Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**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)*

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,731

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

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,731

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

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,820

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

The knowledge of some things as a function of age is a delta function.

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,731

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

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,731

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)*

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,731

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

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,820

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

The knowledge of some things as a function of age is a delta function.

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,731

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

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,820

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

The knowledge of some things as a function of age is a delta function.

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,731

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

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,820

All of the edges of XX'YY' are midsegments of respective triangles made by diagonals of the larger quadrilateral.

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

The knowledge of some things as a function of age is a delta function.

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,731

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

Offline

Pages: **1**