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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

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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!!!)

bbelt711  Aug 16, 2017
edited by Guest  Aug 16, 2017
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#1
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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!!!)

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

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

$$​A=33.94112549695..$$

!

asinus  Aug 16, 2017
edited by asinus  Aug 16, 2017
#2
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How did you get the height asinus ?

Melody  Aug 16, 2017
#4
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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:

$$\overline {AD}\ ^2=\overline{DE}\ ^2+\overline{AE}\ ^2$$

$$s ^2=2h^2$$

$$h^2=\frac{s^2}{2}$$

$$\large h=\frac{s}{\sqrt{2}}$$

!

asinus  Aug 16, 2017
#5
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Thanks asinus :)

Melody  Aug 16, 2017
#3
+91435
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$$h^2+h^2=6^2\\ 2h^2=36\\ h^2=18\\ h=\sqrt{18}\\ h=3\sqrt2$$

So

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

Melody  Aug 16, 2017

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