Find the length XY from the rectangular diagram

In summary, the conversation is about finding the length ##XY## in a given diagram using different methods such as cosine rule, similarity, and pythagoras theorem. The original question was whether the answer ##1## is correct or not. The conversation also includes a discussion about the given angles and sides and how they can help in finding the length. Ultimately, it is concluded that the problem is solvable if the figure is assumed to be a rectangle, and the solution is straightforward using basic geometry principles.
  • #1
chwala
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Homework Statement
Find the length ##XY## by considering the given diagram
Relevant Equations
cosine rule, similarity, pythagoras theorem
1612914126201.png


i saw this question on the internet. The people responding were of the opinion that the length##xy=1##, but no working...first of all is the answer ##1## correct?
i am trying to find the steps to solution, i have managed to find an angle ##57.27^0## by cosine rule, i tend to think that i need to sort of find the angles then calculate the length. My other thinking is to use similarity. Am still trying to work on it. cheers
my working,
1612914443382.png
 
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  • #2
chwala said:
Homework Statement:: Find the length ##XY## by considering the given diagram
Relevant Equations:: cosine rule, similarity, pythagoras theorem

View attachment 277686

i saw this question on the internet. The people responding were of the opinion that the length##xy=1##, but no working...first of all is the answer ##1## correct?
i am trying to find the steps to solution, i have managed to find an angle ##57.27^0## by cosine rule, i tend to think that i need to sort of find the angles then calculate the length. My other thinking is to use similarity. Am still trying to work on it. cheers
my working,
View attachment 277688
Is it given that the figure is a rectangle?
 
  • #3
Mark44 said:
Is it given that the figure is a rectangle?

Yes it is, ...##11## by ##6##
 
  • #4
Is it a given that angle ABX and angle CBX are the same?

If not, I don't see how this is solvable, since these would both be perfectly valid solutions:
1612918026011.png


[ EDIT ] Just checked the convention:

1612918256151.png


OK, so ABX and angle CBX (and their mirrors) are equal.
 
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  • #5
DaveC426913 said:
Is it a given that angle ABX and angle CBX are the same?

If not, I don't see how this is solvable, since these would both be perfectly valid solutions:View attachment 277691

[ EDIT ] Just checked the convention:

View attachment 277692

OK, so ABX and angle CBX (and their mirrors) are equal.
they should be the same...we have isosceles triangles in the mix, and i attempted to use sine rule by considering the 90 degree angle to find other angles...
 
  • #6
chwala said:
they should be the same...we have isosceles triangles in the mix, and i attempted to use sine rule by considering the 90 degree angle to find other angles...
If they are the same - and it seems they are - I suspect you don't need any trig - geometry alone should suffice.
Better than mere isosceles - you've got some right triangles there.

But don't take my word. I may be wrong.
 
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  • #7
chwala said:
Yes it is, ...##11## by ##6##
From what you posted I don't see any information that says the figure is a rectangle, or the the four corner angles are right angles. Are you just assuming that the unlabeled sides are 6 and 11? They could be 6.1 and 11.2. Is there any more to this problem than what is shown in the picture you showed?
 
  • #8
If we assume that an answer can be calculated, then we must conclude that the diagram is a rectangle. In that case, the proposed solution is more complicated than it has to be. (HINT: Very few calculations are needed and any calculations can easily be done in your head.)
We would know that the length of the line AD is 11. We know the angle ABX and can immediately determine the length of line AX. Likewise, we can determine the length of the line YD. The answer is simple to get from that.
 
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  • #9
i still cannot see, am blind here Dave can you show me how?
 
  • #10
Mark44 said:
From what you posted I don't see any information that says the figure is a rectangle, or the the four corner angles are right angles. Are you just assuming that the unlabeled sides are 6 and 11? They could be 6.1 and 11.2. Is there any more to this problem than what is shown in the picture you showed?

yes i did assume...the question attached is posted in its original form. I made an assumption Mark...there is nothing more to the problem than what is given. On the contrary, Dave indicates that it is possible. I would like to know how possible is it to find the length given that it is a rectangle...
 
  • #11
chwala said:
i still cannot see, am blind here Dave can you show me how?
What possible angle for ABX would make the answer immediately obvious?
 
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  • #12
Mark44 said:
... the four corner angles are right angles.
I did not assume this. No angles in the diagram are specified.

Mark44 said:
Are you just assuming that the unlabeled sides are 6 and 11? They could be 6.1 and 11.2. Is there any more to this problem than what is shown in the picture you showed?
OTOH, I think this is assumable. If the other two sides were not 6 and 11, that would be a very sneaky underhanded trick.
 
  • #13
phinds said:
What possible angle for ABX would make the answer immediately obvious?
The angles in the diagram are all specified as equal.

So, the question should be: What possible angle for ABC would make the answer immediately obvious?
 
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  • #14
phinds said:
What possible angle for ABX would make the answer immediately obvious?

i still do not see...you want me to use SOHCAHTOA right...but i only know one side ##6## cm and one angle i.e right angle. You are saying that it is straightfoward? in regards to ##ABX##...
 
  • #15
DaveC426913 said:
The angles in the diagram are all specified as equal.
That's not how it was when I learned geometry. Two equal angles were marked with an arc and a tick mark through the arc. Two other equal angles were marked with two tick marks. Pairs of angles with the same number of tick marks are assumed to be given as equal.

If angles ABX XBC, BCY, and YCD are all congruent (equal), the problem is trivial. The Sine Law and Cosine Law are completely unnecessary.
 
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  • #16
If you have a rectangle, what are the corner angles? Now divide that into two equal angles. Now ABX is a right triangle with one side known and the other angles known. You should be able to figure it out from there with practically no calculations.
 
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  • #17
This really looks to me that the length is a function of the two angles ##\angle## BCD=##\theta## and ##\angle## BCA= ##\alpha## so that the length is a function##f(\theta,\alpha)## since we have no information in the diagram, even though they look the same, that they are.

My suggestion then is to just work it out with a few real angles then generalize it to ##f##
 
  • #18
Mark44 said:
That's not how it was when I learned geometry. Two equal angles were marked with an arc and a tick mark through the arc. Two other equal angles were marked with two tick marks. Pairs of angles with the same number of tick marks are assumed to be given as equal.
Yes, but another convention (sans ticks) is:

1612966939598.png


(as shown in post 4)
 
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  • #19
My training was that identical markings on angles indicated that they were equal. I have seen different symbology used by different people. Furthermore, assuming that the marks on the angles indicate something, I must assume that they indicate equal angles. Otherwise, there is no answer to the problem.
Likewise, there is no answer if the figure is not a rectangle, so I will assume that it is. With those assumptions, the solution is easy. Without those assumptions, there is no solution.
 
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  • #20
Just by visual inspection and not writing anything down.The following is how I would approach it.

Let E be the point of intersection where segments BX and CY intersection. Assume that the figure given is a rectangle. Then we proceed by similar triangles. (I rather not say it which triangles are similar, to give OP a shot at attempting a solution). The only other hint I would give (I assume this is valid) is to construct the perpendicular bisector from the point Y to the segment YX. Then other pieces should pop out at this point.

I would approach this via similar triangles. Sometimes the neat solutions for these types of problems pop out after having all the facts laid out...
 
  • #21
Just seen it. The solution if we assume a rectangle, is short.
 
  • #22
I don't think I've never seen so much overthinking from so many people in so little time.

It's an 11x6 rectangle of which both upper corners are equally bisected by their respective diagonals. Any other interpretation leads to a (comparatively) messy equation as an answer.

1612914126201-png.png

1) We are given that ABCD is a rectangle. This means that all the corner angles (A,B,C,D) are ____ degrees, also known as _____ angles.

(Now, we're just going to concentrate on the left-hand triangle ABX)

2) We are given (after a bit of rummaging around with diagram conventions) that the diagonal BX equally bisects the angle of its origin corner, B. Since we now know the angle of B, this means that the angle ABX is ______ degrees.

3) We know that the internal angular total of any triangle is ______ degrees. This means that the "unknown" angle BXA is actually ______ degrees.

4) This makes the triangle to be of type (equilateral,isosceles,scalene : pick one). Which means that the relationship between the lengths of vertical side AB to the horizontal side AX is _______.

5) Now do steps 2-4 for the right hand triangle DCY.

6) Facepalm and deduct one point when you realize it's just a mirror image of ABX, and you didn't need to do step 5. Anyways,

7) Which means that : since AX and DY have lengths ____ and ____ , and AD is of length 11, the equation for figuring out the length/distance XY is _______________ and the answer is ____.

8) From an imaginary mark, deduct one point if you used units like 'cm' : the problem is defined as dimensionless.
 
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  • #23
MidgetDwarf said:
Just by visual inspection and not writing anything down.The following is how I would approach it.

Let E be the point of intersection where segments BX and CY intersection. Assume that the figure given is a rectangle. Then we proceed by similar triangles. (I rather not say it which triangles are similar, to give OP a shot at attempting a solution). The only other hint I would give (I assume this is valid) is to construct the perpendicular bisector from the point Y to the segment YX. Then other pieces should pop out at this point.

I would approach this via similar triangles. Sometimes the neat solutions for these types of problems pop out after having all the facts laid out...
you may just need to show me, i cannot really see that...how did these internet guys arrive at the solution##XY=1##, they probably must be more intelligent than i :cool:
 
  • #24
1613184226404.png


i think i nailed it...bingo
 
  • #25
It's much simpler than that, assuming that all of the corner angles are equal (and equal to 45°), and that the figure is a rectangle. Given those assumptions, you don't need any of the calculations in the image you attached.
AB = 6, so Ax = 6 as well (isosceles right triangle). CD = 6 and Dy = 6 for the same reason. Since AD = 11, there's an overlap of 1, which is the length of segment XY.
 
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  • #26
chwala said:
i think i nailed it...bingo
Can't argue with success [edit: sure you can]. How did you get the value for the hypotenuse ?
 
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  • #27
hmmm27 said:
Can't argue with success. How did you get the value for the hypotenuse ?

Square root of ## 5.5^2 + 6^2##
 
  • #28
Mark44 said:
It's much simpler than that, assuming that all of the corner angles are equal (and equal to 45°), and that the figure is a rectangle. Given those assumptions, you don't need any of the calculations in the image you attached.
AB = 6, so Ax = 6 as well (isosceles right triangle). CD = 6 and Dy = 6 for the same reason. Since AD = 11, there's an overlap of 1, which is the length of segment XY.
Exactly. That's the hint I gave him in post #11
 
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  • #29
hmmm27 said:
How did you get the value for the hypotenuse ?
Why? The hypotenuse isn't needed.
 
  • #30
Mark44 said:
Why? The hypotenuse isn't needed.

I appreciate your approach ...its good but I think my approach would be better for students who may want to find value of the length segment ##YX## by way of calculations...cheers
 
  • #31
chwala said:
I appreciate your approach ...its good but I think my approach would be better for students who may want to find value of the length segment ##YX## by way of calculations...cheers
I disagree. A simple approach is generally better than one that is more complicated. What's the value in doing calculations that don't need to be done?
 
  • #32
chwala said:
I appreciate your approach ...its good but I think my approach would be better for students who may want to find value of the length segment ##YX## by way of calculations...cheers
The problem was rigged to be solvable without any significant calculations. I'm afraid that your method is not using the principles that the problem was made to exercise. And I am not sure that you understood what many people were trying to tell you about the simpler solution.
 
  • #33
Mark44 said:
I disagree. A simple approach is generally better than one that is more complicated. What's the value in doing calculations that don't need to be done?
FactChecker said:
The problem was rigged to be solvable without any significant calculations. I'm afraid that your method is not using the principles that the problem was made to exercise. And I am not sure that you understood what many people were trying to tell you about the simpler solution.

I strongly agree w/ both of these. @chwala, you seem to have missed the point of the exercise.
 
  • #34
phinds said:
I strongly agree w/ both of these. @chwala, you seem to have missed the point of the exercise.

I understand their approach of using the diagonals ##BX## and ##CY## being equal and using the perpendicular bisector of ##YX##, to come to the conclusion...its very clear to me...ok cheers
 
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  • #35
chwala said:
Square root of ## 5.5^2 + 6^2##
##8.485## is indeed the length of the hypotenuse, but it is not ##\sqrt{5.5^2+6^2}##. How did you arrive at ##8.485## ?
 
<h2>1. How do you find the length XY from a rectangular diagram?</h2><p>To find the length XY from a rectangular diagram, you need to use the Pythagorean theorem. This theorem states that in a right triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides. In this case, the length XY is the hypotenuse, and the other two sides are the length of the rectangle and the width of the rectangle.</p><h2>2. What is the formula for finding the length XY from a rectangular diagram?</h2><p>The formula for finding the length XY from a rectangular diagram is: XY = √(length² + width²). This formula is derived from the Pythagorean theorem and is applicable to any right triangle, including the one formed by the length, width, and length XY in a rectangular diagram.</p><h2>3. Can you find the length XY if the rectangle is not a perfect square?</h2><p>Yes, you can still find the length XY even if the rectangle is not a perfect square. As long as you know the length and width of the rectangle, you can use the formula XY = √(length² + width²) to find the length XY.</p><h2>4. Is it possible to find the length XY if the rectangle is not a right triangle?</h2><p>No, it is not possible to find the length XY if the rectangle is not a right triangle. The Pythagorean theorem only applies to right triangles, so if the rectangle is not a right triangle, the formula XY = √(length² + width²) cannot be used to find the length XY.</p><h2>5. What units should be used for the length and width in the formula to find the length XY?</h2><p>The units used for the length and width in the formula to find the length XY should be consistent. For example, if the length and width are measured in meters, then the length XY will also be in meters. It is important to use the same units throughout the formula to ensure accurate results.</p>

1. How do you find the length XY from a rectangular diagram?

To find the length XY from a rectangular diagram, you need to use the Pythagorean theorem. This theorem states that in a right triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides. In this case, the length XY is the hypotenuse, and the other two sides are the length of the rectangle and the width of the rectangle.

2. What is the formula for finding the length XY from a rectangular diagram?

The formula for finding the length XY from a rectangular diagram is: XY = √(length² + width²). This formula is derived from the Pythagorean theorem and is applicable to any right triangle, including the one formed by the length, width, and length XY in a rectangular diagram.

3. Can you find the length XY if the rectangle is not a perfect square?

Yes, you can still find the length XY even if the rectangle is not a perfect square. As long as you know the length and width of the rectangle, you can use the formula XY = √(length² + width²) to find the length XY.

4. Is it possible to find the length XY if the rectangle is not a right triangle?

No, it is not possible to find the length XY if the rectangle is not a right triangle. The Pythagorean theorem only applies to right triangles, so if the rectangle is not a right triangle, the formula XY = √(length² + width²) cannot be used to find the length XY.

5. What units should be used for the length and width in the formula to find the length XY?

The units used for the length and width in the formula to find the length XY should be consistent. For example, if the length and width are measured in meters, then the length XY will also be in meters. It is important to use the same units throughout the formula to ensure accurate results.

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