Find the Length of AD

[Nunu]

Hi everyone!
So I took a test today and I'm really confused on one of the problems...
What was on my test paper is attached to this post [there was not any additional information, which is why im confused, although the first page of the test did say "all lines that appear tangent, are tangent].

So far the method that I used [Second attachment] is that the tangent lines of each circle create right angles, and the shape that is made by connecting all the lines is a quadrilateral. I split the shape into two triangles, one of them has a right angle so I solved for the Hypotenuse of that using pythagorean theorum, but now I'm confused as to what to do, because I cant solve for the other triangle since its not a right triangle.

Again, the first attachment is what was on the test and the second attachment is what I did so far [I'm not sure if it's the right approach though!]
If anything is illegible, please lmk bc my handwriting isn't the neatest
Any help is appreciated, even if it's just a step that you think of, I'd love to hear it!

What was in the test:

What I did so far:

 

[lev888]

Draw a segment from D to AB which is parallel to BC. See anything useful?

 

[Nunu]

hm, that would create another line that wouldn't equal what I'm looking for, but that would make 2 more right triangles, so I could solve for the new line, and then make another triangle to solve for AD? Let me try:
i'll call the new point that connects D to AB (that is parallel to BC) E. BE is also 7...is DE 18?
if thats the case then I could do 11-7, since its BA-BE to get EA.. so EA is 4?
then we do the pythagorean theorem, so 4^2+18^2=x^2, x=18.4?

 

[LCKurtz]

Yes.

 

[lev888]

x=18.4

If you are asked to round to 1 decimal place.

 

[The Highlander]

Yes

If you are asked to round to 1 decimal place.

Hi All,

Given that B & C actually are the Points of Tangency to the respective circles, even though the OP's sketch doesn't really show them as such (but everyone appears to be assuming that & the problem would be unsolvable if they weren't; it is just a sketch drawn from memory, I suppose ), then constructing the quadrilateral ABCD gives a nice Right Trapezium.

The last Post I looked at involved calculating the Areas of Trapezia and it immediately occurred to me that that might provide a different 'route' to a solution via the Geometry of the figure (rather than the Pythagorean solution adopted above).

So... if you rotate the original figure anti-clockwise through 90° (so that AB & CD become the Bases) then the Area can be easily calculated as 162m² (I have arbitrarily assigned metres to all dimensions).

If you then perform another 90° anti-clockwise rotation (so that BC & AD are now the Bases) I would have expected to be able to equate the Areas and, thence, evaluate AD using the Area Formula (for the Trapezium).

Please have a look at my attachment that illustrates the procedure I've described.

BUT this "technique" produces a significantly different answer for the Length of AD! ?

Where have I gone wrong???

Is it my Geometry or my Algebra or is just my Arithmetic that's at fault? ?

I'm flummoxed! So, after some (too much) time spent trying to figure it out, I just had to ask for others' input! ?

Any explanation(s) would be much appreciated. ?

Thank you very much. ?

 

[lev888]

Hi All,

Given that B & C actually are the Points of Tangency to the respective circles, even though the OP's sketch doesn't really show them as such (but everyone appears to be assuming that & the problem would be unsolvable if they weren't; it is just a sketch drawn from memory, I suppose ), then constructing the quadrilateral ABCD gives a nice Right Trapezium.

The last Post I looked at involved calculating the Areas of Trapezia and it immediately occurred to me that that might provide a different 'route' to a solution via the Geometry of the figure (rather than the Pythagorean solution adopted above).

So... if you rotate the original figure anti-clockwise through 90° (so that AB & CD become the Bases) then the Area can be easily calculated as 162m² (I have arbitrarily assigned metres to all dimensions).

If you then perform another 90° anti-clockwise rotation (so that BC & AD are now the Bases) I would have expected to be able to equate the Areas and, thence, evaluate AD using the Area Formula (for the Trapezium).

Please have a look at my attachment that illustrates the procedure I've described.

BUT this "technique" produces a significantly different answer for the Length of AD! ?

Where have I gone wrong???

Is it my Geometry or my Algebra or is just my Arithmetic that's at fault? ?

I'm flummoxed! So, after some (too much) time spent trying to figure it out, I just had to ask for others' input! ?

Any explanation(s) would be much appreciated. ?

Thank you very much. ?

Please post the trapezoid area formula.

 

[Dr.Peterson]

If you then perform another 90° anti-clockwise rotation (so that BC & AD are now the Bases) I would have expected to be able to equate the Areas and, thence, evaluate AD using the Area Formula (for the Trapezium).

Aren't bases supposed to be parallel?

 

[Nunu]

Any explanation(s) would be much appreciated. ?

Hi the Highlander! Thank you so much for responding.

I did say in my first post in this thread that "all lines that appear tangent, are tangent", which is why everyone who responded assumed that.

So from what I know, I think you went wrong in the third step, when you were solving for x.
When calculating for the area of a trapezoid, base 1 and 2 MUST be parallel; that's what defines a trapezoid. You were using "x" as a base (AD), even though it was not parallel to segment CB. You can't plug the area in like that.
I attached my work as an example, where I calculated for the area of a trapezoid (not the one in my problem! but one with all sides precalculated) both ways, and they were not, in fact, equal.

As you could see, 24 is the area calculated correctly, and by using a segment that is not parallel to the other segment as a base, an incorrect area is produced.
Going back to your work, you can't set 162 as being equal to 1/2(18+x)11 because 1/2(18+x)11≠1/2(7+11)18.

I'm not 100% sure if this is right, and maybe you are the one who is right (if so, I'm sorry!!). Again I'm just a tired 9th grader trying to get by in my Geometry class.

In any case, thank you so much for your feedback!
-Nunu

 

[Nunu]

Please post the trapezoid area formula.

he posted it in his screenshot!
-nunu

 

[lev888]

he posted it in his screenshot!
-nunu

Note that I asked for the formula, not the expression that results after plugging in values. I could not find the formula in the screenshot. I tried to prompt the poster to review the meaning of the formula components, hoping he/she would realize that it uses bases, not sides.

 

[The Highlander]

Aren't bases supposed to be parallel?

You are absolutely right, of course, and so it was just my Geometry that was faulty! I have never paid much attention to Trapezia. We don't teach very much about them in our (Scottish) schools; we pretty much just tell our kids what they are and their specific properties are only mentioned in passing. I just quickly looked up the "formula" for calculating the area of one but I see now that it only "works" if the bases used in the formula are, indeed, parallel!
(Many Thanks to all contributors.)

 

[The Highlander]

Note that I asked for the formula, not the expression that results after plugging in values. I could not find the formula in the screenshot. I tried to prompt the poster to review the meaning of the formula components, hoping he/she would realize that it uses bases, not sides.

Hi lev,

Thank you very much for your input.

I fully understand the approach you adopted and the reasoning behind it but I'm afraid it wouldn't have helped me to have posted the "formula" I was using:-

Area = \(\displaystyle \frac{1}{2}\)(Sum of the Bases) x Height.​

My problem was not recognizing that the Bases must be the parallel sides of the Trapezium.

I could have stared at that formula until doomsday without realizing why it didn't "work" (on the 'rotated' figure) as long as that vital fact was missing from my awareness. ?

Dr.P's comment prompted me to look again at the Geometry of these quadrilaterals (and how that Area formula is derived from it) and that allowed me to see clearly the flaw in my original "proposition". ?

Do appreciate your help, though. ?

 

[The Highlander]

Hi the Highlander! Thank you so much for responding.

I did say in my first post in this thread that "all lines that appear tangent, are tangent", which is why everyone who responded assumed that.

So from what I know,
When calculating for the area of a trapezoid, base 1 and 2 MUST be parallel

Hi Nunu,

I apologize if you felt aggrieved by my criticism of your sketch though I suspect you may have misunderstood exactly what I was referring to.

I did say that it was just a sketch and therefore not required to be precisely accurate and I didn't fail to note your comment that "all lines that appear tangent, are tangent" but I wasn't questioning the tangency of the (roughly ?) straight line to either of the circles.

My "complaint" (and I admit it might well be dismissed as a rather pernickety one) was that your Point B in particular didn't look anything like a Point of Tangency! (If you imagine a line from the centre of the larger circle out to that point, it wouldn't meet the tangent at anything even remotely close to a right angle!)

Even in rough sketches it is good practice to ensure that the critical aspects of your drawings at least visually match what you wish to convey.

Notwithstanding that, I thank you very much for taking the trouble to provide such a full reply and, of course, you did, indeed, identify the problem that was causing my consternation (that the Bases used in calculating the Area of a Trapezium must be the parallel sides).

You are an impressive contributor (for a mere "9th grader" ?) and I suspect you will go on to do very well in your mathematical endeavours.

Keep up the good work! ?

PS: Do ask Santa to bring you a ruler this year! ?

 

[lev888]

Hi lev,

Thank you very much for your input.

I fully understand the approach you adopted and the reasoning behind it but I'm afraid it wouldn't have helped me to have posted the "formula" I was using:-

Area = \(\displaystyle \frac{1}{2}\)(Sum of the Bases) x Height.​

My problem was not recognizing that the Bases must be the parallel sides of the Trapezium.

I could have stared at that formula until doomsday without realizing why it didn't "work" (on the 'rotated' figure) as long as that vital fact was missing from my awareness. ?

Dr.P's comment prompted me to look again at the Geometry of these quadrilaterals (and how that Area formula is derived from it) and that allowed me to see clearly the flaw in my original "proposition". ?

Do appreciate your help, though. ?

If posting the formula didn't help, I would've asked for the definition of trapezoid bases next. I prefer to give small hints to give the person I'm helping every opportunity to work out the rest by her/himself.

 

[The Highlander]

If posting the formula didn't help, I would've asked for the definition of trapezoid bases next. I prefer to give small hints to give the person I'm helping every opportunity to work out the rest by her/himself.

Indeed.

 

[Nunu]

I apologize if you felt aggrieved by my criticism of your sketch though I suspect you may have misunderstood exactly what I was referring to.

Of course not!!! You're right, I should've drawn the diagram to match what information was given.
Again thank you all for the feedback!!!