Easy Maff Question

[Peter Pedant]

ABCD is a trapesium with AB parallel to DC - the diagnonals meet at X - prove DXC and BXA are similar traingles and that

area of DXC = DC2
---------------- ------
araa of AXB AB2

Answer - the similar traingles is easy - Z angles - F angles and corresponding angles
so there fore sinilar

you can say that DC /AB is a ratio let us say 'r' - but how do you show that the altitudes which you need for area are also in the ration 'r' ? - and then the areas are in the ratio r squared which is also DC/AB squared.... ergo - or do you just say - "they are"

The only thing I could manage was
imagine two similar triangles with side ratio 'r' but the altitude NOT r -
and let r approach one - the triangles are now congruent but their areas are not equal .....
which is impossible therefore the ratio of altitudes must also be 'r'

is there a simpler way?

(engtrance exam to secondary school)

[bhg481]

Aren't the altitudes in the same ratio as the bases because the triangles ABX and CDX are similar

[Peter Pedant]

Oh, that goes for any pairs of line inscribed in similar triangles? so long as the angles subtended are OK?

( medians, angle bisectors etc )

[fiction-factory]

Agree on the first bit- except that the justification is best given using alternate angles being equal and vertically opposite angles being equal (or angles in triangle add up to 180 degrees).
Regarding the second bit, are we required to show that the ratio of the areas is equal to (DC)²/(AB)² ?

[Peter Pedant]

hi FF
yes - I cant do indices on AB

[fiction-factory]

I need pen and paper to visualise it, but I think that when you draw your vertical perpendicular line through X you create two more pairs of similar triangles, and can use SOHCAHTOA to show the ratio of the heights and then use area of triangle (BxH)/2 or area of triangle = (1/2)AB sin C rule

[Peter Pedant]

thanks
I think the pair of similar triangles must be it ....

[fiction-factory]

I omitted to mention the Pythagoras step as well.
I really doubt many could do this at age 16 never mind at age 10 now. At age 10 answers I doubt they would be expected to include reference to trig or Pythagoras.
Maybe we are missing something more obvious

[jim360]

Thanks for clearing up the question by the way, for a while I was going berserk thinking that it was a funny triangle that was equal to a square it definitely fits into comfortably...

Agree that proving similarity is two uses of corresponding and one use of opposite angles. So, leaving that to one side...

Isn't it enough to say that, by definition, each side in a similar triangle is in the same ratio to the corresponding side of the original triangle? Therefore the height *must* be in the ratio "r" between both triangles, and the result follows.

In other words, I agree with bhg -- think you're overthinking this. Definition of "similar" saves you.

[fiction-factory]

That must be it, jim.
My only observation though is that the 'easy' bit should be based on alternate rather than corresponding angles (as only Z angles not F angles involved)

[fiction-factory]

If it's age 10 though how would the student be expected to show the ratio of the squared lengths (I didn't do Pythag until later I'm sure)

[Peter Pedant]

many thanks
yes I am not the one taking the exam

oh by the way -
one exam for a secondary school had 27 maff questions to do in 90 mins. Are these exams expected to be completable - or is it do as many as you can

[yes I know every other exam in finishable and it is 'bad training']
thanks for your replies ....

[Peter Pedant]

we did similar triangles in IVa ( nine years )
but nothing like this
just the angles bit and lengths not the same
not even ratios

I was surprised to see a whole question on similar triangles and not a congruent one in sight ( on which we spent weeks)

another one was ABC+ABC+ABC = BBB
find a number that fits

and can you find a number 9 x ABCD = DCBA ?

two of the 27 - I thought it was impossible that kids cd do this in 2 1/2 mins.....

[jim360]

I meant alternate, of course :P

Definition of similar: all sides (and so all length measurements) increase by the same scale factor r. Hence area grows as r^2. as r = AB/CD (or vice versa), then r^2 is the required value. I am sure this can be simplified a little in the language for ten-year-olds, but it seems to me that if they are asking at all then the idea of scale factor must be the thing they're relying on. Everything else is overkill.

[jim360]

For the ABC + ABC + ABC one, clearly BBB is one of 111, 222 etc. Must be bigger than 333 since, when you divide by three, you get a three-digit number all of which digits are different, and the middle of which is the digit in the original number. I think from here it's enough to do trial-and-error on 333 - 999:

333/3 = 111 FALSE.
444/3 = 148 Works (probably could stop here, but worth checking that there is only one unique answer).
555/3 = 185 FALSE.
666/3 = 222 FALSE.
777/3 = 259 FALSE.
888/3 = 296 FALSE.
999/3 = 333 FALSE.


* * * * * *

9 *ABCD = DCBA. Use that sum of the four distinct digits is a multiple of 9 if the actual number is, and A has to be 1 (else 9* ABCD would be a five-digit number). Hence also D = 9, so B+C = 8. Actually 9*1111 = 9999 so that anything bigger than this won't do. Hence B = 0 and C = 8, and so ABCD = 1089.

I mean, I expect you knew how to do these, but I think they are doable for particularly keen and bright ten-year-olds, and I assume that has to be the point of the test.

[maggiebee]

Good job it's maths you're doing PP and not English ha ha

[Peter Pedant]

thx Jim - I had got 148 and 1089

I had tried ABCD has to be less than 1111 and got pained look from the boy wonder in question, and that it had to be 1 and 9 and the referee ( grandmother) then stopped play.

and thanks magz
maff students ( of which I am not one) can do three years at uni and not write a sentence in English - but that of course is another debate.
I suppose it is a good job magz you do english and not maff, pip pip!

[TonyV]

My English teacher held a sort of classroom debate, asking was the most important GCSE subject, after all the suggestions, she explained English was because if you can't read or write, you can't study anything.

[Peter Pedant]

the impasse in theoretical physics in the 1930s was said to be due to the physicists not being able to describe what they had found

maff courses now have English options - now called History of Maff and so on or else they wont turn up ....

[jim360]

It's true that English or equivalent is important to help communicate other ideas with clarity. We're all grateful that PP takes that bit seriously.

[Peter Pedant]

Hi Jim - if only you had done a bit more English during your maff course - you would understand a bit more English
But BA.....

I am sure other contributors whom I thank agree that a BA has to go to a post grad theorist in a maff question

and thanks for your inputs