Proof Help

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

p,q,r,s

Reply 2

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

p,q,r,s

Yup I know I need to find the perpendicular height but I'm still stuck.

Reply 3

MR1999

Original post by MainlyMathsHelp

Got emailed this question by my teacher but I'm struggling with where to start with this.
Screen Shot 2018-03-03 at 19.04.31.png

Tips on how to start these type of questions would be really helpful too. Thanks!

Start with the formula for the area of a parallelogram

Reply 4

Original post by MR1999

Start with the formula for the area of a parallelogram

Yeah a = perpendicular height x base but don't know how to get the p.h.

Reply 5

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?

Reply 6

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?

Reply 7

MR1999

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.

Reply 8

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

\angle BAD = \theta

, then the perpendicular height would just be

|\overrightarrow{AD}| \sin \theta

, thus the area of the parallelogram is

|\overrightarrow{AB}| |\overrightarrow{AD}|\sin \theta = \sqrt{r^2+s^2}\sqrt{p^2+q^2} \sin \theta

Have a go at figuring out what

\sin \theta

Reply 9

username2912782

2FBD65A2-18BE-44FF-967C-3CDF757D0284.jpg.jpeg

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

(edited 6 years ago)

Reply 10

Original post by RDKGames

Probably not.

In which case, you can label

\angle BAD = \theta

, then the perpendicular height would just be

|\overrightarrow{AD}| \sin \theta

, thus the area of the parallelogram is

|\overrightarrow{AB}| |\overrightarrow{AD}|\sin \theta = \sqrt{r^2+s^2}\sqrt{p^2+q^2} \sin \theta

Have a go at figuring out what

\sin \theta

Am I completely missing something because I can't work out what sin theta is

(edited 6 years ago)

Reply 11

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.

(edited 6 years ago)

Reply 12

Original post by psc---maths

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

Sorry this isn't really helpful.

Reply 13

username2912782

564757BE-7484-40F2-8791-5085B41577A5.jpg.jpeg

Try and think of the area of ABCD as the difference of the entire outer shape and the shaded area...

Reply 14