Right Parallelepiped

[Stefany]

Can someone tell me where to start solving this?

I have a right parallelepiped; c (side of the base) = 5 cm; B (base area) = 360 sq. cm; d (diagonal) = 200 cm.
I need to find a and b (sides of the base); S (surface area); V (volume)

All of the formula's I have aren't working because I don't know what a and b are...

 

[Dr.Peterson]

Can someone tell me where to start solving this?

I have a right parallelepiped; c (side of the base) = 5 cm; B (base area) = 360 sq. cm; d (diagonal) = 200 cm.
I need to find a and b (sides of the base); S (surface area); V (volume)

All of the formula's I have aren't working because I don't know what a and b are...

You say a, b, and c are all "sides of the base". Isn't the base a rectangle?

Is the diagonal d the body diagonal (between opposite corners of the entire solid), or a face diagonal? A picture may help.

I'd guess that c is really the height, so the base is ab = 360 cm^2, and d^2 = a^2 + b^2 + 5^2. If so, then you have two equations in a and b, which can be solved by solving the first for b and substituting in the second.

 

[Stefany]

You say a, b, and c are all "sides of the base". Isn't the base a rectangle?

Is the diagonal d the body diagonal (between opposite corners of the entire solid), or a face diagonal? A picture may help.

I'd guess that c is really the height, so the base is ab = 360 cm^2, and d^2 = a^2 + b^2 + 5^2. If so, then you have two equations in a and b, which can be solved by solving the first for b and substituting in the second.

Here is a picture of what I guess it’s suppose to look like.

 

[Stefany]

You say a, b, and c are all "sides of the base". Isn't the base a rectangle?

Is the diagonal d the body diagonal (between opposite corners of the entire solid), or a face diagonal? A picture may help.

I'd guess that c is really the height, so the base is ab = 360 cm^2, and d^2 = a^2 + b^2 + 5^2. If so, then you have two equations in a and b, which can be solved by solving the first for b and substituting in the second.

I’m just a little confused, is it suppose to look like this d^2 = 360^2 + 5^2?

I don’t think I did it right.

 

[Deleted member 4993]

I’m just a little confused, is it suppose to look like this d^2 = 360^2 + 5^2?

I don’t think I did it right.

How did you get that?

d = 200 (given). and

ab = 360 \(\displaystyle \ \to \ \ \) b = 360/a

So:

a^2 + (360/a)^2 + 5^2 = 200^2

Now continue.....

 

[Dr.Peterson]

I’m just a little confused, is it suppose to look like this d^2 = 360^2 + 5^2?

I don’t think I did it right.

No, there is no theorem like that. The Pythagorean theorem, applied twice, yields a^2 + b^2 + c^2 = d^2.

 

[Stefany]

How did you get that?

d = 200 (given). and

ab = 360 \(\displaystyle \ \to \ \ \) b = 360/a

So:

a^2 + (360/a)^2 + 5^2 = 200^2

Now continue.....

Is it suppose to be something like this?

a^2 + 129600/a^2 = 39975^2

 

[Dr.Peterson]

Is it suppose to be something like this?

a^2 + 129600/a^2 = 39975^2

Yes. Keep going ...

 

[Stefany]

Yes. Keep going ...

I did but it looks wrong

 

[Dr.Peterson]

I did but it looks wrong

Please show your work so we can help!

Looking back at the problem, I observe that the diagonal is unexpectedly long, at 200 compared to the height of 5. This implies that one side of the base will be nearly 200 cm, too. The extreme dimensions may be what makes it feel wrong to you.

But you should check that you copied the problem correctly. Please show us what the actual problem says.

 

[Stefany]

This is what I got...

I’m not really sure if the last part is correct, also I don’t have the original problem with me because I copied it down at school.
But this doesn’t seem right at all so I guess it’s 20 instead of 200, I don’t know

 

[Dr.Peterson]

Is it suppose to be something like this?

a^2 + 129600/a^2 = 39975^2

I'd missed something here: that right side shouldn't be squared! It comes from 200^2 - 5^2, and there's no reason to square that again.

So try again with that correction -- and take it to the end, actually calculating a numerical solution.

Then try it again, replacing 200 with 20, in case that's the right number. But I think that may actually be too small.

 

[Stefany]

I'd missed something here: that right side shouldn't be squared! It comes from 200^2 - 5^2, and there's no reason to square that again.

So try again with that correction -- and take it to the end, actually calculating a numerical solution.

Then try it again, replacing 200 with 20, in case that's the right number. But I think that may actually be too small.

Am I suppose to get something like this?

a^4 - 39975.a^2 + 129600 = 0

Because if so, I tried solving that and the numbers I’m getting are huge.

 

[Dr.Peterson]

Am I suppose to get something like this?

a^4 - 39975.a^2 + 129600 = 0

Because if so, I tried solving that and the numbers I’m getting are huge.

Yes, I've told you to go ahead and solve that, and that some numbers will be large, in some sense.

Please do what I say and show us the numbers you get, so I can tell you whether they are right or wrong. Giving us minimal information just wastes time.

 

[Stefany]

Yes, I've told you to go ahead and solve that, and that some numbers will be large, in some sense.

Please do what I say and show us the numbers you get, so I can tell you whether they are right or wrong. Giving us minimal information just wastes time.

???

 

[Dr.Peterson]

The equation you are solving is

[MATH]u^2 - 39,975u + 129,600 = 0[/MATH]​

so that a=1, b=-39,975, and c=129,600. Therefore the discriminant is

[MATH]b^2-4ac = (-39975)^2-4(1)(129,600) = 1,598,000,625-518,400 = 1,597,482,225[/MATH]​

You had a sign error there.

But you have to continue! The solution is therefore

[MATH]u=\frac{39,975\pm\sqrt{1,597,482,225}}{2} = \frac{39,975\pm 39,968.5}{2} =39,971\text{ or }3.24[/MATH]​

There is still one more step after this, which will make the numbers smaller. You'll find the answer is just what I said it would be.

Don't stop prematurely and think the numbers are too large!!

 

[Stefany]

The equation you are solving is

[MATH]u^2 - 39,975u + 129,600 = 0[/MATH]​

so that a=1, b=-39,975, and c=129,600. Therefore the discriminant is

[MATH]b^2-4ac = (-39975)^2-4(1)(129,600) = 1,598,000,625-518,400 = 1,597,482,225[/MATH]​

You had a sign error there.

But you have to continue! The solution is therefore

[MATH]u=\frac{39,975\pm\sqrt{1,597,482,225}}{2} = \frac{39,975\pm 39,968.5}{2} =39,971\text{ or }3.24[/MATH]​

There is still one more step after this, which will make the numbers smaller. You'll find the answer is just what I said it would be.

Don't stop prematurely and think the numbers are too large!!

Is this the next step?

 

[Dr.Peterson]

Is this the next step?

Yes; but check that second answer (I think you missed a digit), and reconsider the [MATH]\pm[/MATH] in view of the nature of the problem.

 

[Stefany]

Yes; but check that second answer (I think you missed a digit), and reconsider the [MATH]\pm[/MATH] in view of the nature of the problem.

Yes, I did miss a digit. It’s suppose to be 199.92 or maybe I could just leave it at 199.9.

Also, since I found a, can I use B = a.b to find b?

 

[Dr.Peterson]

Yes, I did miss a digit. It’s suppose to be 199.92 or maybe I could just leave it at 199.9.

Also, since I found a, can I use B = a.b to find b?

You don't have to ask for permission! You were told to find b, so do it. (And be pleased with what you find!)

I suggest using a few more decimal places in your values for a, in order to avoid a little confusion about b.