hline \end{array}\)

 

or with \boxed{}:

\(\boxed{Box~Example~line~ 1 \\Box~Example~line~ 2 }\)

 

Next line is "\\"

 

laugh

heureka  Jul 29, 2020
edited by heureka  Jul 29, 2020
 #5
avatar+1042 
+7

ah, thanks!

ilorty  Jul 29, 2020

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