Post by: chickenpop on October 26, 2010, 06:24:50 pm
Post by: TrueTears on October 26, 2010, 06:31:19 pm
Post by: 98.40_for_sure on October 26, 2010, 06:34:42 pm
is proving one right angle sufficient for proving a rectangle or square?
Post by: jto1292 on October 26, 2010, 06:40:34 pm
Post by: tram on October 26, 2010, 06:44:39 pm
Mr. tears...
is proving one right angle sufficient for proving a rectangle or square?
I'm not tt but not, not JUST by it's self. If you can prove that there are two parallel sets of side AND that one od the angles is a right angle then yes that is sufficient. If you can also prove that all sies are of equal length, then it is a square.
I'm trying to think of a list of useful geometric properties.
- The diagonals in a rhombus are perpendicuar bisectors of each other
- The mid points of any quadlateral form a paralleogram
- A triangle subtended from the diameter of a circles is a right angled triangle
Hmmmmm geometery is not one of my strong points.... Any care to add what other properties are important to know?
Post by: tram on October 26, 2010, 06:46:25 pm
I reckon you have to prove diagonals bisect for Square and Rect... It depends on actual question
Diagonals only bisent each other in a rhombus, you need further evidence to prove it is a square and perpendicular bisectors are not related to rectangles
Post by: 98.40_for_sure on October 26, 2010, 06:46:44 pm
Quote
This was a straightforward question but many students did not recognise what additional property would lead to a
parallelogram being a rectangle. Many wasted time showing that the opposite pairs of sides had equal length or were
parallel. Some students tried to show that all four angles were right angles, or that all four angles were right angles and
opposite pairs of sides had equal length. A few students correctly showed an adjacent pair of sides were at right angles,
but then wasted time showing that adjacent sides were unequal in length, not realising that a square is a type of
rectangle. Students could also have shown that diagonals have equal length.
Post by: Martoman on October 26, 2010, 06:54:12 pm
It was this question from 08 vcaa exam 1 i think. The assessor report says this:QuoteThis was a straightforward question but many students did not recognise what additional property would lead to a
parallelogram being a rectangle. Many wasted time showing that the opposite pairs of sides had equal length or were
parallel. Some students tried to show that all four angles were right angles, or that all four angles were right angles and
opposite pairs of sides had equal length. A few students correctly showed an adjacent pair of sides were at right angles,
but then wasted time showing that adjacent sides were unequal in length, not realising that a square is a type of
rectangle. Students could also have shown that diagonals have equal length.
All you had to do for that one is show that one side had a right angle. That meant the oppoiste angle must be as well. With 2 right angles and two sides opposite each other the same, the other two angles must be right angles.
Post by: 98.40_for_sure on October 26, 2010, 06:56:00 pm
Though i'm not a man crying, to answer your question, not necessarily. The other sides could have different lengths and angles.
I feel ya, it's just the assessor report leaves me discontent since they didn't make it clear what they actually want us to do with it.
Post by: TrueTears on October 26, 2010, 07:02:02 pm
this is what I posted long time ago.
those are the definitions which DEFINE the shapes, prove them and you have proved the shape.
Post by: itolduso on October 26, 2010, 07:07:29 pm
Post by: tram on October 26, 2010, 10:01:32 pm
http://vcenotes.com/forum/index.php/topic,17229.msg176466.html#msg176466
this is what I posted long time ago.
those are the definitions which DEFINE the shapes, prove them and you have proved the shape.
cheers TT :)
Post by: kyzoo on October 26, 2010, 10:03:21 pm
Post by: tram on October 26, 2010, 10:19:37 pm
All the proofs are the same? I dunno, it's jsut all right angles, shapes, and the occasional application of linaer independence
hmmmm vector proffs while aren;t conceptually hard are one of my weak areas.... i can do albedraic proofs (e.g. the trig proofs we often get) but the whole spacial thing just messes with my head.... i just don't see the thigns i need to....
Post by: TrueTears on October 26, 2010, 10:22:45 pm
and yes geometry definitely aint my fave area of maths :)
Post by: Martoman on October 26, 2010, 10:45:56 pm
how do you prove the altitudes of any triangle are concurrent?
That is a very hard question
Post by: theuncle on October 27, 2010, 10:08:57 am
given
hence,
sorry i suck at latex but you get the picture!
Post by: googoo on October 27, 2010, 01:24:43 pm
now how do you prove the medians of any triangle are concurrent?
this is in the study design
Post by: tram on October 27, 2010, 04:33:39 pm
Post by: Martoman on October 27, 2010, 06:20:55 pm
or http://img827.imageshack.us/f/median.jpg/
same diff
Post by: itolduso on October 27, 2010, 07:07:46 pm
Post by: Martoman on October 27, 2010, 07:11:43 pm
Post by: itolduso on October 27, 2010, 07:13:52 pm
if you have two or more perp bisectors it must be an equilateral triangle. I'd think then they would be.
??
Post by: Martoman on October 27, 2010, 07:22:26 pm
Post by: itolduso on October 27, 2010, 07:29:24 pm
Post by: Martoman on October 27, 2010, 07:44:06 pm
Post by: itolduso on October 27, 2010, 07:54:28 pm
Shouldn't they? If you have one perp bisector you have a isosoles. If you have two then all sides are equal.
now i understand what you are saying. a perp bisector of a side of a triangle does not necess. start from the opposite vertex.
Post by: Elnino_Gerrard on October 28, 2010, 08:24:13 pm
Post by: jasoN- on October 28, 2010, 08:25:20 pm
Post by: TrueTears on October 28, 2010, 08:26:53 pm
So for a square, i dint quite understand what we have to prove. could someone please explain? :(http://vcenotes.com/forum/index.php/topic,17229.msg176466.html#msg176466
Post by: jto1292 on October 28, 2010, 08:27:42 pm
Post by: Elnino_Gerrard on October 28, 2010, 08:34:21 pm
Post by: jto1292 on October 28, 2010, 08:35:19 pm
Post by: Elnino_Gerrard on October 28, 2010, 08:44:50 pm
Post by: jto1292 on October 28, 2010, 08:46:59 pm
Post by: jasoN- on October 28, 2010, 09:09:48 pm
when doing dynamics and resolving forces can i write this:
resolving // to plane: __________
resolving _|_ (perpendicular symbol) to plane: ______
Post by: jto1292 on October 28, 2010, 09:12:23 pm
i don't want to create another thread but question:
when doing dynamics and resolving forces can i write this:
resolving // to plane: __________
resolving _|_ (perpendicular symbol) to plane: ______
I don't see why not... Isn't that how you normally resolve forces?
Post by: jasoN- on October 28, 2010, 09:14:02 pm
sometimes i see people draw in a small j versus i graph, indicating perpendicul and parallel axis and then go resolving in the 'i' direction etc.
i don't like this method
Post by: 98.40_for_sure on October 28, 2010, 09:15:08 pm
equating forces // to plane ... equating forces |_ to plane (i colour in the little right angle thingie)
or
equating forces vert to plane ... equating forces horiz to plane
Post by: jto1292 on October 28, 2010, 09:15:23 pm