Quadrilaterals Help

denis_gylaev · 2015-08-19 08:33:38

Let ABCD and BEDF be two 2 times 3 rectangles that overlap, as shown. Find the area of the overlap.

ABCD is a square. Parallel lines m, n, and p pass through vertices A, B, and C, respectively. The distance between m and n is 12, and the distance between n and p is 17. Find the area of square ABCD

Let P be a point inside rectangle ABCD such that PA = 1, PB = 7, and PC = 8. Find PD.

Please provide a solution if possible. Thanks

bobbym · 2015-08-19 11:38:27

Hi;

For the last one I like

Bob · 2015-08-19 19:49:31

hi denis_gylaev

Another member asked about Q1 and Q2 recently. Look here:

http://www.mathisfunforum.com/viewtopic.php?id=22451

I'm still working on Q3.

Bob

Bob · 2015-08-19 23:16:29

Q3.

A similar question came up a long time ago here:

http://www.mathisfunforum.com/viewtopic.php?id=14832

We never did find a simple (grade 9)solution using Pythagoras but the method of post 12 was considered to be the simplest at the time.

My software, Sketchpad, assigns letters to points in alphabetic order, so your point P is my point E on the diagram.

Rotate the whole diagram 90 around point B. Rotated positions for A, D and E are labelled A', D' and E'.

Triangle BEE' is isosceles with EBE' = 90 so you can calculate EE'. Then using the cosine rule angle AEE'. Add 45 to get AEB.

Now you can use the cosine rule again to get AB, the side of the square.

From there you can calculate angle BAE and hence AED.

You now know two sides and the included angle in AED so one more use of the cosine rule will give ED.

Bob

denis_gylaev · 2015-08-20 01:25:21

Thanks for the help, I will look through the posts, if there is anything I dont understand I will ask it here

Bob · 2015-08-20 01:53:11

OK. I don't know if bobbym has a better method for Q3. Mine seems to go on forever!

Bob

denis_gylaev · 2015-08-20 03:20:06

Can you provide a solution for number 1? I don't think that 4 sqrt7 is correct. Also, for number 3 is there anyother way to do it besides using tan, cos or sine because I haven't learned those yet

Bob · 2015-08-20 04:17:23

I think that is the answer to Q3.

I've had another look at Q1 and I've come up with a much neater method. You do need Pythagoras' theorem though. Is that OK ?

Here's my diagram:

Because both rectangles are 3 x 2, there is a symmetry about the diagram which means BH = BG = DH = DG. So the overlap, BGDH, is a rhombus, which means the diagonals bisect each other at 90 degrees.

Call the middle of the rhombus J.

In triangles BDC and BJH, one angle is 90 and another is common to both, so these triangles are similar.

=> BJ : JH = 3 : 2

By pythag. BD = root(13) => BJ = root(13)/2 => JH = 2/3 of root(13)/2 = root(13)/3

So area BGDH = 4 x 1/2 x root(13)/3 x root(13)/2 = 13/3

Q3. A solution not involving trig ? That was the problem with the linked thread. That problem came at the end of a grade 9 chapter on pythag as the final question. bobbym came up with a solution using only pythag but it involved solving 4 simultaneous equations with 4 unknowns. We all felt that was a bit tough for grade 9.

I'll keep looking. Where did the problem come from ? That might give me a clue about what is expected.

Bob

denis_gylaev · 2015-08-20 05:28:15

The problems are from an Introduction To Geometry Book

Sorry I meant that 4 sqrt7 for the last one is wrong whoops

Last edited by denis_gylaev (2015-08-20 05:42:29)

bobbym · 2015-08-20 06:20:17

Hi;

Sorry I meant that 4 sqrt7 for the last one is wrong whoops

You are correct! I just wanted to see if you were awake. I am now going to use that excuse every time I get one wrong...

The new and improved correct answer is now 4 and P does not appear to need to be inside the rectangle!

Bob · 2015-08-20 08:00:24

hi bobbym,

And are you going to reveal your method ? Or is that another test to see if we're awake ?

I have a method using pythag but the algebra is fairly horrid and some bug has crept in causing complex solutions! Hopefully I'll sort it out soon.

LATER EDIT.

That's funny. I cannot get around this. I've also tried by my earlier method and that leads to an impossible ARC-COS.

Bob

Last edited by Bob (2015-08-20 18:52:16)

bobbym · 2015-08-20 18:58:19

Hi Bob;

I am pretty sure of the answer of 4 because it was obtained empirically in two different ways.

The Colonel wrote:

If you can not work your way back when you have the answer to a proof or at least a convincing argument it shows that you have no understanding of the question.

That is where I am now. I have the answer through EM but can not work backwards to a satisfying method.

Bob · 2015-08-20 20:40:09

hi bobbym,

I am unable to make a diagram. Are you able to post one ?

Bob

bobbym · 2015-08-20 21:39:43

I have a solution, one answer is definitely 4. There may be others, I have not explored that.

The algebra is very large and would not fit in a post. Geogebra on the other hand, can do this problem with a high degree of certainty in 10 minutes.

My diagram will not be much help but it is how I worked through the problem.

Start with point P at the origin, this will simplify the calculations. Draw three circles of radii 1,7 and 8 with P as the center. Pick a random point on the smallest green circle and call it A. Pick a random point on the blue circle, call it B. This is side AB. Drop perpendiculars until D is drawn. The labeling of the points will quickly get out of hand and the expressions will grow accordingly.

I have to get offline now, good luck.

Bob · 2015-08-20 23:08:43

Thanks bobbym,

I've done a similar thing, but with the ABCD square constraint. There is only freedom to move A and B independently. C and D are fixed, relative to A and B. Your quadrilateral doesn't look like a square in the image. Also, the question says P is inside the square:

Let P be a point inside rectangle ABCD such that PA = 1, PB = 7, and PC = 8. Find PD.

My algebraic approach looked nastier than it turned out to be, because I was able to do lots of simplifying. I can post it if you would like. It leads to a quadratic in 's' the side of the square ... with complex solutions.

My method in post 4 also failed as COS(AEE') calculates as 5root(2)/4.

Bob

Bob · 2015-08-21 01:05:31

I'm still not convinced but how about this?

Draw lines through P parallel to the sides of the square.

Let s be the length of a side and x and y be the distances as shown in the diagram.

Using Pythagoras on each right angled triangle:

Now it is possible to 'solve' the algebra even if real values for s, x and y don't exist. According to my working they take complex values. But I may have made an error.
Maybe someone else can solve using three of the triangles to find these unknowns as show me where I'm going wrong.

Bob

bobbym · 2015-08-21 04:48:15

Hi;

Also, the question says P is inside the square:

It does not appear to matter but I did it that way too.

Can you post the equations that lead to complex results and I will check them?

Bob · 2015-08-21 06:22:10

Thanks,

Bob

bobbym · 2015-08-21 07:35:39

Hi;

No real solutions exist for those equations.

denis_gylaev · 2015-08-21 11:40:38

Hi sorry for the late reply, I managed to figure out the 3 problems I posted, I don't know which ones you guys dont understand so ill post answers to all 3 of them, if you guys need solutions feel free to ask.

1) 13/3
2) 433
3) 4

Also, all of these can be solved without using trig

Bob · 2015-08-21 18:55:00

hi denis_gylaev

Those are also my answers (and no trig. only pythag).

But I have found a curiosity with Q3. That answer (4) pops out by pythag, but if I try to find a square with those properties it doesn't seem to exist.

So I'd be very interested in your method for that one.

Bob

denis_gylaev · 2015-08-25 05:36:14

Hi, sorry I was very sick with a fever for the last couple of days. Here is the solution for Q3:

$78843207e542d25628ba35912ec513828d3a1e66.png$

Let the line through P parallel to \overline{AB} intersect \overline{BC} and \overline{AD} at T and V, respectively. Let the line through P/parallel to \overline{AD} intersect \overline{AB} and \overline{CD} at S and U, respectively. Then by Pythagoras on right triangles PAV, PBT, PCT, and PDV, PA^2 = AV^2 + PV^2, PB^2 = BT^2 + PT^2, PC^2 = CT^2 + PT^2, PD^2 = DV^2 + PV^2, so PA^2 + PC^2 = AV^2 + CT^2 + PV^2 + PT^2 and PB^2 + PD^2 = BT^2 + DV^2 + PV^2 + PT^2. But quadrilaterals ABTV and CDVT are rectangles, so AV = BT and CT = DV. Hence, PA^2 + PC^2 = PB^2 + PD^2. Substituting the given information, we get 1^2 + 8^2 = 7^2 + PD^2, so PD^2 = 1^2 + 8^2 - 7^2 = 16, which means PD ={4}.

Someone please put math tags in for me and sorry about the diagram being too big

denis_gylaev · 2015-08-25 05:38:17

Also, can someone show me how to do these 2 problems please? A solution would be nice

The interior angle measures of a pentagon form an arithmetic progression. The difference between the largest and smallest angle measures is $44^\circ$. Find the measure of the smallest angle, in degrees.

Two regular pentagons and a regular decagon, all with the same side length, can completely surround a point. An equilateral triangle, a regular octagon, and a regular n-gon, all with the same side length, also completely surround a point. Find n.

Bob · 2015-08-25 06:06:26

hi denis_gylaev

Your method for calculating AD is exactly the same as mine.

Are you familiar with complex numbers ?

Complex numbers will also fit Pythagoras theorem so the method works whatever.

When I use pythag to calculate AV, AB and AS I get complex numbers not real ones. bobbym has confirmed this. If I use the cosine rule method described in post 4 the arcos is impossible to compute because it is not ≤ 1. I also cannot actually draw the square with those constraints. I conclude that the questioner forgot to check that this square can actually exist in the real world.

Pentagon question.

arithmetic progression means the angles are x, x + d, x + 2d, x + 3d, x + 4d

The angle sum for a pentagon is 540, so you can add up those angles and set equal to 540.

Also (x + 4d) - x = 44, so you can find 'd'. And hence x, the smallest angle.

Fitting together problem.

Two pentagons and one decagon meeting at a point means 108 + 108 + 144 = 360, which it does, so the fit is good.

An equilateral + an octagon mean the angle total so far is 60 + 135 = 195.
So the other polygon must fit in the space left, which means its internal angle is 360 - 195 = 165.
So the external angle is 180 - 165.

And (external angle) times n = 360. you can solve this for n.

Bob

denis_gylaev · 2015-08-25 07:08:15

Thanks bob, Ill get back to you with the answers for the 2 problems
I am getting 86 and 24, can you check that they are right? Thanks

Last edited by denis_gylaev (2015-08-25 07:14:31)

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#1 2015-08-19 08:33:38

Quadrilaterals Help

#2 2015-08-19 11:38:27

Re: Quadrilaterals Help

#3 2015-08-19 19:49:31

Re: Quadrilaterals Help

#4 2015-08-19 23:16:29

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#14 2015-08-20 21:39:43

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