# Thread: Could use some help with a rhombus question

1. ## Could use some help with a rhombus question

Sigh... I'm doing independant study to get some much needed math credits and I'm starting to get a feeling that the work material was designed for the sole purpose of testing my sanity. The key questions (they're like mini-tests) at the end of each lesson aren't so much fair as they are mind blowing curveballs from math- . It's like the people who wrote the lessons gave up at the end and figured that instead of thinking up challenging relevant questions, they'd just toss in a bunch of random nonsense that's impossible to answer given what's actually taught in the lesson.

Oh well, enough complaining. I'm determined, and it's true that you can't keep a good man down. All that's needed is a little help from you, the hero. The one who's going to explain to me how I can prove quadrilateral W(-0.5, 1.5) X(4, 1) Y(3.5, 5.5) Z(-1, 6) is infact, a rhombus.

2. Originally Posted by mathdonkey
... how I can prove quadrilateral W(-0.5, 1.5) X(4, 1) Y(3.5, 5.5) Z(-1, 6) is infact, a rhombus.
Hi,

a quadrilateral with 4 equal sides is a rhombus. Therefore calculate the length of the sides. If they are all equal then the quadrilateral is a rhombus.

Hint: If you have 2 points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ then the distance between these 2 points is calculated by:

$d(P_1P_2) = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

3. Originally Posted by earboth
Hi,

a quadrilateral with 4 equal sides is a rhombus. Therefore calculate the length of the sides. If they are all equal then the quadrilateral is a rhombus.
Really??? That's it?

I looked at this site and assumed more would be needed!

4. Originally Posted by mathdonkey
Really??? That's it?

I looked at this site and assumed more would be needed!
I'm awfully sorry but math is not that complicated then you believe it is. Even the mentioned site states nothing but that the sides are all equal.

Am I a hero now?

5. Originally Posted by earboth
I'm awfully sorry but math is not that complicated then you believe it is. Even the mentioned site states nothing but that the sides are all equal.

Am I a hero now?
Definately.

I guess if all sides in a quadrilateral are equal then all the other properties must be true too?

6. Hello, mathdonkey

If you think that this is a difficult problem, you ain't seen nothing yet!

Prove that quadrilateral: W(-½, 1½), X(4, 1), Y(3½, 5½), Z(-1, 6) is a rhombus.
It's always a good idea to make a sketch . . .

Code:
          Z   |
*   |             Y
(-1,6) |             *
|          (3½,5½)
|
|
|
W |
* |               X
(-½,1½)|               *
|             (4,1)
- - - - + - - - - - - - - - -
|

Do you know what makes a quadrilateral a rhombus?
. . That's right . . . it has four equal sides.

Does WXYZ has four equal sides?
. . How can we determine if: . $WX \:=\: XY\:=\:YZ \:=\:ZW$ ?
That's right . . . the Distance Formula.

Go for it!

7. Originally Posted by mathdonkey
Definately.

I guess if all sides in a quadrilateral are equal then all the other properties must be true too?
That's really a perfect guess.

Probably you have found out that all sides of your quadrilateral have the same length with $l = \frac12 \cdot \sqrt{82} \approx 4.5277$

8. Originally Posted by Soroban
If you think that this is a difficult problem, you ain't seen nothing yet!
Haha, now that I know how to do it, it's actually quite easy!

I just over analyzed the problem. Blame it on my math phobia.

9. NB: After you’ve made your sketch, as Soroban did, you do not have to use the distance formula for all four sides! Just use the distance formula on any two adjacent sides, say WX and WZ. Then all you need to do next is show that the vectors $\vec{\mathrm{WX}}$ and $\vec{\mathrm{ZY}}$ are equal, i.e. the sides WX and ZY are equal and parallel. That is all.

Everyone here seems so one-tracked about this problem that they seem to have forgotten that the rhombus has other properties too. Recall that a rhombus is a parallelogram. So another way of proving that a quadrilateral is a rhombus is to show that it is a parallelogram with equal sides. Showing that it has four equal sides is only way – but not necessarily the easiest way. Don’t be too narrow-minded when you are solving math problems.

In fact, you can also show that it’s a rhombus without using the distance formula and messing around with squares and square roots at all. Once you’ve shown that $\vec{\mathrm{WX}}$ and $\vec{\mathrm{ZY}}$, the fact that the other two sides are also equal and parallel will follow; hence WXYZ is a parallelgram. To show that it’s a rhombus, take the dot product of $\vec{\mathrm{WY}}$ and $\vec{\mathrm{XZ}}$ and show that it is 0, i.e. the diagonals are perpendicular. The rhombus is the only parallelogram whose diagonals intersect perpendicularly.