Question

PLEASE CHECK ANSWER

Answer 1

but I'm really confused as to how to do this

Answer 2

im sorry that stuff is so confusing to me

Answer 3

Is 2x+8 the formula for the base and 6x the formula for the diagonal?

Answer 4

Just trying to get clarification before I help out.

Answer 5

I believe so

Answer 6

Yeah it is

Answer 7

So, the dependency between the diagonals of a parallelogram to the sides is given by:

\[p^2 + q^2 = 2 (a^2 + b^2)\]

where p and q are the diagonals and a and b are the sides.

Answer 8

So, it would be 6x for p and q?

Answer 9

No, that would only be true in the case of a square. What we need to do is use the Pythagorean theorem so solve for the other diagonal.

Answer 10

the key in this problem is not to bother taking calculations too far.

So, 2x+8 is the length of the hypotenuse of the upper triangle (BC and the point in the middle). Do you see that?

Answer 11

Yes

Answer 12

Sorry, having trouble with page and trying to type in formula. lol

Answer 13

It's fine lol I just want to know how to solve this

Answer 14

(6x)^2 + (Y)^2 = (2x+8)^2
Y^2 = (2x+8)^2 - (6x)^2 (this is half the diagonal, so we will double it)
2Y^2 = 2(2x+8)^2-2(6x)^2 (this is the whole diagonal of the other side)

make sense?

Answer 15

the first line is the p-theorem formula. and then solving to the end to find the diagonal.

Answer 16

Uh.. Do I solve the last one?

Answer 17

No, we just hang on it. Also, I forgot that 6x was the whole diagonal... so it's actually 3x instead (half of the diagonal). Next we need to look at the other side... thankfully it is also equal to 2x+8.

Answer 18

So, now we just fill in that formula that I gave you before about the dependency between diagonals and sides and solve for x.

Answer 19

Sorry it is taking so long. Not too much more to go though.

Answer 20

It's okay.

Answer 21

So, do 6x is 3x now? Did you divide it by two? Or just because it's half of 6

Answer 22

Right I divided it by two (6x / 2 = 3x)
so we start with \[a^2+b^2 = p^2+q^2\]

Answer 23

\[(2x+8)^2 + (2x+8)^2 = p^2 + q^2\]

Answer 24

Do I solve both? If so, do I just multiply 8 * 8 in both equations, then add 2?

Answer 25

Sorry, was hoping to edit what I wrote, the left side should be multiplied by 2
\[p^2+q^2 = 2(a^2+b^2)\]

Answer 26

Not quite that simple. We need to handle the right side of the equation.
a^2 = (6x)^2
b^2 = our 2Y^2 that we got when we solved the P-theorem. We plug those in and then solve for x.

Answer 27

but I thought 6x was 3x now?

Answer 28

I'm sorry Chester. I meant that p^2 = (6x)^2 and q^2 was what we solve before. <sigh> this is a touch problem to help with on the computer. Lots of book work.

Answer 29

p and q represent the diagonals.

Answer 30

Okay, I'm confused now, can you write it out as an equation ?

Answer 31

like the equation I would need to solve the problem?

Answer 32

Let's try and identify all of what we know about the figure.

Let a and be be the length of the sides
Let p and q be the length of the diagonals.
We know \[p^2+q^2 = 2(a^2+b^2)\]
We also know that \[a=b\]
\[a = 2x+8\]
\[b=2x+8\]
\[p=6x\]
q = other diagonal.

Answer 33

how do we find the the diagonal?

Answer 34

other**

Answer 35

We found the other diagonal by using the Pythagorean theorem.

Answer 36

\[2((2x+8)^2+(2x+8)^2)=(6x)^2+q^2\]

Answer 37

Oh okay, so what's the next step? Do we start solving now?

Answer 38

Right after we substitute for q...which I am just getting ready to put in.

Answer 39

\[q^2 = 2(2x+8)^2-2(6x)^2\]

Answer 40

Finally,

\[2[(2x+8)^2+(2x+8)^2]=(6x)^2+2(2x+8)^2-2(6x)^2\]

Answer 41

Now, you can solve for x... lots of like terms thankfully.

Answer 42

Okay, so do I do, 2 *2 + 8^2 ?

Answer 43

Yep.

Answer 44

Sorry for the long description. Hard to capture everything in small segments to explain it.

Answer 45

It's fine. So that = 68