CONSTRUTION:Construt a line segment of length √3 units

bpb101

hey Folks. i Was wondering would somebody tell us how to construct a line segment of √3 units.This Question Appeared in the 2011 SEC Hl (project maths strand 2, for the 24 pilot schools ) exam paper.
I looked at the marking scheme but i can't understand the theory behind it.

They seem to say draw a line segment AC of length 2 units and draw a 2 unit arch from A and the another 2 unit arch from side C and draw a perpendicular line down where they meet. This does work out but can somebody explain it, and how one would do it for √2 units



"Using only a comps and straight edge"

Thanks,

Chikablam

Start off by drawing your number line. Mark 0 on it, then make another mark, of any length. This will be your length of "1".
Take your compass, and measure the distance between 0 and 1, then mark "2", by putting one end of the compass on "1", then marking.
Now, draw a line straight up from 1. With your compass, measure the distance 2. Now, put the pointy end on 0, then mark off the other end on the line straight up from 1.
Now, you have a distance 1, along the number line and a distance 2 from "0" to directly above 1.
From Pythagoras, we know the distance between the number line and the mark on the line going up from 1 must be the root of 3, since 1^2 + 3 = 2^2.
Then you simply measure the distance of root 3 with the compass, and mark it on the number line

MathsManiac

The version in the marking scheme is just a simplified version of what Chikablam has described. The reason it gives a segment of the required length is Pythagoras' theorem, as Chikablam describes.

It should be noted that Chikablam has not described how you use only a compass and straight edge to "draw a line straight up from 1" (i.e., erect a perpendicular). The version in the scheme achieves this easily.

Try to think about how you would do root 2. It might be helpful to note that, if a square has side 1 unit, it's diagonal is root 2...

bpb101

Thanks , ive got both them working now

evolving_doors

draw rectangle of 3x1 (therefore area is 3units sqrd)

then convert to a square of equal area (see attached picture (sorry I dont have the time to do my own so i pinched it!)
squaringrectangle.jpg

If area of a square is 3 therefore sides must be √3




shortcut is to draw any square and assert that the area is 3units sqrd... so therefore the sides must be √3... (the examiner will scratch his head wondering what mark to award, so it could be a bit risky)

bnt

I think the key to the question is trigonometry. Sin 60° = √3/2, and constructing a 60° angle is as easy as drawing an equilateral triangle. The solution described by the OP is just that: drawing an equilateral triangle with all three sides 2 units in length. Drop a perpendicular line as described, you are creating two right-angled triangles, and so basic trigonometry applies.



In the above picture AB = BC = CA = 2. Cosine = opposite / hypotenuse and so AD / AB = √3/2 . Since AB = 2, AD = √3 /2 x 2 = √3.



---

For √2 ... think Pythagoras.

locum-motion

Armelodie wrote: »

draw rectangle of 3x1 (therefore area is 3units sqrd)

then convert to a square of equal area (see attached picture (sorry I dont have the time to do my own so i pinched it!)
squaringrectangle.jpg

If area of a square is 3 therefore sides must be √3

I never learned this 'squaring a rectangle' thing in school (or if I did, I don't remember), so I'm trusting that the theorem or whatever is valid. Assuming that it is, then this is an elegant way to answer the question.
I've tried to prove the theorem for myself, but haven't succeeded yet.

Armelodie wrote: »

shortcut is to draw any square and assert that the area is 3units sqrd... so therefore the sides must be √3... (the examiner will scratch his head wondering what mark to award, so it could be a bit risky)

I'd say risky is an understatement!

locum-motion

I think bringing Sins and Coses into it complicates it unnecessarily.

bnt wrote: »

I think the key to the question is trigonometry. Sin 60° = √3/2, and constructing a 60° angle is as easy as drawing an equilateral triangle. The solution described by the OP is just that: drawing an equilateral triangle with all three sides 2 units in length. Drop a perpendicular line as described, you are creating two right-angled triangles, and so basic trigonometry applies.



In the above picture AB = BC = CA = 2. Cosine = opposite / hypotenuse and so AD / AB = √3/2 . Since AB = 2, AD = √3 /2 x 2 = √3.



---

For √2 ... think Pythagoras.

bnt

locum-motion wrote: »

I think bringing Sins and Coses into it complicates it unnecessarily.

Basic trigonometry is more complicated than "squaring a rectangle"? :eek:

Knowing that Sin 60° = √3/2 is not asking much. This method actually explains the solution that the OP was given - the bit about drawing arcs of length 2.

PS: please don't quote the whole of a long post just to add one line. How much of that was relevant to what you wrote? Very little, if any.

evolving_doors

locum-motion wrote: »

I never learned this 'squaring a rectangle' thing in school (or if I did, I don't remember), so I'm trusting that the theorem or whatever is valid. Assuming that it is, then this is an elegant way to answer the question.
I've tried to prove the theorem for myself, but haven't succeeded yet.
QUOTE]

Sorry it was from my tech drawing days i learned that one...never bothered with the proof..

also remember the Wheel of Theodorus was a good one to get any √

A little easier on the eye too...click the link below (ignoe the attachment as it's coming up black)

Wikipedia page

There was also something about a hexagon of sides 1 had a height √3 but that's much the same as BNT had posted earlier i think..

locum-motion

bnt wrote: »

Basic trigonometry is more complicated than "squaring a rectangle"? :eek:

...

Oh, I know it's complicated (See my other thread). I didn't mean to imply that it wasn't. Sorry.

im invisible

but its not....

(i remember that squaring the rectangle from tech. graphics, but never knew how it worked)

i wonder would they have accepted if you constructed a unit square, get a diagonal, draw a perpendicular from one end of this, mark off one unit, and join to the other end of the diagonal?
(using the usual method of extending lines and marking arcs to get perpendiculars?)

evolving_doors

im invisible wrote: »

i wonder would they have accepted if you constructed a unit square, get a diagonal, draw a perpendicular from one end of this, mark off one unit, and join to the other end of the diagonal?
(using the usual method of extending lines and marking arcs to get perpendiculars?)

Shur why wouldn't they, one you show enough construction lines and label everything the examiner should have enough to go on if they aren't familiar with that method.
the method you quoted there is the start of the Theodorus Wheel...keep repeating the process and you get root 4, root 5 etc