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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
avatar+1710 
+1

Just curious, Why can't we use trig?

 Jul 12, 2019
 #2
avatar
+1

I haven’t learnt trigonometry yet

 Jul 12, 2019
 #3
avatar+1710 
+1

Are you DurgneyDabs? I can't really tell- try logging back in.

tommarvoloriddle  Jul 12, 2019
 #5
avatar+39 
0

Yup... I am. Forgot to login.

DungeyDabs  Jul 12, 2019
 #6
avatar+1710 
+1

ok its fine

tommarvoloriddle  Jul 12, 2019
 #4
avatar+111396 
+3

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

 

 

cool cool cool

 Jul 12, 2019
 #7
avatar+39 
-1

Is there any way to do this problem without trig?????

 Jul 12, 2019
 #8
avatar+39 
-1

forgot to login in again

DungeyDabs  Jul 12, 2019

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