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Find the area of a parallelogram with sides of length 6 and 8, and with an interior angle with measure 45 degrees. (No trigonometry please!!!)

 Aug 16, 2017
edited by Guest  Aug 16, 2017
 #1
avatar+8079 
+1

Find the area of a parallelogram with sides of length 6 and 8, and with an interior angle with measure 45 degrees. (No trigonometry please!!!)

 

Basic line a = 8
Sidelines s = 6
Height \(h=\frac{s}{\sqrt{2}}\)

 

\(A=a\times h=a\times\frac{s}{\sqrt{2}}=8\times\frac{6}{\sqrt{2}}=\frac{48}{\sqrt{2}}\)

\(​A=33.94112549695..\)

laugh  !

 Aug 16, 2017
edited by asinus  Aug 16, 2017
 #2
avatar+99388 
+1

How did you get the height asinus ?    

Melody  Aug 16, 2017
 #4
avatar+8079 
+2

How did you get the height asinus ?   Thank's Melody!

 

parallelogram \(ABCD\)

 \(\overline{AB}=a\\ \overline{BC}=s\\ triangle\ BAD=45°\)

\(perpendicular\ from\ D\ to\ a\ to\ point\ E\ is\ h.\)

\(AED\ is\ an\ isosceles\ right\ triangle.\)  \((\overline {DE}=h)=\overline{AE}\)

 

Set of Pythagoras:

\(\overline {AD}\ ^2=\overline{DE}\ ^2+\overline{AE}\ ^2\)

\(s ^2=2h^2\)

\(h^2=\frac{s^2}{2}\)

\(\large h=\frac{s}{\sqrt{2}}\)

laugh  !

asinus  Aug 16, 2017
 #5
avatar+99388 
+1

Thanks asinus :)

Melody  Aug 16, 2017
 #3
avatar+99388 
+2

 

 

 

 

\(h^2+h^2=6^2\\ 2h^2=36\\ h^2=18\\ h=\sqrt{18}\\ h=3\sqrt2\)

 

So

\(area=base \times height\\ A=8\times 3\sqrt2\\ A=24\sqrt2\;\;units^2\\\)

.
 Aug 16, 2017

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