The Student Room Group

Scroll to see replies

Original post by CanineVet
I found the general and credit I very easy but II was okay apart from the last one where I realised my problem it said cosA was 0.5 but I put cos0.5 silly me ahah I was wondering was I so close yet not right ahaha! And I believe it is half the sum of the parallel sides and the distance between is multiplied so 1/2(a+b)xh? I can't remember the actual formula just the technique aha! See for 7 I got 4 but I didn't know how to prove it, I eventually just tried it with every number like 1,2,3,4 till I got 4 but I didn't know if there was a more valid way to prove it? :smile:


Posted from TSR Mobile

Honestly, after so many exams I have no idea which "number 7" you mean. You're going to have to explain it for me to remember.
Reply 201
Original post by Princepieman
Honestly, after so many exams I have no idea which "number 7" you mean. You're going to have to explain it for me to remember.


Hahaha sorry it was the one on jacks diet how many months will it take for him to be under 73 kg


Posted from TSR Mobile
Original post by CanineVet
Hahaha sorry it was the one on jacks diet how many months will it take for him to be under 73 kg


Posted from TSR Mobile


The answer to that is may because you just do after 1 month, then after 2 months etc untill you get an answer below 76
Original post by CanineVet
I found the general and credit I very easy but II was okay apart from the last one where I realised my problem it said cosA was 0.5 but I put cos0.5 silly me ahah I was wondering was I so close yet not right ahaha! And I believe it is half the sum of the parallel sides and the distance between is multiplied so 1/2(a+b)xh? I can't remember the actual formula just the technique aha! See for 7 I got 4 but I didn't know how to prove it, I eventually just tried it with every number like 1,2,3,4 till I got 4 but I didn't know if there was a more valid way to prove it? :smile:


Posted from TSR Mobile


What I did was work out the percentage difference between the two, then times by 4 to get past that percentage. Pretty good method even if I say so myself.

You guys know how much I was dreading this but it went amazing ! So happy hoping for >95 % :top:



Posted from TSR Mobile
Original post by CanineVet
Hahaha sorry it was the one on jacks diet how many months will it take for him to be under 73 kg


Posted from TSR Mobile


Ah, I thought it asked which month he would reach his goal in, not how many months? Well, I put he would reach it somewhere in the month of april. For proof, I just showed it like I would for a compound interest question e.g. end of january: 90 * 0.93; end of February: etc. Don't know if i'm right.
Original post by Princepieman
Ah, I thought it asked which month he would reach his goal in, not how many months? Well, I put he would reach it somewhere in the month of april. For proof, I just showed it like I would for a compound interest question e.g. end of january: 90 * 0.93; end of February: etc. Don't know if i'm right.


Yes, i thought this too, i also worked it out using the same method as you :biggrin:
Original post by Bromaldehyde
What I did was work out the percentage difference between the two, then times by 4 to get past that percentage. Pretty good method even if I say so myself.

You guys know how much I was dreading this but it went amazing ! So happy hoping for >95 % :top:



Posted from TSR Mobile


How would know what your percentage is when they don't state it on the certificate?
Original post by Pennyarcade
Yes, i thought this too, i also worked it out using the same method as you :biggrin:


Argh, we'll never know which is right until the papers get given to the teachers. I would be so disappointed if I got that wrong!
Original post by Princepieman
How would know what your percentage is when they don't state it on the certificate?


The school gets a full breakdown of your results :smile:
Original post by Princepieman
Argh, we'll never know which is right until the papers get given to the teachers. I would be so disappointed if I got that wrong!


the way we both done it is correct as i have done it in a pastpaper before.
Original post by TheFOMaster
The school gets a full breakdown of your results :smile:


That's made me quite uneasy now...:frown:
Reply 211
Original post by Princepieman
Ah, I thought it asked which month he would reach his goal in, not how many months? Well, I put he would reach it somewhere in the month of april. For proof, I just showed it like I would for a compound interest question e.g. end of january: 90 * 0.93; end of February: etc. Don't know if i'm right.


Sorry it was which month, I said April but it may be may because he started in January and one month from that is February thus it may be May :s-smilie: ah well :tongue: I did compound interest too just repeated it till I got my results.


Posted from TSR Mobile
It's April. The first month was January as it stated that Jack was starting the diet at the first of January.
Reply 213
Original post by Bromaldehyde
It's April. The first month was January as it stated that Jack was starting the diet at the first of January.


It must be April actually because it said during which month and at the start of April was 75 and start of may was 70 so yeah April phew! Aha :smile:


Posted from TSR Mobile
Original post by Princepieman
How did everyone find the credit math exam? The non-calculator was surprisingly easy and most of the calculator paper was straightforward, apart from the last question asking to prove the quadratic by using the cosine rule. Also, I completely blanked on the formula to find the area of a trapezoid so I just found the area of the two triangles at the sides and added that to the square in the middle. Can any of you remind me of what the formula is again?


I'll do you one better. I'll show you how it's derived!

Suppose we have a completely RANDOM trapezoid. I've chosen this one at random.

Now, let's do what you did. Let's split it up into 2 triangles and a rectangle (we can't assume height is the same size as length, so it's not a square necessarily) by drawing vertical lines.

So our left triangle - it has height hh, and I'm going to call its length xx.

Our right triangle has the same height, but is shorter - I'll call it length yy

Now our rectangle also has height, h, but I'll call this length b1b_1

The entire bottom side of our trapezoid MUST be the lengths of the two triangles plus the length of the square, right? So I'll call this longer length b2b_2 then b2=b1+x+yb_2 = b_1 + x + y

Now we know the area of a triangle, it's 1/2 base x height, yeah? So we have two different bases: x and y.

Area of the two triangles must be: 12xh+12yh\frac{1}{2}xh + \frac{1}{2}yh, and our rectangle area is simply b1hb_1h

So to get the total area of the trapezoid, we add them all together like you did in the exam:

A=12xh+12yh+b1hA = \frac{1}{2}xh + \frac{1}{2}yh + b_1 h
This is the area of a trapezoid, but it's not very pretty is it? Let's fix that: Take out h as a common factor, since it's in all of them!
A=h(12x+12y+b1)A = h(\frac{1}{2}x + \frac{1}{2}y + b_1)
Now the next step is trickier to see, but we can actually take out 1/2 as a common factor, let's see if you see the trick:
A=12h(x+y+2b1)A =\frac{1}{2}h(x + y + 2b_1) - I've taken a half out as a factor of each of the terms, which means I have to double everything inside the bracket to keep it correct. You can see that this is true simply by expanding the bracket out - you get what you had at the start again!

Now the next step is easy:
A=12h(x+y+b1+b1)A = \frac{1}{2}h(x + y + b_1 + b_1) - we've just rewritten 2b_1 is all. You know, like 2x = x + x, same deal here.

Remember what we said at the start? We said that the longest side of the trapezoid is b2=x+y+b1b_2 = x + y + b_1. We have that inside the bracket! So let's substitute it in, yeah?

A=12h(b1+b2)A = \frac{1}{2}h(b_1 + b_2). And there we go. That's the area of a trapezoid by using the length of the two parallel lines, but we can make it a little bit prettier by moving things around a little and putting the 1/2 back in the bracket:

A=(b1+b22)hA = (\frac{b_1 + b_2}{2})h

TADA! The area of a trapezoid as you were taught at standard grade! :biggrin:

You don't actually have to remember very many "area of..." formulas. You could honestly just remember the area of a square - and then you can prove the area of a lot of other shapes (the triangle, then the trapezoid for example) from that, as long as you can take some simple little logical steps! I had no idea what the area of a trapezium was, but just decided I'd work it out there. :tongue:
(edited 10 years ago)
Original post by Hype en Ecosse
x


You deserve a cookie.
Reply 216
Original post by Hype en Ecosse
I'll do you one better. I'll show you how it's derived!

Suppose we have a completely RANDOM trapezoid. I've chosen this one at random.

Now, let's do what you did. Let's split it up into 2 triangles and a rectangle (we can't assume height is the same size as length, so it's not a square necessarily) by drawing vertical lines.

So our left triangle - it has height hh, and I'm going to call its length xx.

Our right triangle has the same height, but is shorter - I'll call it length yy

Now our rectangle also has height, h, but I'll call this length b1b_1

The entire bottom side of our trapezoid MUST be the lengths of the two triangles plus the length of the square, right? So I'll call this longer length b2b_2 then b2=b1+x+yb_2 = b_1 + x + y

Now we know the area of a triangle, it's 1/2 base x height, yeah? So we have two different bases: x and y.

Area of the two triangles must be: 12xh+12yh\frac{1}{2}xh + \frac{1}{2}yh, and our rectangle area is simply b1hb_1h

So to get the total area of the trapezoid, we add them all together like you did in the exam:

A=12xh+12yh+b1hA = \frac{1}{2}xh + \frac{1}{2}yh + b_1 h
Take out h as a common factor, since it's in all of them!
A=h(12x+12y+b1)A = h(\frac{1}{2}x + \frac{1}{2}y + b_1)
Now the next step is trickier to see, but we can actually take out 1/2 as a common factor, let's see if you see the trick:
A=12h(x+y+2b1)A =\frac{1}{2}h(x + y + 2b_1) - I've taken a half out as a factor of each of the terms, which means I have to double everything inside the bracket to keep it correct. You can see that this is true simply by expanding the bracket out - you get what you had at the start again!

Now the next step is easy:
A=12h(x+y+b1+b1)A = \frac{1}{2}h(x + y + b_1 + b_1) - we've just rewritten 2b_1 is all. You know, like 2x = x + x, same deal here.

Remember what we said at the start? We said that the longest side of the trapezoid is b2=x+y+b1b_2 = x + y + b_1. We have that inside the bracket! So let's substitute it in, yeah?

A=12h(b1+b2)A = \frac{1}{2}h(b_1 + b_2). And there we go. That's the area of a trapezoid, but we can make it a little bit prettier by moving things around a little and putting the 1/2 back in the bracket:

A=(b1+b22)hA = (\frac{b_1 + b_2}{2})h

TADA! The area of a trapezoid as you were taught at standard grade! :biggrin:

You don't actually have to remember very many "area of..." formulas. You could honestly just remember the area of a square - and then you can prove the area of a lot of other shapes (the triangle, then the trapezoid for example) from that, as long as you can take some simple little logical steps! I had no idea what the area of a trapezium was, but just decided I'd work it out there. :tongue:


My mind is breaking in two! Beats my rhyme: "half the sum of the parallel side, time the distance between them. That's how you get the area; of a trapezium!" Sing it to a jack in the box theme :tongue:


Posted from TSR Mobile
Original post by Hype en Ecosse
I'll do you one better. I'll show you how it's derived!

Suppose we have a completely RANDOM trapezoid. I've chosen this one at random.

Now, let's do what you did. Let's split it up into 2 triangles and a rectangle (we can't assume height is the same size as length, so it's not a square necessarily) by drawing vertical lines.

So our left triangle - it has height hh, and I'm going to call its length xx.

Our right triangle has the same height, but is shorter - I'll call it length yy

Now our rectangle also has height, h, but I'll call this length b1b_1

The entire bottom side of our trapezoid MUST be the lengths of the two triangles plus the length of the square, right? So I'll call this longer length b2b_2 then b2=b1+x+yb_2 = b_1 + x + y

Now we know the area of a triangle, it's 1/2 base x height, yeah? So we have two different bases: x and y.

Area of the two triangles must be: 12xh+12yh\frac{1}{2}xh + \frac{1}{2}yh, and our rectangle area is simply b1hb_1h

So to get the total area of the trapezoid, we add them all together like you did in the exam:

A=12xh+12yh+b1hA = \frac{1}{2}xh + \frac{1}{2}yh + b_1 h
This is the area of a trapezoid, but it's not very pretty is it? Let's fix that: Take out h as a common factor, since it's in all of them!
A=h(12x+12y+b1)A = h(\frac{1}{2}x + \frac{1}{2}y + b_1)
Now the next step is trickier to see, but we can actually take out 1/2 as a common factor, let's see if you see the trick:
A=12h(x+y+2b1)A =\frac{1}{2}h(x + y + 2b_1) - I've taken a half out as a factor of each of the terms, which means I have to double everything inside the bracket to keep it correct. You can see that this is true simply by expanding the bracket out - you get what you had at the start again!

Now the next step is easy:
A=12h(x+y+b1+b1)A = \frac{1}{2}h(x + y + b_1 + b_1) - we've just rewritten 2b_1 is all. You know, like 2x = x + x, same deal here.

Remember what we said at the start? We said that the longest side of the trapezoid is b2=x+y+b1b_2 = x + y + b_1. We have that inside the bracket! So let's substitute it in, yeah?

A=12h(b1+b2)A = \frac{1}{2}h(b_1 + b_2). And there we go. That's the area of a trapezoid by using the length of the two parallel lines, but we can make it a little bit prettier by moving things around a little and putting the 1/2 back in the bracket:

A=(b1+b22)hA = (\frac{b_1 + b_2}{2})h

TADA! The area of a trapezoid as you were taught at standard grade! :biggrin:

You don't actually have to remember very many "area of..." formulas. You could honestly just remember the area of a square - and then you can prove the area of a lot of other shapes (the triangle, then the trapezoid for example) from that, as long as you can take some simple little logical steps! I had no idea what the area of a trapezium was, but just decided I'd work it out there. :tongue:

WOW! That all made sense! Thanks so much for going through the trouble of showing me this :biggrin:. I've always just remembered the formula and that was it but now I understand how one gets to that formula - wish I had you as a math teacher, because that was quite cool. Is this how math progresses or is it the same mundane memorisation game it was in standard grade?
Original post by Princepieman
WOW! That all made sense! Thanks so much for going through the trouble of showing me this :biggrin:. I've always just remembered the formula and that was it but now I understand how one gets to that formula - wish I had you as a math teacher, because that was quite cool. Is this how math progresses or is it the same mundane memorisation game it was in standard grade?


Maths involves more understanding as you progress up the levels. Higher Maths requires more understanding than SG; AH involves more than Higher. Tertiary level maths is almost 100% understanding. This is why many mathematicians hate secondary school mathematics - it's the exact opposite to the subject that they love. :tongue:

You're not going to do proofs at Higher, but that proof I did there didn't involve any skills past first year! You can certainly chose to prove a lot of what you learn. They teach you a tool at Higher Maths called "completing the square" and you can use that to prove the quadratic formula, which is always a fun proof.
AH Maths involves a lot of "show that...", "prove that...", "from first principles..." questions. It's really fun and that's the stuff I enjoyed doing, but there is still a hefty bit of memorisation involved.

Hopefully ukd is still around (:teehee:) if you end up taking up AH Maths - he's really good at encouraging you to understand the material you're simply asked to accept. I know a little bit of maths past AH level simply because ukd pushed me. I don't do anything related to maths at all at uni!

You know how you have to know two trig identies for credit? tanx=sinxcosx,sin2x+cos2x=1\tan x = \frac{\sin x}{\cos x}, \sin ^2x + \cos ^2x = 1, and you can prove both of them with stuff you'll have learned at Standard Grade. They're both proved using the same logic. Do you want to give it a try? We can give you a hint if you get stuck. :wink:
Original post by Hype en Ecosse
Maths involves more understanding as you progress up the levels. Higher Maths requires more understanding than SG; AH involves more than Higher. Tertiary level maths is almost 100% understanding. This is why many mathematicians hate secondary school mathematics - it's the exact opposite to the subject that they love. :tongue:

You're not going to do proofs at Higher, but that proof I did there didn't involve any skills past first year! You can certainly chose to prove a lot of what you learn. They teach you a tool at Higher Maths called "completing the square" and you can use that to prove the quadratic formula, which is always a fun proof.
AH Maths involves a lot of "show that...", "prove that...", "from first principles..." questions. It's really fun and that's the stuff I enjoyed doing, but there is still a hefty bit of memorisation involved.

Hopefully ukd is still around (:teehee:) if you end up taking up AH Maths - he's really good at encouraging you to understand the material you're simply asked to accept. I know a little bit of maths past AH level simply because ukd pushed me. I don't do anything related to maths at all at uni!

You know how you have to know two trig identies for credit? tanx=sinxcosx,sin2x+cos2x=1\tan x = \frac{\sin x}{\cos x}, \sin ^2x + \cos ^2x = 1, and you can prove both of them with stuff you'll have learned at Standard Grade. They're both proved using the same logic. Do you want to give it a try? We can give you a hint if you get stuck. :wink:


I'll probably have to try that later, right now I'm stuffing my face with Ben and Jerry's because of how exhausted I am. It's also a little treat for finishing the math exams :teehee:

Latest