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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+26396 
+2

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

Let P2=(x2y2) Let P4=(x4y4) 

 

P1=a(sin(φ)cos(φ))P2=P1+3a(cos(φ)sin(φ))P2=a(sin(φ)cos(φ))+3a(cos(φ)sin(φ))P2=a(sin(φ)+3cos(φ)cos(φ)3sin(φ))x2=a(sin(φ)+3cos(φ))|x2=11a(sin(φ)+3cos(φ))=11a=11(sin(φ)+3cos(φ))(1)

 

P3=2a(cos(φ)sin(φ))P4=P33a(sin(φ)cos(φ))P4=2a(cos(φ)sin(φ))3a(sin(φ)cos(φ))P4=a(2cos(φ)3sin(φ)2sin(φ)3cos(φ))|y4|=a(2sin(φ)+3cos(φ))|y4=13a(2sin(φ)+3cos(φ))=13a=13(2sin(φ)+3cos(φ))(2)

 

(1):a=11(sin(φ)+3cos(φ))(2):a=13(2sin(φ)+3cos(φ))a=11(sin(φ)+3cos(φ))=13(2sin(φ)+3cos(φ))11(sin(φ)+3cos(φ))=13(2sin(φ)+3cos(φ))11(2sin(φ)+3cos(φ))=13(sin(φ)+3cos(φ))22sin(φ)+33cos(φ)=13sin(φ)+39cos(φ)22sin(φ)13sin(φ)=39cos(φ)33cos(φ)9sin(φ)=6cos(φ)tan(φ)=23


a= ?

a=11(sin(φ)+3cos(φ))sin(φ)=tan(φ)1+tan2(φ)|tan(φ)=23sin(φ)=231+49sin(φ)=21313cos(φ)=11+tan2(φ)|tan(φ)=23cos(φ)=11+49cos(φ)=31313a=1121313+331313a=1121313+91313a=11111313a=1313a=13

 

The shaded area:

The shaded area=11136a2|a=13The shaded area=1113613The shaded area=513The shaded area=65

 

laugh

 Jul 29, 2020
 #3
avatar+1094 
+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+26396 
+2

Hallo ilorty,

 

i use:

 

Example line 1Example line 2

 

or with \boxed{}:

Box Example line 1Box Example line 2

 

Next line is "\\"

 

laugh

heureka  Jul 29, 2020
edited by heureka  Jul 29, 2020
 #5
avatar+1094 
+7

ah, thanks!

ilorty  Jul 29, 2020

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