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Given that the diagonals of a rhombus are always perpendicular bisectors of each other, what is the area of a rhombus with side length sqrt89 units and diagonals that differ by 6 units?

 

Thanks so much!

 Mar 22, 2018
edited by AnonymousConfusedGuy  Mar 22, 2018

Best Answer 

 #4
avatar+10441 
+1

Given that the diagonals of a rhombus are always perpendicular bisectors of each other, what is the area of a rhombus with side length sqrt89 units and diagonals that differ by 6 units?

 

 Mar 23, 2018
 #1
avatar+10441 
0

How many units of length does a???

 Mar 22, 2018
 #3
avatar+1438 
+1

Sorry, fixed it

AnonymousConfusedGuy  Mar 22, 2018
 #2
avatar+1206 
+1

A quadrilateral is a rhombus if and only if the diagonals are perpendicular bisectors of each other.

This is an "if and only if" proof, so there are two things we have to prove:

One way to prove this is to use congruent triangles.

However, note that this essentially runs through the proof of one of the isosceles triangle theorems which we have already proved. Note the SSS reason after we established the reflexive side. We do not have to do it again. We can simply refer to it.

For the converse, 

there is the congruent tirangle proof.

But, again we are running through a proof of one of the isosceles triangle theorems. We could simply refer to it instead of proving it again.

 

One could also use the fact that a point is equidistant from two given points if and only if it is on the perpendicular bisector of the line segment between them.

 Mar 22, 2018
 #4
avatar+10441 
+1
Best Answer

Given that the diagonals of a rhombus are always perpendicular bisectors of each other, what is the area of a rhombus with side length sqrt89 units and diagonals that differ by 6 units?

 

Omi67 Mar 23, 2018
 #5
avatar+1438 
+1

Thank you!

AnonymousConfusedGuy  Mar 23, 2018

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