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Agnishom
2013-03-13 23:25:51

I shall make sure about it.

Even if I get 80/90 it would miss A1 by 2 marks

bob bundy
2013-03-13 21:58:34

You are welcome.

80/90 sounds good to me.  Let us know when you get the results.

And next time, allow a little more time before you post for help.

Bob

Agnishom
2013-03-13 20:28:18

Thanks

Today, there wasn't plenty of time in the exam.
So, I couldn't solve a problem of 4 marks
and perhaps I have done plenty of silly mistakes
Hopefully, I shall atleast get 80 out of 90

bob bundy
2013-03-13 19:24:02

hi

see diagram.

Construct midpoints L and M as shown.  DL ia parallel to AB etc.

Let BED have area = 3k

Using parallel rule the following also have area 3k

FMD, BDM, MLB, MLD, ALD, AML.  => LDC = 3k too.

ABC = 12k

ABD = ADC by median rule => ADC = 6k = ABD

ABF = ABD (parallel rule) => BFD = AFD

To do the rest I need to show that BF:FD = 2:1

At the moment I cannot see how to do that without using similarity so I'll post this and keep trying.

Bob

bob bundy
2013-03-13 18:53:30

arhh, tricky.  You should have said at the start.

Firstly, good luck with the exam.

Secondly, I'll try to find a permitted way to do these, but it may not be in time for you.

Bob

Agnishom
2013-03-13 12:11:29

Thanks but this is quite complicated.

We are not allowed to use trigonometry or similarity
The main properties we are supposed to use are:
1. Triangles on the same base and between same parallels have equal areas
2. A Median divides a triangle into two triangles of equal areas

bob bundy
2013-03-13 05:35:31

Q4

Consider ABFC and EDFB

These shapes are similar including the position of  F (because angle BAF = angle DEF)

So FD = half BF and FE = half FA

let angle BFE (=DFA) = x

ar(BFE) = half BF.FE.sin x = half (2 FD).(half AF) sin x = half FD.AF sin x = ar(FAD)

Q6 Let ar(FDE) = k.  express all other areas in terms of k

BFE = 2k = AFD

BAF = 2 AFD = 4k  => ABD = 6k => ADC = 6k

=> AFC = 2k + 6k  = 8k

=> ar(FDE) = 1/8 ar(AFC)

Bob

bob bundy
2013-03-13 04:04:22

Q2.

area ABE = half AB.BE.sin120 = half (2 BE)BE sin60 {sin 60 = sin 120}

= 2 (half BE.BE sin60) = 2 ar(BDE)

=> ar(BDE) = half ar(BAE)

Bob

bob bundy
2013-03-13 03:58:37

Q4 eludes me at the moment but assuming it is true then Q5 follows:

Comparing AFD and FDE

they have a common base FD and the height of one is half the other so ar(AFD) = 2 ar(FDE)

Using Q4 => ar(BFE) = 2 ar(FED)

Bob

bob bundy
2013-03-13 03:52:53

ABC and BDE are similar (both equilateral)

D is midpoint of BC => distances in BDE are half the equivalent distances in ABC.

So area = half base x height => area BDE is half area ABC  Q1

Also height of equilaterals are in same ratio

so comparing ABC and BCE

they have the same base BC and one has a height half the other => area BCE = half area ABC  Q3.

Still looking at the rest

Bob

bob bundy
2013-03-13 03:46:15

OK. I'm back and looking now.

Bob

Agnishom
2013-03-13 03:37:01

Thanks but I am having problems to show that ΔBEC is half of ΔABC

Please show atleast 1 and 5

P.S: Its already 10 o' clock here, so

bob bundy
2013-03-13 03:29:03

hi

Have to be a quick post now as I've got to do a job outside before it gets dark.

Q1.  The small equilateral has sides half the big one.  As area depends on two sides that means the small is 1/4 of the bigger.

More later.

Bob

Agnishom
2013-03-13 03:14:22

In the given figure:
1. ΔABC is equilateral
2. ΔBDE is equilateral
3. D is the midpoint of BC

Prove that:
1. ar( ΔBDE ) = 1/4*(ar(ΔABC))
2. ar( ΔBDE ) = 1/2*(ar(ΔBAE))
3. ar( ΔABC ) = 2*ar(ΔBEC)
4. ar( ΔBFE ) = ar(ΔAFD)
5. ar( ΔBFE ) = 2*ar(ΔFED)
6. ar( ΔFED ) = 1/8*(ar(ΔAFC))

Hint: EC and AD are joined
1.BE || AC
2.DE || AB

P.S.: I have my maths exam tomorrow