**HL Maths P2 Before/After

WestIRL

MattHelders wrote: »

I let y=0 and for one and x=0 for the other.

I then got the perpendicular distance from those points to the line opposite and used simultaneous equations

Got k=1.

Not sure if its right though

Well at the start it says k > 1....

MattHelders

WestIRL wrote: »

Well at the start it says k > 1....

Bollocks. I'm going to leave this thread before I start panicking

hamal

MattHelders wrote: »

I let y=0 and for one and x=0 for the other.

I then got the perpendicular distance from those points to the line opposite and used simultaneous equations

Got k=1.

Not sure if its right though

yeh i did that but the realised it couldnt be right cos that would hav meant that the line hit the x axis at (2,0) which was where the other line hit it!

cocopopsxx

M&S* wrote: »

I honestly don't see how half of that is justified as maths?

True true. Bloody horrific paper, going to bring me down LOADS!

Cathalog

Was 1 (b) a parallellogram? I got "no".

MattHelders

hamal wrote: »

yeh i did that but the realised it couldnt be right cos that would hav meant that the line hit the x axis at (2,0) which was where the other line hit it!

Should have copped that. Do you reckon I will just get zero for that?

SazKav

Think most are feeling pretty good about that paper, thank god, I was absolutely dreading it... delighted its over, well done to everyone Once one is down, there is about 20 seconds of relief before the panic of the next sets in! Ughhhh

c28omzk7ihsxv0

FnCK I forgot to get the percentages in the distribution table! AND I GOT ALL THE DMAN FIGURES RIGHT!

How many marks would I lose?

Luno

Cathalog wrote: »

Was 1 (b) a parallellogram? I got "no".

I also got no

hamal

Cathalog wrote: »

Was 1 (b) a parallellogram? I got "no".

no it says prove that it is.. i just showed that the distance of the opp sides were equal .'. its a parallelogram

Epsi

Thought it was ridiculous. Completely different from what I though it would be. Got an A2 ( in higher level) on paper 1 and now my result is going to be dragged down into the mud. Will Probably only get an b2 now. A large number of people who also sat the exam in the same hall found in equally difficult , past and honors alike.

GarIT

MattHelders wrote: »

For the question on the 60 degree angle. I just bisected the 60 degree angle twice to get the 15 degree one. Is that alright?

I worked it out along the side in a really complicated way. Basically I worked out mathmatically that in a right angles triangle if the angles attatched to the right angle are in the ratio of 1:5 so will the sides. So therefore the other two angles =90 so split into 1:5 you get 15 and 75 so then just draw one line 10 cm and one line 2 and you get 15 degrees.

hamal

MattHelders wrote: »

Should have copped that. Do you reckon I will just get zero for that?

I honestly dont know, i didnt get any further than that either so hopefully we'll get some credit

MattHelders

hamal wrote: »

no it says prove that it is.. i just showed that the distance of the opp sides were equal .'. its a parallelogram

It said determine I think.

I got the distance between 2 of the sides were equal.

But the distance between the other 2 weren't so I said no

DepoProvera

hamal wrote: »

no it says prove that it is.. i just showed that the distance of the opp sides were equal .'. its a parallelogram

Also did this.. Also my 3 ways to see if parallelogram were : get slopes of each side and use tan0 formula to see if opposite angles were equal/ if the transversals bisected/if opposite sides were equal..

Cathalog

hamal wrote: »

Cathalog wrote: »

Was 1 (b) a parallellogram? I got "no".

no it says prove that it is.. i just showed that the distance of the opp sides were equal .'. its a parallelogram

Got the paper here in front of me: "Using one of the methods you described, determine whether the quadrilateral with vertices (-4, -2), (21, -5), (8, 7) and (-17, 10) is a parallelogram."

it could be yes or no!

Cathalog

DepoProvera wrote: »

hamal wrote: »

no it says prove that it is.. i just showed that the distance of the opp sides were equal .'. its a parallelogram

Also did this.. Also my 3 ways to see if parallelogram were : get slopes of each side and use tan0 formula to see if opposite angles were equal/ if the transversals bisected/if opposite sides were equal..

I used slope, tan formula, equal pairs of sides

DepoProvera

MattHelders wrote: »

It said determine I think.

I got the distance between 2 of the sides were equal.

But the distance between the other 2 weren't so I said no

Distances were equal they were root 634 and root 300 and something

hamal

MattHelders wrote: »

It said determine I think.

I got the distance between 2 of the sides were equal.

But the distance between the other 2 weren't so I said no

maybe ur right then, although i did the same and found that all opp sides wer equal somethin like root634 for 2 of them and root313 for the others i think
ah well just hope for some nice attempt methods seeing as i had the right method

hamal

DepoProvera wrote: »

Distances were equal they were root 634 and root 300 and something

exactly what i got root313

Cathalog

hamal wrote: »

MattHelders wrote: »

It said determine I think.

I got the distance between 2 of the sides were equal.

But the distance between the other 2 weren't so I said no

maybe ur right then, although i did the same and found that all opp sides wer equal somethin like root634 for 2 of them and root313 for the others i think
ah well just hope for some nice attempt methods seeing as i had the right method

Damn, I did the same. And i got 4 different values... But one was root 6hundred and sumthin and the other was root 3hundred something.
I'd say your correct

DepoProvera

hamal wrote: »

maybe ur right then, although i did the same and found that all opp sides wer equal somethin like root634 for 2 of them and root313 for the others i think
ah well just hope for some nice attempt methods seeing as i had the right method

No i got these values aswell.. I assume the others mixed up which were sides of the parallelogram

WestIRL

GarIT wrote: »

I worked it out along the side in a really complicated way. Basically I worked out mathmatically that in a right angles triangle if the angles attatched to the right angle are in the ratio of 1:5 so will the sides. So therefore the other two angles =90 so split into 1:5 you get 15 and 75 so then just draw one line 10 cm and one line 2 and you get 15 degrees.

See this is what the problem is. I'm not picking on you in particular, as I was the same before I devoted the whole weekend to Paper2. You're looking for the hardest way to get the answer. Paper2 is actually the easier of the two papers, it's just that it seems TOO EASY- so you go about over complicating things when really it's just a case of using the techniques taught in the book. For instance if you have looked over contructions you would have known how to bisect an angle of 60 twice to get 15.

Again I'm not criticising you, I was exactly the same and made a **** of my mocks Paper2 by trying to go for the complicated method. And that's not to say you made a **** of this paper I'm just taking that question as an example.

WestIRL

Cathalog wrote: »

Got the paper here in front of me: "Using one of the methods you described, determine whether the quadrilateral with vertices (-4, -2), (21, -5), (8, 7) and (-17, 10) is a parallelogram."

it could be yes or no!

I plotted the vertices and got the slope of two opposite sides. They were parralel.

DepoProvera

That circle and tangent question was hard: I tried getting the equation of the circle to no avail, tried Pythagoras on the triangle, tried slopes.. So hard id say I got some attempt marks though

c28omzk7ihsxv0

DepoProvera wrote: »

Distances were equal they were root 634 and root 300 and something

Both proved were root 634. Which proves the question.

GarIT

WestIRL wrote: »

See this is what the problem is. I'm not picking on you in particular, as I was the same before I devoted the whole weekend to Paper2. You're looking for the hardest way to get the answer. Paper2 is actually the easier of the two papers, it's just that it seems TOO EASY- so you go about over complicating things when really it's just a case of using the techniques taught in the book. For instance if you have looked over contructions you would have known how to bisect an angle of 60 twice to get 15.

Again I'm not criticising you, I was exactly the same and made a **** of my mocks Paper2 by trying to go for the complicated method. And that's not to say you made a **** of this paper I'm just taking that question as an example.

I do DCG so I know all about angles. It just never dawned on me. Technically I'm right, because a/sinA=b/sinB if you let a=1 and b=5 you get sinA=15 and SinB=75, then I just did the lengths in the ratio of 1:5 and then my formula proved it.

I think I got nearly full marks on the paper so I should be grand.

patrickkelly

Anybody get Alpha= 44Degreees and Beta=100Degrees for part (i) of question 8?

hamal

patrickkelly wrote: »

Anybody get Alpha= 44Degreees and Beta=100Degrees for part (i) of question 8?

alpha 44 and beta 76 for me

Helloxoxo

GarIT wrote: »

MattHelders wrote: »

For the question on the 60 degree angle. I just bisected the 60 degree angle twice to get the 15 degree one. Is that alright?

I worked it out along the side in a really complicated way. Basically I worked out mathmatically that in a right angles triangle if the angles attatched to the right angle are in the ratio of 1:5 so will the sides. So therefore the other two angles =90 so split into 1:5 you get 15 and 75 so then just draw one line 10 cm and one line 2 and you get 15 degrees.

Yeah I just dropped a perpendicular from the point and that gives 30degrees then I bisected that angle to get 15