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MATHS - HEIGHT OF A TRIANGLE! URGENT!!!!

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_sophie_ | 21:47 Tue 09th Feb 2010 | Jobs & Education
19 Answers
Ok, i need this in for tomorrow.
I have a right angled triange, and I am asked to find the height (using my answer to part a. of the question which asked for the area).
The area is 149.7m squared..
The hypotenuse of the triangle (opposite the right angle) is 15m.
That's all the information i have and i need to work out the height of this!!
HELP ME.
hhaa, thanks.
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7.5m? at a guess , I'm rubbish at geometry. But i'll guess at anything lol
The area of a triangle is 0.5*(width of the base * height)
Does that now help you know what to do?
The sides of the triangle are a and b. Area is 149.7 = (a x b)/2
From Pythagoras 15^2 = a^2 + b^2 (^2 = squared).

You now have 2 equations, so can substitute. Off you go! Your next problem is, is a or b the height?
Question Author
ahhh! i hate maths sometimes!
im trying to find out the height, buildersmate, but thanks anyway!
sddsddean, i can see how this would work but my values for a and b are x! a is the height, which i am trying to work out and i have no value for the base of my triangle.
You've overcomplicated it, sd.
I repeat, the area of a triangle is one half of the width across the base times the vertical height. It matters not if it's a right-angled triangle or not (which this one happens to be) - it's always the same formula.
Question Author
ahh never mind!
couldn't care less about maths anyway. hahaha :)
"using my answer to part a. of the question which asked for the area"

Post that question as well.
Question Author
i was waiting for that one haha!
basically there is a triangle, base is 25m, other two sides are 20m and 15m. i used the cosine rule to work out the area (god help me if that's wrong too, which it probably is).
and the second part of the question asks me to find the height another triangle. this 'other triangle' is only part of the original triangle, and so i do not have a value for the base of my new triangle.

probably makes no sense but if you could see the piece of paper i have infront of me you would get it!!!! hahaha.
The trouble is, we don't know which way up this triangle is. I'm assuming the hypothenuse is flat on the ground, with the right-angle forming the apex above it - in which case the answer to the vertical height is 2*149.7/15 or about 20 metres.
Question Author
the hypotenuse is not on the ground, it is at a slant to the right of the triangle.

honestly, never mind! it doesn't matter haaa. probs me being stupid again :)
Confusion reigns.
This first triangle has sides 15, 20 and 25. The area of that is 150 square metres.
You've now got another right-angled triangle of hypotenuse 15m - the first one was 25m.
The 2nd ones height is in a straight ratio of the first - the ratio being 15/25 or 0.6.
If the height of your first one was 15, the height of your second one is 15*0.6 or 9
But if height first one was 20, the 2nd one's height is 20*0.6 or 12m.
I still don't know which way up it is.
just wing it sophie like i always did, i always got over 65% even guessing all the answers, except once. lol
There is no solution to this if the area is 149.7 m². If the hypotenuse is 15m then the base and height much both be less than 15m. One combination using Pythagoras is a base of 9 and a height of 12. But given that area = half base x height then there is no possible solution that gives an area anywhere near as big as 149.7. For example a 9, 12, 15 triangle has an area of only 54.
-- answer removed --
Precisely my confusion, F30.
The first right-angled triangle had sides of 25. 20 and 15. That has an area of 150, which Sophie had calculated as 149.7, presumably because of rounding error in the cosine table she used (it didn't actually need cosines to works that out).
We then discover the existence of a second triangle that has a hypotenuse of 15 and are asked to work out the height. Hence my last answer - it's surely a simple ratio job.
I suspect this has long-since past its 'best before' date.
6 . 23 and a half feet ?
hee hee - buildersmate is a hoot!!!
you must also have been given an angle
you clid, I clid, we all clid for Euclid

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