+64 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!!!)
+14995 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 \(h=\frac{s}{\sqrt{2}}\)
\(A=a\times h=a\times\frac{s}{\sqrt{2}}=8\times\frac{6}{\sqrt{2}}=\frac{48}{\sqrt{2}}\)
\(A=33.94112549695..\)
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+14995 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:
\(\overline {AD}\ ^2=\overline{DE}\ ^2+\overline{AE}\ ^2\)
\(s ^2=2h^2\)
\(h^2=\frac{s^2}{2}\)
\(\large h=\frac{s}{\sqrt{2}}\)
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