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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

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,371

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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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

*Last edited by Agnishom (2013-03-12 04:39:49)*

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,371

OK. I'm back and looking now.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,371

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,371

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,371

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,371

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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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

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,371

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

*Last edited by bob bundy (2013-03-12 20:24:31)*

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,371

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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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

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,371

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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I shall make sure about it.

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

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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