Stuck on HL maths paper 2 question

EvM

I'm having a bit of trouble with a paper 2 question from the edco sample papers. It's question 9, sample paper D, where you have to find the area of an irregular pentagon. I have all the angles and sides, so it should be easy, but I just can't think of a way to divide it so I can work out the area of the individual triangles instead. Thanks!

Leaving Cert Student

Divide into the triangles as you have done, get all sides and angles then use .5abSINC for each triangle?

EvM

Leaving Cert Student wrote: »

Divide into the triangles as you have done, get all sides and angles then use .5abSINC for each triangle?

If I try working it out like I divided them up though, I'm not sure how to calculate the area of the middle triangle as I don't know a full angle of it because I've cut into each of its angles. Is there any way to know in what ratio I've cut each of it's angles?

Leaving Cert Student

Could you not use the cosine rule to get angle on middle triangle??

little sis...

This actually came up in my summer exam!

And I think I got it right.

what you do is split it into 3 triangles but not the way you have done there. draw a line from the top middle corner to bottom left corner and another line from same top middle corner to bottom right corner.

you'll see that you'll have 2 isosceles triangles on the right and left so those angles can be worked out and then you will be able to find the top angle of the triangle in the middle as a result.

then find the lengths of the lines you drew by using cosine rule (as you know the opposite angle and lengths of other two sides in other 2 triangles).

then use half ab sinC formula for all 3 triangles.


hope thats not confusing
it probably is sorry

there are many other ways though ...

peekachoo

Easier way;

Divide the polygon into a square and a triangle. Square has 360 degrees and triangle has 180. Add those two and take the angles you have from 540. Answer

little sis...

peekachoo wrote: »

Easier way;

Divide the polygon into a square and a triangle. Square has 360 degrees and triangle has 180. Add those two and take the angles you have from 540. Answer

OP got the angle already I believe

peekachoo

little sis... wrote: »

OP got the angle already I believe

ohh I just read the question on the picture...genius strikes again :pac:

EvM

Leaving Cert Student wrote: »

Could you not use the cosine rule to get angle on middle triangle??

Oh yes you're right! Damn, I didn't even think of that.. Must be the tiredness .. Thanks a lot for the help .

teotihuacan

find the lengths of the two pencil lines you have drawn using the cosine rule.....once you have these lengths, its just a case of using the Sine rule to find the new angles you have formed.

calculate all of these first and write them in to make your life easier, then as Leaving Cert Student said, use (0.5)absinC to get the area!

simples.

Monsieur Folie

Change one of your lines, the one furthest to the left. Change it to a line joining the top centre point to the bottom left point and now you have, as someone said above, two isosceles triangles to work on, which will help you get enough information to work on the middle triangle.

Leaving Cert Student

EvM wrote: »

Oh yes you're right! Damn, I didn't even think of that.. Must be the tiredness .. Thanks a lot for the help .

No problem

Leaving Cert Student

peekachoo wrote: »

Easier way;

Divide the polygon into a square and a triangle. Square has 360 degrees and triangle has 180. Add those two and take the angles you have from 540. Answer

?