Vector & Square Root Question for GCSE Maths

[tomtomtom1]

I have attached a copy of a vector question which i cannot do, i do not even understand what the question is asking can someone help?

On a different note i seem to have a lot of trouble with simplifying square roots for example what is the square root of 2704/ is there any way to find the root easily?

Thanks

 

[Nick O]

The vector question:

Remember that the defining parts of a vector are its length and direction. You can produce a vector of the same length and direction (but not position) as BC using some combination of AB and CD. Hint: How does the distance from A to the midpoint compare to the distance from B to C? What about the angles? Can you prove it?

The roots question:

I use a calculator, to be honest. But, in a pinch, I would decompose the number into prime factors like so:

sqrt(2704) = sqrt(2*2*2*2*13*13) = sqrt(2^4 * 13^2) = 4*13 = 52

Finding the factors is a bit time consuming though.

 

[Chestermiller]

On a different note i seem to have a lot of trouble with simplifying square roots for example what is the square root of 2704/ is there any way to find the root easily?

Thanks

Factor the 2704 into its lowest factors. Start with 2.

Chet

 

[tms]

I have attached a copy of a vector question which i cannot do, i do not even understand what the question is asking can someone help?

It's a regular hexagon. What does that tell you about the lengths of the sides?

On a different note i seem to have a lot of trouble with simplifying square roots for example what is the square root of 2704/ is there any way to find the root easily?

How about a calculator? If you want a manual method, you could start here: http://en.wikipedia.org/wiki/Methods_of_computing_square_roots

 

[haruspex]

I would decompose the number into prime factors like so:

sqrt(2704) = sqrt(2*2*2*2*13*13) = sqrt(2^4 * 13^2) = 4*13 = 52

Finding the factors is a bit time consuming though.

It can be done in the head like this:
2704 = 2700+4 = 4*(27*25+1) = 4*((26+1)(26-1)+1) = 4*((26²-1²)+1)...

 

[tms]

Of course, the quick and lazy way to factor a number is with a package such as Maxima or GNU Octave:

(%i1) factor(2704);

                                     4   2
(%o1)                               2  13

 

[tomtomtom1]

The vector question:

Remember that the defining parts of a vector are its length and direction. You can produce a vector of the same length and direction (but not position) as BC using some combination of AB and CD. Hint: How does the distance from A to the midpoint compare to the distance from B to C? What about the angles? Can you prove it?

The roots question:

I use a calculator, to be honest. But, in a pinch, I would decompose the number into prime factors like so:

sqrt(2704) = sqrt(2*2*2*2*13*13) = sqrt(2^4 * 13^2) = 4*13 = 52

Finding the factors is a bit time consuming though.

I am trying to prove that the distance from A to the mid point is equal to the midpoint to any other corner, what i have done is found the angles at the midpoint (360/6) to get 60 degrees, i have alos found the interior angles which is 120 degrees and the exterior angles which are 60 degrees but i cannot prove the length of the mid point is equal to the midpoint to all other points?

see attached.

 

[Nick O]

Would it help to note that the triangles inside are all equiangular, and therefore equilateral? Every side of every triangle has the same length.

Edit: If we want to prove this rigorously, I would start with the fact that |FC| = 2|AB| (given), therefore |FO| = |AB|. But, I doubt that this is necessary as long as you have a convincing (and correct!) argument.

 

[tms]

I am trying to prove that the distance from A to the mid point is equal to the midpoint to any other corner,

The question asks you to relate the length of one side to the lengths of two other sides. Just consider the definition of a regular polygon, and the answer should be obvious.

 

[verty]

The question asks you to relate the length of one side to the lengths of two other sides. Just consider the definition of a regular polygon, and the answer should be obvious.

But let's pretend that it isn't a regular hexagon, this makes the question more interesting. Essentially you must tackle this like a geometry problem, start from what you know and fill in anything you can add with geometric reasoning. Eventually you will find a way forward.