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**harrychess****Member**- Registered: 2014-04-04
- Posts: 33

Please Help! I need someone to explain how to work this problem.

The diagonals of a trapezoid are perpendicular and have lengths 8 and 10. Find the length of the median of the trapezoid.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi

I am getting

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,909

hi harrychess,

By using Sketchpad and measuring, I'm getting this:

AC = 10, BD = 8. AB parallel to DB.

I can move E about, and still get the same result for FG, which is interesting as the whole shape changes.

Using those right angled triangles, it's not too hard to show that the area of ABCD = 0.5 x 8 x 10.

So if I could calculate h, the height of the trapezium, then the length of FG would be easy.

But I cannot see how to get h just yet. Still thinking ....................................

Bob

Children are not defined by school ...........The Fonz

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,909

Ok. Extra bits on diagram:

h = HD = CI. The angles marked with a dot are equal (alternate angles). Say, alpha.

So using the area I have already worked out and

where a and b are the lengths of the parallels, you can work out FG.

Bob

*Last edited by Bob (2014-07-31 00:49:14)*

Children are not defined by school ...........The Fonz

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**Lenchen****Member**- Registered: 2015-08-11
- Posts: 2

Can somebody explain the solution to this problem without using trig?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Of course trig is the best way, the thematic way, but if you must avoid it you could use the computer. Geogebra can solve it if you are not too concerned with rigor and just want the answer.

Here is some info on this:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Lenchen****Member**- Registered: 2015-08-11
- Posts: 2

I would like a way to prove it using just algebra and geometry

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Did you make a drawing first?

This drawing covers the cases and it is easy to get the coordinates of each point. You want the distance of the red line.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,909

hi Lenchen,

Welcome to the forum.

Trig works because of the properties of similar triangles, so what you ask should be possible. My answer is in three parts, with three diagrams. To keep each part as simple as I can I will just show what is needed in each diagram.

Part one: calculate the area ABCD.

I have boxed in the shape. This box has an area of 8 x 10. The sides of the shape cut across the middle of four rectangles, splitting the box into 4 pairs of right angled triangles. So the area of ABCD must be half the box, ie. 40.

Part two:

GF is perpendicular to AB.

triangles BEF and DEG are similar, so

Similarly

and

If DE = 8k and CE = 10k then CD = root(164)k Sorry, couldn't avoid Pythagoras here.

Part 3.

H is the foot of the perpendicular on AB.

angle DBH = EDC (by parallels)

Consider triangles BDH and DCE

They have one angle equal by above and one right angle so they are similar.

Therefore

The area of a trapezium is

where a and b are the lengths of the parallels and h the perpendicular distance between them. Also the median will be the average of a and b so

Bob

ps. I have been asked to help with a lot of geometry recently, including the same question more than once. I'm not complaining; I like to be helpful; but I was wondering whether it would be helpful to members if I created a post consisting of a geometry contents list with links to the relevant posts. What do you think of this idea?

Bob

Children are not defined by school ...........The Fonz

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**llee97****Member**- Registered: 2020-06-15
- Posts: 1

If you move one of the diagonal out parallel to the original one using bob's picture (HD // AC, HD = AC, AH = DC, BD ⊥ DH ), so that the two diagonals and the extended long base form a right triangle, and the new extended base length is DC + AB, you may use Pythagorean to find sqr ( DC + AB) = sqr(diagonal 1) + sqr (diagonal 2). I did not know how to embed the imagine here.

*Last edited by llee97 (2020-06-15 09:04:24)*

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,909

hi llee97

Welcome to the forum.

Wow! It was a few years ago when I did that. I've had to reconstruct the diagram. I'm not getting what you mean.

(HD // AC, HD = AC, AH = DC, BD ⊥ DH )

so I'm stuck at the moment. If I manage to make your diagram, I'll post it for you.

Bob

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

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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