Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

hi markosheehan,

Here's the sort of calculation that should do the job:

Each year interest is gained at 3% so a multiplier for the change in principal is m = (1+3/100) = 1.03

REVISED ANSWER:

He is taking out 20000 from his pot each year of retirement for 25 years then start with a pot = Q.

At the end of retirement year 1 he has (Q-20000)m

At the end of retirement year 2 he has (Qm - 20000m - 20000)m

and so on. After 25 years this will be equal zero from which you can work out Q.

Qm^24 - 20000m^25 - .... -1 = 0

So

Hope that helps.

Bob

hi Wangotango

Define S to be where the plane CPQ intersects the x axis (AD).

S lies in the plane of ABCD and in the plane of PQRC so CP extended must go through S

This method (from championshipgirl) is much simpler than my original method. But it does show that many methods may all lead to the same result and shows the unity of geometry, vector algebra and coordinate geometry. Neat!

Bob

hi markosheehan

You need to use the extension rather than the natural length. The 'restoring' force is proportional to the extension but in the opposite direction so that gives you an equation like this:

That is the defining equation for SHM. I think, but you'll need to check me on this, that k = 500.

I notice that this is just part (i) If the subsequent parts involve velocities, positions and times, I expect I can do these too.

(Used to be my favourite equation until I discovered this:

)

Bob

Here's one I did recently:

And this is the code for it:

`[img]http://i.imgur.com/vwe2bk6.gif[/img]`

Bob

hi Aeronaesis

Which is a contradiction, because the area of a circle doesn't equal base times height.

True, but I don't think that 'proves' the impossibility. Say a rectangle was 6 π long and 6 wide; it would have an area of 36 π . This is the same as a circle with radius 6.

Bob

hi bobbym,

From memory it was

A company logo consists of a circle with an isosceles triangle inscribed inside so that the three vertices lie on the circumference. Prove that it is impossible for the area of the triangle to be exactly half that of the circle.

I have restored that above.

Bob

hi 010595

Welcome to the forum.

There are several construction techniques that allow a triangle to be 'converted' into an equal area square.

It would be relatively easy to make that a square with twice the area of the triangle.

It has been shown to be impossible to construct a square equal in area to a circle,

https://en.wikipedia.org/wiki/Squaring_the_circle

Bob

Well done!

Bob

hi dazzle1230

This problem has been asked before here:

http://www.mathisfunforum.com/viewtopic … 83#p365383

Bob

hi basketballstar123

This wasn't my proof, but here's how I know this to be true. The circle centred on G defines the point H. AHB is 90 at it is the angle at the circumference subtended by the diameter AB **. AB is parallel to CD so angle AHB = HBD (use x + y = 90).

Bob

** http://www.mathisfunforum.com/viewtopic.php?id=17799 post #6

hi Zeeshan 01

When you divide by 10 the point moves one place.

So that's why you need to move the point 5 places.

Bob

hi lofi

Welcome to the forum.

The circle is inscribed in a square with sides 8 cm. What is the area of the shaded part in square centimeters? Express your answer in terms of pi?

I think I can help here without the diagram.

You have a square 8 by 8 so its area is 64.

The circle fits inside so that it just touches the sides of the square. So its diameter is 8 and so radius 4. area = pi x 4squared.

I'm guessing the shaded bit is the part inside the square but outside the circle.

So subtract the second area from the first.

Bob

hi basketballstar123

Welcome to the forum.

This question has been posted before:

http://www.mathisfunforum.com/viewtopic.php?id=22435

Bob

hi anniep418

Q2.

tan(54)=a/5

Right angled triangle! That should be a/2.5

There are 10 right angle triangles.

The base and height are 'a' and 2.5. Make sure you're calculating the area of the right triangle.

Q3. That answer looks ok to me. You've said 10.609 for one triangle. Say which triangle that is and put in the steps that lead to that answer.

Q4. If you call one side 's' then go through what you usually do to calculate the area of one right angled triangle and times by 14. That will give you an equation that you can solve for s.

Q5.

I split an octagon into 8 triangles with angles of 135deg. Then into right angle triangles of 45-45-90.

You have split the octagon into 8 triangles meeting at the centre. Good! 360/8 = 45 so the angle at the top of one triangle is 45. That makes the two equal angles at the base (180-45)/2 = 67.5 So you need to re-calculate this area. I thought the answer you posted at the start was correct.

Q6. Here's a diagram. I think the dotted line is 14, so AD is 7. You will need to use trigonometry to work out AB.

Q7. From the picture it looks like the pool sides are made from equal length sections. From one white square to the next is what I mean by a section. They look to be the same size. There are 4 around the half octagon at one end and 4 again at the other end. Between the two half octagons there are 2 sections per side making a rectangular shape in the middle. If one section is length 's' the the octagon has sides 's' and the rectangle is 2s one way and let's call it W for the width of the pool. You could measure this separately or, if you wanted to be clever, calculate it using the angles of an octagon. I suggest the first as less stressful. Your teacher wants you to put in every step of the method so imagine you're explaining this to a six year old who has got to go out with a tape measure and do the job. Anything you leave out means your young friend will get stuck with how to complete the task.

Hope that helps,

Bob

Thanks thickhead. You may have missed my edit. I agree D is the answer. But what do they mean by the slant height is 10 ?

Bob

When the GCSE exam was devised in the UK, one of the requirements was that questions should be 'realistic'. So questions are supposed to use values that could really occur in real life. So the Toblerone box is a favourite as you don't get many triangular prisms in real life and this is one that will be familiar to the candidates. From their website it looks like the company is a UK one so I'm not sure how much of the world has the joy of this product.

As this forum is read all over the world, maybe they'll offer me a job in the world-wide advertising department As I haven't eaten one for years I cannot comment on the tastiness of the product. If they read this perhaps they'll sent some samples

Bob

hi Itsmeali

This looks like a CompuHigh question. They like to pack a lot of questions in by giving the information once. It doesn't quite work here because an equilateral end will have a base size and slant length that are the same. Let's assume the triangles have a side length of 6 and the prism is 8 long. (Hope that's what they intend)

The surface area consists of three identical rectangles and two triangular ends. Lateral area should just mean the rectangles, so (6x8)x3 for that.

The volume is (area of an end) x the length so that would be (0.5 x 6 x 6(√ 3)/2) x 8.

I think this question may have been posted before so I'll have a search.

Bob

edit: Yes, it was posted here http://www.mathisfunforum.com/viewtopic.php?id=22608 but Lisa didn't ask about Q1-5. Some of the measurements on this sheet lead to impossible shapes, so we may have some fun here

hi Sunnysonny

Welcome to the forum.

Sorry, I haven't got one in the cupboard. In the UK there a plenty of shops that sell them. I've also seen giant ones at airports.

Couldn't find sizes on their website, but, given how much free advertising they are getting, I'm sure they'll be happy to tell us.

I've sent them an email.

Bob

hi thickhead,

Wow! Excellent proof. I would never have thought of that. Have a gold medal:

Bob

hi jacks,

Yes, I got that value to, just by calculation in Excel. But I guess you want an analysis of how to get this by algebra etc. I've been working on this, on and off, for two days and haven't got a method yet. My first attempt was to try and simplify the trig. using various trig. identities. Just went round in circles with that. Then I made a diagram with a 14 point polygon inscribed in a circle, and tried to find a method using the distances in right angled triangles plus similar triangles, angles in semi circle etc. I think that is the way to go with this, but a solution still alludes me. If I still cannot get anywhere by the end of today, I'll post my diagram. Maybe someone else can spot what I'm missing.

Bob

hi anniep418

When it said 'surface area' I was thinking just the area of the water surface. But I see now it wants the surface area of the liner. That means you would need to include the area of the sides. It looks like the sides are made up of equal straight sections so once you've measured one (with a tape measure? do they really want you to say that? ) you can multiply by the depth for the area of one, and then multiply by the number of them (8 around the octagon plus 4 making the rectangular sides) to get the total area of the sides.

Add to your answer what 'a', 'l' and 'w' stand for, and introduce a letter for the depth. Explain the formula for the octagon ... how it is made from triangles etc. Hopefully that should cover it.

Bob

ps. Is this from 'CompuHigh' by any chance?

hi anniep418

Welcome to the forum.

Q1,2,3,5 look correct to me.

For Q7. Looking from above the pool is made up of three shapes: half an octagon at each end and a rectangle joining them. So the surface area would be the area of an octagon plus a rectangle. So call the lengths by suitable letters and write out the correct formulas for the areas. The volume will just be area of the top times the depth of the pool. It would be more complicated if the pool had a shallow end and a deep end, but the picture shows the pool is a prism and you're told it's full to the top.

Hope that helps,

Bob

hi ninjaman

i dont understand the ruler/protractor method.

I just meant try to draw it. You could start with any line length b. Make an angle of C at one end. Label that end with letter C and the other with letter A. At this stage you won't know how long to make CB so just make it a long line and don't write on B yet. Set a compass to radius c and draw an arc from A to cut CB. Three things could happen: (1) c isn't long enough to reach CB. That means the triangle is impossible. (2) c just touches CB once. In that case the triangle is right angled at B. (3) Most likely case; the protractor cuts CB in two places. Either place makes a possible triangle so you'll have two answers, B' and B''.

Construction is not an acceptable way to get the answer if the question was to test your ability with trig. but it helps to at least imagine it as it gives you the 'tip off' that two answers are possible. Both calculation methods will yield two answers, so there's a sensible match between construction and calculation. (Just as well, or we'd have to reject the calculation method as an inadequate way to get an answer. )

Bob

hi ninjaman

There are two ways I can think of.

(1) As you know angle C, and sides c and b, you could use the sine rule instead to get angle B. Once you know C and B you can work out A and hence use the sine rule again to get side a.

(2) Write the cosine equation as a quadratic in side a, and use the quadratic formula to solve for a.

note: (1) When you so asin() to get angle B, be prepared for two answers, one acute and one obtuse; hence two answers for a.

(2) If you choose the quadratic method you'll get two answers for your quadratic.

If you try a ruler and protractor construction, you'll see why this happens.

Bob

This sounds like CompuHigh to me. Maybe they have given more information earlier on the sheet.

Bob