Re: Exam question assistance please


"mark" <aboard_epsilon@xxxxxxxxx> wrote in message
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On 14 Nov, 15:50, "Dave Baker" <D...@xxxxxxxx> wrote:
"Dave Baker" <D...@xxxxxxxx> wrote in message

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"mark" <aboard_epsi...@xxxxxxxxx> wrote in message
in that case then ...it would be "c" which is the best .because it has
the most metal in it .

You might want to rethink that before the examiner hears or he'll be
after
you for the remedial geometry class I'm afraid.

PS. Here's a clue. Why is the area of any triangle, symmetrical or
otherwise, always 1/2 base x height?
--
Dave Baker - Puma Race Engines

the exam question is flawed ...
if you wanted to show differences

he should of set out out with the same amount of metal spars doing the
their thing in each diagram ..

He did. That was in fact the whole point of the question!


c has a multitude of spars ...despite being all at the wrong angle it
will win over the others ...because of the shear (sic) amount of metal in
it .

Well don't say I didn't try to give you a hint. The spars consist of
triangles, all facing the same way and with their bases touching each other.
To find the total area of metal in the triangles in each of the three cases.
Area of each triangle = 1/2 base x height. As the bases touch then the sum
of the base length is the length of the beam - in all three cases. The
height is the depth of the beam - in all three cases. Therefore the total
area of metal = 50% of the cross sectional area of the beam - in all three
cases.

It doesn't matter what shape the triangles are in such an arrangement. The
area will always be half that of the beam. The triangles can even be non
symmetrical without affecting the solution and in fact it still doesn't
matter if they aren't all the same shape.
--
Dave Baker - Puma Race Engines


.

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