Ahh!

The Preacher

Junior Member
Joined
Sep 13, 2005
Messages
53
I'm so sorry to be asking so many questions. I thought I was starting to get it. I don't think you're wasting your time though, I'm actually starting to understand some of it!

The shorter diagonal of a rhombus has the same length as a side. Find the area of the rhombus if the longer diagonal is 12 in. long.

They didn't give me any angles, other lengths, or anything, and I have absolutely no idea where to start! Could you please give me a starting point?

The answer is in radical form, so I'm guessing it involves me figuring out something with triangles. Thanks in advance for any assistance.

God bless,
-The Preacher
 
The shorter diagonal of a rhombus has the same length as a side. Find the area of the rhobus if the longer diagonal is 12 in. long.

Draw a picture. Mark all segments/sides/diagonals that are congruent. Think: Triangles.

Hope that helps.
 
That is your old friend, the equilateral triangle. Short diagonal, and two sides all equal.
 
Gene said:
That is your old friend, the equilateral triangle. Short diagonal, and two sides all equal.

problem78dc.png


All I can think of is that that middle line = 12. I know that the other diagonal will make an equilateral triangle with the sides, but how in the world am I supposed to find the length of it?

EDIT: I don't want you to do my work for me. I just don't know what to do. If you can tell me what and (if I need help) how to do it, without giving me the answer, that'd be dandy. Thanks, I just need to learn this stuff.
 
The Preacher said:
Gene said:
That is your old friend, the equilateral triangle. Short diagonal, and two sides all equal.

problem78dc.png


All I can think of is that that middle line = 12. I know that the other diagonal will make an equilateral triangle with the sides, but how in the world am I supposed to find the length of it?

Let's call the length of a side y. The long diagonal = 12, so that each half of it is 6.
Let x be the (altitude) perpendicular distance from one vertex to the long diagonal. It's also true that the short diagonal equals a side. From that, we know that y = 2x.

The Pythagorean theorem then can be set up as 6² + x² = y².

See if replacing y with 2x helps you solve that equation.
 
Agh.

I labeled all the sides and stuff on my rhombus to help me think about it, but I just can't figure out the numbers.

Is there a formula to figure out what numbers to plug into the Pythagorean Theorem there? I understand the relationships... I just don't know what numbers are supposed to relate in that way.

I feel like I'm asking y'all to do my thinking for me, but I just don't know what to think.
 
The Preacher said:
Agh.

I labeled all the sides and stuff on my rhombus to help me think about it, but I just can't figure out the numbers.

Is there a formula to figure out what numbers to plug into the Pythagorean Theorem there? I understand the relationships... I just don't know what numbers are supposed to relate in that way.

I feel like I'm asking y'all to do my thinking for me, but I just don't know what to think.

The equation is 6² + x² = y², and I suggested you replace y with 2x, so you'll have
6² + x² = (2x)². What numbers do you need?
 
TchrQbic said:
The Preacher said:
Agh.

I labeled all the sides and stuff on my rhombus to help me think about it, but I just can't figure out the numbers.

Is there a formula to figure out what numbers to plug into the Pythagorean Theorem there? I understand the relationships... I just don't know what numbers are supposed to relate in that way.

I feel like I'm asking y'all to do my thinking for me, but I just don't know what to think.

The equation is 6² + x² = y², and I suggested you replace y with 2x, so you'll have
6² + x² = (2x)². What numbers do you need?

I don't know. If the next step is telling me the answer, then just don't tell me.

I probably have to multiply something by some imaginary number.

It's (duh) 36 + \(\displaystyle x^2\) = (\(\displaystyle 2x)^2\) I'm putting that down because maybe looking at it will help me think. I dunno.

Wait...x has to be smaller than six. Wow, that narrows it down a lot...I can't believe it takes me so long to think of stuff like this.
 
Gene said:
Perhaps all you need is
(2x)²=4x², not 2x²

Where did you get the 4? What's it there for?

I've tried substituting 1, 2, 3, 4, and 5 for x, but none of them equal out. Those are all the numbers lower than six, and x has to be smaller than 6! What am I doing wrong?

Am I making this more complicated than it has to be?
 
(2x)²=(2*x)(2*x)=2*2*x*x=4x²
Just combine the x² terms on one side and the 36 on the other side and solve for x. It wont be an integer which is why your trial-and-error approach didn't work.
 
I'm not sure if we're on the same page. This whole time I've been trying to find half of the shorter diagonal. Are you trying to show me how to find the whole diagonal, or half?

You're usually patient on the boards, but if I'm frustrating you, I can try to figure this out on my own, Gene. Either way, I appreciate it.




:oops:
 
I'm using Qbic's description and variables which you used in
"its (duh) x^2+36=(2x)²"
After correcting (2x)² = 4x² just combine the x² terms on one side and the 36 on the other side and solve for x.
 
Gene said:
I'm using Qbic's description and variables which you used in
"its (duh) x^2+36=(2x)²"
After correcting (2x)² = 4x² just combine the x² terms on one side and the 36 on the other side and solve for x.

Dangit, I don't get it. I solved it wrong and got two, but it's because I isolated the \(\displaystyle x^2\) terms on one side and got \(\displaystyle 3x^2\) on the right side, and kept trying to solve it like it was \(\displaystyle (3x)^2\).

I just don't get this. =[

Argh!

Is it 36=3x\(\displaystyle ^2\)?

How do I get x? I'm sorry, it's been so long since I've done algebra. I keep thinking it's the square root of something. I'm getting very frustrated with this.
 
The shorter diagonal of a rhombus has the same length as a side. Find the area of the rhombus if the longer diagonal is 12 in. long.

They didn't give me any angles, other lengths, or anything, and I have absolutely no idea where to start! Could you please give me a starting point?

The answer is in radical form, so I'm guessing it involves me figuring out something with triangles. Thanks in advance for any assistance.

1--By definition, the figure is 2 equilateral triangles joined together.
2--With the long disgonal being 12, the altitude of each equilateral triangle is 6.
3--Therefore, the triangle sides are 6/cos(30º).= 6/sqrt3/2 = 12/sqrt3
4--The area of each triangle is therefore A = 6x12/sqrt3 =72/sqrt3
5--The area of the rhombus is therefore 144/sqrt3
 
Sorry, I'm going to have to give you the answer. I find it very hard to believe you can't solve
3x²=36
Divide by 3
x²=12
Take the
sqrare root of boyh sides
x=+sqrt(12) =
+2*sqrt(3)

PS. My ISP went ill for a while so I missed some posts. I'll leave this as I had it typed even though it is redundent now. No response needed.
----------------
Gene
 
Gene said:
Sorry, I'm going to have to give you the answer. I find it very hard to believe you can't solve
3x²=36
Divide by 3
x²=12
Take the
sqrare root of boyh sides
x=+sqrt(12) =
+2*sqrt(3)

PS. My ISP went ill for a while so I missed some posts. I'll leave this as I had it typed even though it is redundent now. No response needed.
----------------
Gene

Yeah. I'm not that good at math. I'm sorry.
 
The Preacher said:
Gene said:
I'm using Qbic's description and variables which you used in
"its (duh) x^2+36=(2x)²"
After correcting (2x)² = 4x² just combine the x² terms on one side and the 36 on the other side and solve for x.

Dangit, I don't get it. I solved it wrong and got two, but it's because I isolated the \(\displaystyle x^2\) terms on one side and got \(\displaystyle 3x^2\) on the right side, and kept trying to solve it like it was \(\displaystyle (3x)^2\).

I just don't get this. =[

Argh!

Is it 36=3x\(\displaystyle ^2\)?

How do I get x? I'm sorry, it's been so long since I've done algebra. I keep thinking it's the square root of something. I'm getting very frustrated with this.

That equation...you can just divide both sides by 3
which makes; 12 = x^2
and that will become
X = √12
"That is if your equation is right"
 
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