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

]]>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

]]>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

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

]]>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

]]>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

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

]]>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

]]>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

]]>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

]]>Please show atleast 1 and 5

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

]]>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

]]>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

]]>