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**demha****Member**- Registered: 2012-11-25
- Posts: 195

I need help with a lot of these math problems. Not sure exactly how to get them. Just want help understanding how to get the answer:

For 1 through 10, what is the area and volume of the given shape, if the length of one side of the base is 6, the height is 8, and the slant height is 10? (Not all shapes will require all three numbers.)

The shape is a right prism with:

1. an equilateral triangle as the base

A Lateral area: 90; Volume: 288

BLateral area: 192; Volume: 64

CLateral area: 120; Volume: 240

DLateral area: 144; Volume: 72 sqrt(3)

E Lateral area: 176; Volume: 18*sqrt(15)

F Lateral area: 100; Volume: 288

2. a rectangular base with a width of 3

A Lateral area: 192; Volume: 144 sqrt(3)

BLateral area: 144; Volume: 144

CLateral area: 90; Volume: 240

DLateral area: 120; Volume: 288

E Lateral area: 192; Volume: 64

F Lateral area: 144; Volume: 18*sqrt(15)

3. a square base

A Lateral area: 176; Volume: 240

BLateral area: 90; Volume: 64

CLateral area: 120; Volume: 144

DLateral area: 288; Volume: 144 sqrt(3)

E Lateral area: 144; Volume: 18*sqrt(15)

F Lateral area: 192; Volume: 288

4. a rectangular base with a width of 5

A Lateral area: 288; Volume: 96

BLateral area: 120; Volume: 248

CLateral area: 240; Volume: 144

D Lateral area: 192; Volume: 48

E Lateral area: 176; Volume: 240

F Lateral area: 100; Volume: 64

5. the base is an isosceles triangle with a height of 8 and a base of 3 (sides of 6)

A Lateral area: 120; Volume: 18*sqrt(15)

BLateral area: 100; Volume: 96

CLateral area: 192; Volume: 144

DLateral area: 176; Volume: 18*sqrt(15)

E Lateral area: 288; Volume: 288

F Lateral area: 240; Volume: 64

The shape is a pyramid with:

6. a rectangular base with a width of 4

A Lateral area: 120; Volume: 98

B Lateral area: 192; Volume: 76

C Lateral area: 240; Volume: 51

DLateral area: 176; Volume: 33

E Lateral area: 288; Volume: 75

F Lateral area: 100; Volume: 64

7. a square base

A Lateral area: 120; Volume: 96

BLateral area: 90; Volume: 64

CLateral area: 176; Volume: 144

DLateral area: 192; Volume: 35

E Lateral area: 288; Volume: 288

F Lateral area: 240; Volume: 75

8. a rectangular base with a width of 3

A Lateral area: 192; Volume: 144

BLateral area: 90; Volume: 48

CLateral area: 276; Volume: 64

DLateral area: 176; Volume: 144

E Lateral area: 92; Volume: 96

F Lateral area: 62; Volume: 24

9. a rectangular base with a width of 5

A Lateral area: 100; Volume: 48

BLateral area: 240; Volume: 112

CLateral area: 176; Volume: 96

DLateral area: 110; Volume: 80

E Lateral area: 288; Volume: 144

F Lateral area: 90; Volume: 64

10. a rectangular base with a width of 7

A Lateral area: 240; Volume: 64

BLateral area: 188; Volume: 96

CLateral area: 176; Volume: 144

DLateral area: 130; Volume: 112

E Lateral area: 144; Volume: 215

F Lateral area: 100; Volume: 128

I want to try and do the first 10 and once done, there is a second 10.

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~Abraham Lincoln

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**Fruityloop****Member**- Registered: 2009-05-18
- Posts: 120

It seems you get two equations with two variables.

One equation for the lateral area has two variables, say x and y.

One equation for the volume is another equation with x and y.

5. the base is an isosceles triangle with a height of 8 and a base of 3 (sides of 6)

I'm a little unclear with this.

The eclipses from Algol (an eclipsing binary star) come further apart in time when the Earth is moving away from Algol and closer together in time when the Earth is moving towards Algol, thereby proving that the speed of light is variable and that Einstein's Special Theory of Relativity is wrong.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,273

hi demha

This page shows many prisms with the cross section:

http://www.mathsisfun.com/geometry/prisms.html

I would make a sketch of each one; writing on the lengths given.

The formula for the volume of a prism is

The lateral area will require two calculations:

and

I don't know what is meant by the slant height for a prism.

Let's sort the prisms out before going on to the pyramids.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

Ok, from the link I understood how to get the area. So the base they give me is 6 as the base and the height is 8. So i have done:

A = .50 x 6 x 8 = 24

So now I have the area, I times it with height to get: V = 24 x 8 = 192

Now I'm not sure if I'm doing it EXACTLY right. I'm trying to figure out the answer for #1. To me it seems as if I got the lateral area which I feel is wrong:

The shape is a right prism with:

1. an equilateral triangle as the base

A Lateral area: 90; Volume: 288

B Lateral area: 192; Volume: 64

C Lateral area: 120; Volume: 240

D Lateral area: 144; Volume: 72 sqrt(3)

E Lateral area: 176; Volume: 18*sqrt(15)

F Lateral area: 100; Volume: 288

*Last edited by demha (2013-09-18 21:14:26)*

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,273

hi demha,

The height = 8 measurement cannot be for the triangle because it is equilateral with side = 6. There isn't enough space inside it for a height of 8. You can get it's height by using your earlier work on trig etc.

The 8 must refer to the height of the prism. Stand it on its triangular base and it is 8 high.

In my diagram I cannot make the triangle look equilateral that way round so I've shown it with the triangle as the front and then the 8 becomes its length.

So work out the height of the triangle, then its area, then the volume.

The rectangular sides are easy as they are all the same length and width.

Looking at those answers, I think the 'lateral' area may mean just the rectangles, not the triangular ends.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

What equation exactly would I be using exactly? would it be the

a^2 x b^2 = c^2

or would it be

A = (1/2)bh

"The thing about quotes on the Internet is you cannot confirm their validity"

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**bob bundy****Moderator**- Registered: 2010-06-20
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Use both. Pythag to get the height of the triangle and then the area of triangle formula.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

a^2 + b^2 = c^2

6^2 + 8^2 = c^2

36 + 64 = c^2

100 = c^2

c = 10

A = (1/2)bh

A = (1/2)(6)(8)

A = (1/2)48

A = 24

10 x 24 = 240. Is that the volume? \(^o^)/

"The thing about quotes on the Internet is you cannot confirm their validity"

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**bob bundy****Moderator**- Registered: 2010-06-20
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No! See new version of diagram.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

I'm really confused now. I'm not sure which numbers to use in which equations to get the height & area.

I would really appreciate you explaining it briefly and if you could, you can do like the previous lesson, create a similar problem and solve it while explaining.

*Last edited by demha (2013-09-19 05:56:23)*

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

I've done some further researching and I think I am understanding it more now. First I need to get the area of the triangle shape. I believe that would be:

6 x 3 = 18.

Now:

V = A x H

V = 18 x 8 | v = 288

Is that correct?

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**bob bundy****Moderator**- Registered: 2010-06-20
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OK. Let's say we have a prism which has an equilateral triangle for its base, side 4 cm, and it is 9 cm high.

What is its volume?

And the lateral area?

and if the ends should be included (I'm not sure; you'll have to look at the possible answers)

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

Just a quick questions, how does the first part (answer for the height) work out? I mean how did you get

2\sqrt{3}

in the end?

Also, isn't this another way to get the volume?

area of the front triangle x the height?

*Last edited by demha (2013-09-19 10:59:10)*

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,273

I have split the equilateral triangle in half so that there is a right angled triangle. The hypotenuse is 4 and one side is half that, ie 2

I want the height of the triangle so I used Pythagoras' theorem.

Now I can work out the area of the triangle.

Also, isn't this another way to get the volume?

area of the front triangle x the height?

??? That is exactly the method I am using. :)

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

Ok let me try it.

Volume:

h = sqrt{6^2 - 3^2}

h = sqrt{27}

h = sqrt{9 x 3}

h = 3 {sqrt 3}

area = 1/2 x (6) x 3 {sqrt 3} = 9 {sqrt 3}

volume = 9 {sqrt 3} x 8 = 72 {sqrt 3}

I think I got it this time

Lateral Area:

6 x 8 = 48

3 x 48 = 144

Now for #1, I am looking at D as my answer.

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**bob bundy****Moderator**- Registered: 2010-06-20
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I agree with those two answers. Hurrah!

The next one should be easier as it's a rectangular base.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

2. a rectangular base with a width of 3

A Lateral area: 192; Volume: 144 sqrt(3)

B Lateral area: 144; Volume: 144

C Lateral area: 90; Volume: 240

D Lateral area: 120; Volume: 288

E Lateral area: 192; Volume: 64

F Lateral area: 144; Volume: 18*sqrt(15)

So the base of the rectangle has sides of 3 and 6. So:

Volume:

H = 3 x 6

H = 18

H = sqrt {6 x 3}

H = 6 {sqrt 3}

A = 1/2 x 6 x 6 {sqrt 3}

A = 18 {sqrt 3}

V = 18 {sqrt 3} x 8= 144 {sqrt 3}

Lateral Area:

6 x 8 = 48

3 x 48 = 144

3. a square base

A Lateral area: 176; Volume: 240

BLateral area: 90; Volume: 64

CLateral area: 120; Volume: 144

DLateral area: 288; Volume: 144 sqrt(3)

E Lateral area: 144; Volume: 18*sqrt(15)

F Lateral area: 192; Volume: 288

Since it's a square base, all sides are equal, which means all sides measure 6. So:

Volume:

A = 6 x 6 = 36

V = 36 x 8 = 288

Lateral Area:

6 x 8 = 48

3 x 48 = 144.

Not sure what I'm doing wrong.

*Last edited by demha (2013-09-20 02:44:01)*

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,273

hi demha

2. a rectangular base with a width of 3

A Lateral area: 192; Volume: 144 sqrt(3)

B Lateral area: 144; Volume: 144

C Lateral area: 90; Volume: 240

D Lateral area: 120; Volume: 288

E Lateral area: 192; Volume: 64

F Lateral area: 144; Volume: 18*sqrt(15)

So the base of the rectangle has sides of 3 and 6. So:

Volume:

H = 3 x 6 This is the area of the end rectangle (18). Pythagoras and square roots are unnecessary as you have the area at this point.

H = 18

H = sqrt {6 x 3}

H = 6 {sqrt 3}

A = 1/2 x 6 x 6 {sqrt 3}

A = 18 {sqrt 3}

V = 18 {sqrt 3} x 8= 144 {sqrt 3}

Lateral Area:

The sides of the prism are (i) a rectangle 6 by 8 (ii) a rectangle 3 by 8 (iii) another rectangle 6 by 8 and finally (iv) another rectangle 3 by 8. What you have worked out here is not correctly the way to get the lateral area.

6 x 8 = 48

3 x 48 = 144

3. a square base

A Lateral area: 176; Volume: 240

BLateral area: 90; Volume: 64

CLateral area: 120; Volume: 144

DLateral area: 288; Volume: 144 sqrt(3)

E Lateral area: 144; Volume: 18*sqrt(15)

F Lateral area: 192; Volume: 288

Since it's a square base, all sides are equal, which means all sides measure 6. So:

Volume:

A = 6 x 6 = 36

V = 36 x 8 = 288 Correct!

Lateral Area:

6 x 8 = 48 So far so good......

3 x 48 = 144.How many lateral faces are there?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

2. a rectangular base with a width of 3

A Lateral area: 192; Volume: 144 sqrt(3)

B Lateral area: 144; Volume: 144

C Lateral area: 90; Volume: 240

D Lateral area: 120; Volume: 288

E Lateral area: 192; Volume: 64

F Lateral area: 144; Volume: 18*sqrt(15)

So the base of the rectangle has sides of 3 and 6. So:

Volume:

A = 3 x 6 This is the area of the end rectangle (18). Pythagoras and square roots are unnecessary as you have the area at this point.

A = 18

V = 18 x 8= 144

Lateral Area:

So would it be:

2 x 6 x 8 = 96

2 x 3 x 8= 48

96 + 48 = 144 again... which means answer should be B?

A Lateral area: 176; Volume: 240

BLateral area: 90; Volume: 64

CLateral area: 120; Volume: 144

DLateral area: 288; Volume: 144 sqrt(3)

E Lateral area: 144; Volume: 18*sqrt(15)

F Lateral area: 192; Volume: 288

Since it's a square base, all sides are equal, which means all sides measure 6. So:

Volume:

A = 6 x 6 = 36

V = 36 x 8 = 288

Lateral Area: (there are FOUR lateral faces for a square )

6 x 8 = 48

4 x 48 = 192.

Answer is F.

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

(literally same as #2)

4. a rectangular base with a width of 5

A Lateral area: 288; Volume: 96

BLateral area: 120; Volume: 248

CLateral area: 240; Volume: 144

D Lateral area: 192; Volume: 48

E Lateral area: 176; Volume: 240

F Lateral area: 100; Volume: 64

Volume:

A = 5 x 6

A = 30

V = 30 x 8

V = 240

Lateral Area:

2 x 6 x 8 = 96

2 x 5 x 8 = 80

96 + 80 = 176

Answer: E

5. the base is an isosceles triangle with a height of 8 and a base of 3 (sides of 6)

A Lateral area: 120; Volume: 18*sqrt(15)

B Lateral area: 100; Volume: 96

C Lateral area: 192; Volume: 144

D Lateral area: 176; Volume: 18*sqrt(15)

E Lateral area: 288; Volume: 288

F Lateral area: 240; Volume: 64

Would I be using the same method as #1 to get the volume & lateral area for this?

*Last edited by demha (2013-09-20 09:50:58)*

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

This is what I created for #5. Am I correct with this image?

Sorry if it's so huge I'm trying to figure out how to work this image upload thing

*Last edited by demha (2013-09-20 11:26:59)*

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,273

hi demha

A Lateral area: 192; Volume: 144 sqrt(3)

B Lateral area: 144; Volume: 144

C Lateral area: 90; Volume: 240

D Lateral area: 120; Volume: 288

E Lateral area: 192; Volume: 64

F Lateral area: 144; Volume: 18*sqrt(15)

So the base of the rectangle has sides of 3 and 6. So:

Volume:

A = 3 x 6 This is the area of the end rectangle (18). Pythagoras and square roots are unnecessary as you have the area at this point.

A = 18

V = 18 x 8= 144

Lateral Area:

So would it be:

2 x 6 x 8 = 96

2 x 3 x 8= 48

96 + 48 = 144 again... which means answer should be B?

Correct. Yes, you had the right answer before but not for the right reason. I thought it important that you understand the method; not just get the answer by luck.

A Lateral area: 176; Volume: 240

BLateral area: 90; Volume: 64

CLateral area: 120; Volume: 144

DLateral area: 288; Volume: 144 sqrt(3)

E Lateral area: 144; Volume: 18*sqrt(15)

F Lateral area: 192; Volume: 288

Since it's a square base, all sides are equal, which means all sides measure 6. So:

Volume:

A = 6 x 6 = 36

V = 36 x 8 = 288

Lateral Area: (there are FOUR lateral faces for a square big_smile)

6 x 8 = 48

4 x 48 = 192.

Answer is F.

Correct!

4. a rectangular base with a width of 5

A Lateral area: 288; Volume: 96

BLateral area: 120; Volume: 248

CLateral area: 240; Volume: 144

D Lateral area: 192; Volume: 48

E Lateral area: 176; Volume: 240

F Lateral area: 100; Volume: 64

Volume:

A = 5 x 6

A = 30

V = 30 x 8

V = 240

Lateral Area:

2 x 6 x 8 = 96

2 x 5 x 8 = 80

96 + 80 = 176

Answer: E

Correct!

5. the base is an isosceles triangle with a height of 8 and a base of 3 (sides of 6)

A Lateral area: 120; Volume: 18*sqrt(15)

B Lateral area: 100; Volume: 96

C Lateral area: 192; Volume: 144

D Lateral area: 176; Volume: 18*sqrt(15)

E Lateral area: 288; Volume: 288

F Lateral area: 240; Volume: 64

Would I be using the same method as #1 to get the volume & lateral area for this?

Yes, that's right. And you diagram looks good to me too.

Sorry if it's so huge I'm trying to figure out how to work this image upload thing

Well the image upload worked so that's ok. You could have cropped some white space to save on the image size. If you're using Paint you can move the right and bottom borders in, then rotate the image 180 and shrink the other two borders.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

5. the base is an isosceles triangle with a height of 8 and a base of 3 (sides of 6)

A Lateral area: 120; Volume: 18*sqrt(15)

B Lateral area: 100; Volume: 96

C Lateral area: 192; Volume: 144

D Lateral area: 176; Volume: 18*sqrt(15)

E Lateral area: 288; Volume: 288

F Lateral area: 240; Volume: 64

Now for #1 I was using the base and half the base (which was 6 & 3) so I am assuming I do the same here. So it would be:

h = sqrt{3^2 - 1.5^2}

h = sqrt{9 - 2.25}

h = sqrt{6.75}

h = .45 {sqrt 15}

a = 1/2 x 3 x .45 {sqrt 15}

a = 0.675

v = 0.675 x 8 = 5.4 {sqrt 15}

And that looks terribly wrong...

Lateral Area:

3 x 8 = 24

3 x 24 = 72

"The thing about quotes on the Internet is you cannot confirm their validity"

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**bob bundy****Moderator**- Registered: 2010-06-20
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Look back to your diagram. The triangle has sides 6,6 and 3. Make a right angle by finding the midpoint of the base. The right angled triangle has sides of 6, 1.5 and h, where h is the height of the triangle.

Then half base x h and finally x by 8.

The lateral area isn't correct either because only one of the lateral rectangles is 3 by 8.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**demha****Member**- Registered: 2012-11-25
- Posts: 195

Ok I will use Pythagorean theorem to find the height.

h^2 + 1.5^2 = 6^2

h^2 + 2.25 = 36

h^2 = 33.75

h = 5.80

v = 1.5 x 5.80 x 8

v = 69.6

I still don't know what's going wrong. Could you give me an example of how to find the area of this triangle?

Also, I had an idea for getting the lateral area:

2 x 6 x 8 = 96

1 x 3 x 8 = 24

96 + 24 = 120

L = 120

*Last edited by demha (2013-09-21 02:08:11)*

"The thing about quotes on the Internet is you cannot confirm their validity"

~Abraham Lincoln

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