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**bob bundy****Moderator**- Registered: 2010-06-20
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Ok, let's see. L = 120 looks OK for the lateral area. So that suggests answer A. You have a decimal answer and A has a root 15. I expect you can get that too by using exact square roots.

h^2 + 1.5^2 = 6^2

This is a good start so how about writing 6 as 4 x 1.5 ?

If you make h^2 the subject of this and leave the 1.5 squared as it is, I think you will find that root 15 creeps in soon

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: 186

Ok, so:

h^2 + 1.5^2 = 4^2 x 1.5^2

h^2 + 1.5^2 = 16 x 1.5^2

h^2 + 1.5^2 = 24^2

is that right so far?

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**Fruityloop****Member**- Registered: 2009-05-18
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16 x 1.5^2 != 24^2, 16^2 x 1.5^2 = 24^2.

Try moving the 1.5^2 to the right side of the equation and see what you can do.

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
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hi demha

h^2 + 1.5^2 = 4^2 x 1.5^2

h^2 + 1.5^2 = 16 x 1.5^2 Ok so far.

but don't do this next step.

h^2 + 1.5^2 = 24^2

Leave the 1.5^2 exactly as it is. Make h^2 the subject and factorise the 1.5^2. When you take the square root that element will become just 1.5 once more and the required root 15 will appear. By doing the working this way, you keep clear of an approximate, decimal answer.

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
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h^2 + 1.5^2 = 4^2 x 1.5^2

h^2 + 1.5 = 16 x 1.5

I am really confused I have no idea what I'm doing.

"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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Don't panic. It's a little algebraic trick. Once you've seen it done, it will stick in your brain and you'll be able to use it again as necessary.

The problem with doing the question your way:

h^2 + 1.5^2 = 6^2

h^2 + 2.25 = 36

h^2 = 33.75

h = 5.80

is you end up with an approximate decimal answer, and a quick look at the multiple choice answers tells us that's not what they want.

So we have to find a way to avoid that. The 1.5 is where the difficulty starts, so avoid doing anything with it:

Rearrange:

Factorise the 1.5 to get it 'out of the way'.

Now square root both sides

You don't need to work out a decimal value as you now have the **root 15** in your answer but, just for your own satisfaction, try working out that number. You'll see that you had the correct answer before; but not written in the way the question requires.

Now you have h, you should be able to work out the half base x height for the triangle and multiply by 8 for the volume.

Make sure you use **root 15** for the rest of the working and I recommend that you write the final expression with the 1.5 as it is (ie. Don't evaluate the area, leave it as a calculation to be done) and then use the 8 as 2 x 2 x 2 to 'convert' each 1.5 (x2) into 3. You see, I'm still trying to avoid any decimals, even at this stage.

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: 186

Math is my only weakest subject I'm not too good at it. It takes me time to understand how to do things : /

so if h = 1.5 {sqrt 15}, then

a = 1/2 x 3 x 1.5 {sqrt 15}

a = 2.25 {sqrt 15}

v = 8 x 2.25 {sqrt 15}

v = 18 {sqrt 15} FINALLY. first 5 done and another 15 to go o_O

So now for the second set of 5 questions:

!we should keep in mind!

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.)

~~~

It seems as if the slant height was meant for these pyramid shapes.

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

D Lateral 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

B Lateral area: 90; Volume: 64

C Lateral area: 176; Volume: 144

D Lateral 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

B Lateral area: 240; Volume: 112

C Lateral area: 176; Volume: 96

D Lateral 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

B Lateral area: 188; Volume: 96

C Lateral area: 176; Volume: 144

D Lateral area: 130; Volume: 112

E Lateral area: 144; Volume: 215

F Lateral area: 100; Volume: 128

*Last edited by demha (2013-09-21 23:27:17)*

"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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hi demha,

Hurrah! The prisms are finally done correctly.

The volume formula for any pyramid (including a cone) is

So very much like the prisms but with that 1/3 element because the pyramid tapers towards the vertex.

So you need to sketch the shape; and work out the base area. They all seem to be rectangles or squares so that should make it easier.

The lateral areas will all be triangles. The bottom of the triangle will be part of the base.

I'm having trouble myself making sense of what they mean by the slant height.

Do you think you can try Q6 volume while I look at the lateral areas. If you have a diagram that shows what is meant by the slant height please show me.

EDIT: I've found out but it hasn't helped. I've asked some other members what they think.

Bob

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**demha****Member**- Registered: 2012-11-25
- Posts: 186

Well this is what my lesson says and shows:

"Regular Pyramid Area Theorem

The area L of any regular pyramid with a base that has perimeter P and with slant height l is equal to one-half the product of the perimeter and the slant height.

Formula: L = (1/2)(P)l"

"A pyramid is a polyhedron with a single base and lateral faces that are all triangular. All lateral edges of a pyramid meet at a single point, or vertex. A regular pyramid is a pyramid that has a base that is a regular polygon and with lateral faces that are all congruent isosceles triangles.

At any rate, the equations for area and volume are just like all the others we've done--just plug in the numbers.

If a regular pyramid has a square base with a length and width of 3, and slant height of 5, and a height of 4 then what is the area of the pyramid, and what is the pyramid's volume.

L = (1/2)(P)l

L = (1/2)(12)5

L = 30

V = (1/3)Bh

V = (1/3)9(4)

V = 12

"

So I guess the Slant Height is used for finding the Lateral Area. I have a question though, is the Perimeter the same thing as the area of the Base?

And I would try to do #6 my self but I am visiting my grandparents at the moment so it will have to wait for now

*Last edited by demha (2013-09-22 05:54:50)*

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Hi demha and Bob

Q6 is about to give me a headache. Based on the given information, the volume should be divisible by 8, and that is possible only if the answer is E. But in that case I can't get the sides of the base such that the lateral area is 100.

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**bob bundy****Moderator**- Registered: 2010-06-20
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demha wrote:

I have a question though, is the Perimeter the same thing as the area of the Base?

No. The perimeter is the distance all the way round the shape.

I'm happy with what you have been taught for a regular polygon base. But the rectangles are not regular polygons. It looks to me like the person who set these questions may have slipped up. For me Q6 lateral area is not possible. For Q7 I get one letter for the volume but a different letter for the lateral area. Very odd .....

Let's see if Stefy has a different opinion.

Bob

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**bob bundy****Moderator**- Registered: 2010-06-20
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Stefy wrote:

But in that case I can't get the sides of the base such that the lateral area is 100.

For all these questions (1-10) you're supposed to take the height = 8, one side of the base = 6 and the slant height = 10.

But my first difficulty is: if you know the side and the height, then Pythag gives the slant height and a rectangle cannot have the same slant height for two different sides.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
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The problem with Q6 is it is over-determined.

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Hi Bob

Have you seen this bit:

If a regular pyramid has a square base with a length and width of 3, and slant height of 5, and a height of 4 then what is the area of the pyramid, and what is the pyramid's volume.

L = (1/2)(P)l

L = (1/2)(12)5

L = 30V = (1/3)Bh

V = (1/3)9(4)

V = 12

???

*Last edited by anonimnystefy (2013-09-22 07:34:13)*

Here lies the reader who will never open this book. He is forever dead.

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**bob bundy****Moderator**- Registered: 2010-06-20
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Yes, I have!!!! And the formula given in the lesson only works for a pyramid with a regular polygonal base. So why have questions with rectangles?

Have you tried Q7 yet ? (assume that the slant height is correct)

Bob

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**anonimnystefy****Real Member**- From: The Foundation
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Well, they seem to be assuming that (side of base)^2+(height)^2=(slant height)^2, which is never true...

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**bob bundy****Moderator**- Registered: 2010-06-20
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But use the formulas anyway (volume and lateral area).

Bob

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**anonimnystefy****Real Member**- From: The Foundation
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Then I'm getting A as the answer for Q7.

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**bob bundy****Moderator**- Registered: 2010-06-20
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Thank you so much. It was driving me mad because I wasn't getting the right volume ... but, somehow, just having you say that has re-aligned my brain correctly to answer A too. Oh joy! Just got to square these formulas with the warped non Euclidean space that they occupy and I'm there. Many, many thanks for your input. I think I'm OK to proceed now.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
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Huh? I'm not sure what to make of your reply. I don't know how you will proceed when their questions are incorrect.

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**demha****Member**- Registered: 2012-11-25
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It's interesting to see two math geniuses at work

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

D Lateral area: 176; Volume: 33

E Lateral area: 288; Volume: 75

F Lateral area: 100; Volume: 64

So this is what I did. First I got the area of the rectangle:

a = 6 x 4

a = 24

v = 1/3 x 24 x 8

v = 64

That's what I'm getting for the volume.

As for the lateral area, would I have to do:

2 x 6 x 8

2 x 4 x 8

Get the answers for these, add them together to get the perimeter and then move on?

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

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**anonimnystefy****Real Member**- From: The Foundation
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It truly is. The way you two tumble down those problems is amazing!

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**bob bundy****Moderator**- Registered: 2010-06-20
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hi demha,

Glad you enjoyed our discussion about these problems. We have identified a big error with the work that Compu High are teaching you. The teaching page is available as a 'demo' so I have been able to read it all. The volume calculations are fully correct but the slant height section is WRONG.

Because I am a teacher of maths, I cannot just let you do these questions without showing you the correction. I'll do that first. Then you'll want to get their answers so you get full marks. That's OK. I have used their method and Q6 to Q10 all have a 'correct' answer. So hopefully, at the end, you will understand the work properly and be able to get full marks. I would just add, this is not the only criticism I have of this on-line course but I won't burden you with all my objections. Let's get on.

Correction: If you buy some wire so you can make a wire-frame version of these pyramids, you could proceed as follows. (see diagram)

(i) Cut two lengths **a** and two widths **b**. Join them to make the base rectangle.

(ii) Join the diagonals to find the middle and make a temporary wire for the height **h**. This fixes the vertex.

**At this point the five points of the pyramid are all fixed.**

(iii) You can cut four wires to join the base to the vertex and the model is complete.

**It is not necessary is say how long the red lines for the slant heights are because they are already there, whatever size they need to be to join the midpoints of the base to the vertex.** Just choosing a length for these red lines is not an option. Their lengths can be calculated using Pythagoras' theorem. Furthermore, with a rectangular base the two slant heights will not be the same, so giving a single slant height value is only possible where the base is a regular polygon. A rectangle is not a regular polygon.

This is why I have had such a problem with the slant height values given. The pyramids described cannot exist because those slant height values won't work!

Sorry for the rant, but I felt I had to get that off my chest. I feel better. If that's leaving you worried and confused I apologise, but I have to take account of the possibility that other members of the forum may be reading this thread and I don't want them being taught incorrectly.

So now, back to the problems. You can, of course, ignore all I have just said, and concentrate on getting the 'right' answers. (Anyway, once you have a volume, you'll know which answer to choose, without having to calculate the lateral areas at all. This is another criticism whoops, I'd better move on!)

The formula for the volume is

You have done Q6 volume correctly.

Their formula for lateral area is

So what you need to do is

(i) Add a + b + a + b to get the perimeter (distance around the base).

(ii) Calculate 1/2 of this and multiply by the slant height.

You can be fairly confident you have the right result if you can choose a single multi-choice answer that fits both the volume and area you have worked out.

Good luck.

Bob

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**demha****Member**- Registered: 2012-11-25
- Posts: 186

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

D Lateral area: 176; Volume: 33

E Lateral area: 288; Volume: 75

F Lateral area: 100; Volume: 64

Volume:

a = 6 x 4

a = 24

v = 1/3 x 24 x 8

v = 64

Lateral Area:

6 + 4 + 6 + 4 = 20

20 x .50 = 10

10 x 10 = 100

Answer: F

7. a square base

A Lateral area: 120; Volume: 96

B Lateral area: 90; Volume: 64

C Lateral area: 176; Volume: 144

D Lateral area: 192; Volume: 35

E Lateral area: 288; Volume: 288

F Lateral area: 240; Volume: 75

Volume:

a = 6 x 6

a = 36

v = 1/3 x 36 x 8

v = 96

Lateral Area:

6 + 6 + 6 + 6 = 24

24 x .50 = 12

12 x 10 = 120

Answer: A

8. a rectangular base with a width of 3

A Lateral area: 192; Volume: 144

B Lateral area: 90; Volume: 48

C Lateral area: 276; Volume: 64

D Lateral area: 176; Volume: 144

E Lateral area: 92; Volume: 96

F Lateral area: 62; Volume: 24

Volume:

a = 6 x 3

a = 18

v = 1/3 x 18 x 8

v = 48

Lateral Area:

3 + 6 + 3 + 6 = 18

18 x .50 = 9

9 x 10 = 90

Answer: B

9. a rectangular base with a width of 5

A Lateral area: 100; Volume: 48

B Lateral area: 240; Volume: 112

C Lateral area: 176; Volume: 96

D Lateral area: 110; Volume: 80

E Lateral area: 288; Volume: 144

F Lateral area: 90; Volume: 64

Volume:

a = 5 x 6

a = 30

v = 1/3 x 30 x 8

v = 80

Lateral Area:

6 + 5 + 6 + 5 = 22

22 x .50 = 11

11 x 10 = 110

Answer: D

10. a rectangular base with a width of 7

A Lateral area: 240; Volume: 64

B Lateral area: 188; Volume: 96

C Lateral area: 176; Volume: 144

D Lateral area: 130; Volume: 112

E Lateral area: 144; Volume: 215

F Lateral area: 100; Volume: 128

Volume:

a = 6 x 7

a = 42

v = 1/3 x 42 x 8

v = 112

Lateral Area:

7 + 6 + 7 + 6 = 26

26 x .50 = 13

13 x 10 = 130

Answer: D

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**demha****Member**- Registered: 2012-11-25
- Posts: 186

Last 10 to go!

I have an isosceles triangle with a height of 4 and a base of 6:

11. What is the area? - Answer: C

A 19

B 35

C 12

D 16

E 22

F 54

This triangle just became the base of a regular prism, with a height of 8:

12. What is the lateral area? - Answer: D

A 105

B 28

C 35

D 128

E 56

F 12

13. What is the volume? **need help solving

A 95

B 54

C 32

D 23

E 96

F 10

14. What is the area of the largest rectangular side? - Answer: D

A 95

B 54

C 32

D 48

E 96

F 10

*Last edited by demha (2013-09-23 03:31:44)*

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

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