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# Parallelogram Problem ​

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The diagonals of a parallelogram intersect at a 45 degree angle. If the diagonal lengths are 8 cm and 10 cm, what is the area of the parallelogram. Would appreciate an answer without trig.

Thanks

Jul 12, 2019

#1
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Just curious, Why can't we use trig?

Jul 12, 2019
#2
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I haven’t learnt trigonometry yet

Jul 12, 2019
#3
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Are you DurgneyDabs? I can't really tell- try logging back in.

tommarvoloriddle  Jul 12, 2019
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Yup... I am. Forgot to login.

DungeyDabs  Jul 12, 2019
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ok its fine

tommarvoloriddle  Jul 12, 2019
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Well.....if we can just use a "slight" bit of trig.....

The diagonals of a parallelogram bisect each other.....so.........these diagonals form four cogruent triangles

Each of these triangles will have two sides with lengths of 4 and 5 and an included angle between these sides of 45°

And the area of one of these triangles  =  (1/2) (product of these sides) * sine of the included angle

So.....the area of one of the triangles  =  (1/2) (4 * 5) * sin (45°)

[ sin (45°  =    √2 / 2 ]....so we have

(1/2) ( 4 * 5) ( √2/2)

(1/4) (20) * √2

And since we havefour of these triangles....the total area is

4 (1/4) * (20) * √2  =

20√2  units^2   Jul 12, 2019
#7
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Is there any way to do this problem without trig?????

Jul 12, 2019
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