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# Proof Help watch

1. Got emailed this question by my teacher but I'm struggling with where to start with this.

Tips on how to start these type of questions would be really helpful too. Thanks!
2. (Original post by MainlyMathsHelp)
Got emailed this question by my teacher but I'm struggling with where to start with this.

Tips on how to start these type of questions would be really helpful too. Thanks!
First figure out what you require in order to calculate the area of a parallelogram.
Then find those things that you require in terms of
3. (Original post by RDKGames)
First figure out what you require in order to calculate the area of a parallelogram.
Then find those things that you require in terms of
Yup I know I need to find the perpendicular height but I'm still stuck.
4. (Original post by MainlyMathsHelp)
Got emailed this question by my teacher but I'm struggling with where to start with this.

Tips on how to start these type of questions would be really helpful too. Thanks!
5. (Original post by MR1999)
Yeah a = perpendicular height x base but don't know how to get the p.h.
6. (Original post by MainlyMathsHelp)
Yeah a = perpendicular height x base but don't know how to get the p.h.
Well if you know about the cross product then this is doable in one line.
What's the background on this? Do you know the dot product?
7. (Original post by RDKGames)
Well if you know about the cross product then this is doable in one line.
What's the background on this? Do you know the dot product?
Nope, never heard of it. Do you know if that's an AS topic?
8. (Original post by MainlyMathsHelp)
Nope, never heard of it. Do you know if that's an AS topic?
I don't think the cross product is even A level and the dot product is A2 content.
9. (Original post by MainlyMathsHelp)
Nope, never heard of it. Do you know if that's an AS topic?
Probably not.

In which case, you can label , then the perpendicular height would just be , thus the area of the parallelogram is

Have a go at figuring out what is

10. Does this help? (Think about areas of component shapes)
11. (Original post by RDKGames)
Probably not.

In which case, you can label , then the perpendicular height would just be , thus the area of the parallelogram is

Have a go at figuring out what is
Am I completely missing something because I can't work out what sin theta is
12. (Original post by MainlyMathsHelp)
Am I completely missing something because I can't work out what sin theta is
OK. I suggest you follow psc---maths' approach. My one requires use of the dot product which isn't in AS after looking at it.

His approach deals with inscribing the parallelogram inside a rectangle, then taking away areas of pieces between the parallelogram and the rectangle, from the area of the whole rectangle, to be left with the area of the parallelogram.
13. (Original post by psc---maths)

Does this help? (Think about areas of component shapes)

14. Try and think of the area of ABCD as the difference of the entire outer shape and the shaded area...
15. (Original post by psc---maths)

Try and think of the area of ABCD as the difference of the entire outer shape and the shaded area...
Ahh i realised when I tried it the first time I made a couple of stupid errors (multiplied p and r instead of adding). Thank you very much!
16. psc_maths has the best option for the proof. this is a proof for determinant of a matrix, which you would see at further maths, but is really what's required/expected here.

There is a much more laborious, algebraic/trig proof (and most likely several other proofs), but I have attached one which although laborious, should be within AS capability:show that 8ma0.pdf

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