anonymous
  • anonymous
Which chart best compares a rectangle and a rhombus according to their side and diagonal properties?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
@ganeshie8 Can you help?
anonymous
  • anonymous
@PaxPolaris can you help?

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anonymous
  • anonymous
I'm bad with this.
PaxPolaris
  • PaxPolaris
|dw:1348676604835:dw| have you been able to eliminate any of the options so far....
PaxPolaris
  • PaxPolaris
@BIGDOG96 Were you able to eliminate any of the options at all for any reason?
anonymous
  • anonymous
2 and 4...right?
PaxPolaris
  • PaxPolaris
yes if you're looking at sides ... Rectangles: opposite sides equal Rhombus: All equal so... we're left with 1 and 3
anonymous
  • anonymous
Yep
PaxPolaris
  • PaxPolaris
now look at diagonals ... in a rectangle do the diagonals bisect each other ... (cut each other in half)
PaxPolaris
  • PaxPolaris
?
anonymous
  • anonymous
Yes
PaxPolaris
  • PaxPolaris
and are they perpendicular to each other?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
So option 3 right?
PaxPolaris
  • PaxPolaris
perpendicular means they intersect at 90 degree angles...? |dw:1348677416596:dw| do all 4 angles made by the intersecting diagonals look the same ?
anonymous
  • anonymous
Yes
PaxPolaris
  • PaxPolaris
|dw:1348677680210:dw| are you saying angle 1 and angle 2 have the exact same measure?
anonymous
  • anonymous
N
anonymous
  • anonymous
No
PaxPolaris
  • PaxPolaris
right ... in a rectangle: diagonals are congruent to each other ... and bisect each other .... But they are NOT necessarily perpendicular to each other if they were perpendicular ... the rectangle would have to be a Square
anonymous
  • anonymous
Right
anonymous
  • anonymous
Gotcha
PaxPolaris
  • PaxPolaris
|dw:1348678135474:dw| In a Rhombus diagonal are perpendiular ... AND they bisect each other .... the diagonals just don't have to be congruent to each other. if they were congruent the rhombus would have to be a square. so option 1.
anonymous
  • anonymous
Thanks
PaxPolaris
  • PaxPolaris
no problem

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