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**harrychess****Member**- Registered: 2014-04-04
- Posts: 34

Help please! An answer and explanation will do.

In quadrilateral ABCD, AB=BC=13, CD=DA=24, and angle D=60. Points X and Y are the midpoints of BC and DA respectively. Compute XY^2 (the square of the length of X).

*Last edited by harrychess (2014-07-22 14:52:24)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,303

hi harrychess,

Hmm. Working on it.

So far:

BD is a line of symmetry for the kite. AC is perpendicular to BD and equal to 24. ADB = 30.

In my diagram if Z is at the intersection of XY and BD, it looks like XZ = ZY, but I cannot see why just yet.

Hopefully more will follow when I've caught up from jet lag.

Bob

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,303

I think this will do it.

The shape is a kite and BD is the line of symmetry. AC is at right angles to BD.

Z is the point where BD cuts XY. I have also reflected XY in the line BD to create points X' and Y'.

Because of the symmetry YY' = XX' = 12, and both these lines are perpendicular to BD **. Thus X'YY'X is a rectangle and so XZ = XY.

Thus you can get XY by first using Pythag on triangle YY'Z.

Bob

** edit. Certainly YY' = 12 by equilateral triangles. I'm thinking I may be assuming I have 'proved' XX' is also 12 when I may not yet have done so. Just improving my proof now.

*Last edited by bob bundy (2014-07-26 00:47:50)*

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,343

Hi Bob

How would you use Pythag on YY'Z? It's not a right triangle.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,303

hi Stefy,

Yes, you're right. The above is flawed. I'm working on a corrected version. At the moment X'YY'X is an isosceles trapezium. I'm sure I had it sorted a while back but I cannot re-trace my steps. I had a way of showing XZ = ZY and that forces the trapezium to become a rectangle. Then you probably have to use cosine rule.

Also DYC = 90. This should help but I need angle YCX.

Thinking still ........................................

Bob

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,303

OK. Like this:

AC = 24. Use cosine rule to calculate ABC and hence BCD. Subtract 30 and you have YCB. Then use cosine rule on XYC.

Checking ..........................

Bob

*Last edited by bob bundy (2014-07-26 01:30:55)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,303

A better version.

Because YD = 12 and CD = 24, that makes CYD a 30-60-90 triangle.

Angle ACB can be found thus:

cosine ACB = 12/13 and thus sine ACB = 5/13.

cosine BCD = cosine(BCA + 60) = cosBCA.cos60 - sinBCAsin60 = (12/13 - 5root3/13)/2 = (12 - 5root3)/26

***find sine BCD. Running out of time now, I'll fill this in later.

cosine XCY = cosine(BCD - 30) = cos BCD.cos30 + sinBCD.sin30

Then use cosine rule on XCY.

Note: I've used compound angle formulas to avoid the 'inaccuracy' of using a calculator for these angles.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,343

Haven't checked your new solution yet, but XX'YY' is not an isosceles trapezium, it's a parallelogram.

EDIT: Actually, yes, it's both, but calling it a parallelogram is more restrictive.

EDIT #2: And, yes, it is also a rectangle

*Last edited by anonimnystefy (2014-07-26 03:07:34)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
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It is my hope to prove it is a rectangle. So far I have only proved the isosceles trapezium. I am about to tidy up and finish off my method. It will take a little while.

What this space.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,343

Well, the opposite sides are equal and parallel...

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,303

Stefy wrote:

Well, the opposite sides are equal and parallel...

From my diagram and analysis YY' is parallel to XX' (as both are perpendicular to BD) and X'Y = XY' (by reflection). I don't yet have evidence that X'Y is parallel to XY' nor that X'X = YY'.

But I have simplified my calculations:

Let BCA = a and XCY = b

By simple trig cos(a)=12/13 and sin(a) = 5/13

Therefore, cos(b) = cos(a+30) = cos(a)cos(30) - sin(a)sin(30)= (12root3 -5)/26

Then XY^2 = 144x3 +169/4 -6root3(12root3-5) = 310.2115

This is confirmed by my Sketchpad drawing.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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All of the edges of XX'YY' are midsegments of respective triangles made by diagonals of the larger quadrilateral.

Here lies the reader who will never open this book. He is forever dead.

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**bob bundy****Moderator**- Registered: 2010-06-20
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Arhhh. You say that like it is a theorem I should know. I don't recall ever meeting that one, but that's not surprising as I don't bother to learn theorems. But it was easy to prove, so I'll add it to the list of theorems that I won't bother to learn.

Then it quickly follows that the shape is a rectangle. I expect you can get XY^2 using that, but I don't think I'll bother now. I'll leave it to the interested reader.

Bob

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