help with proportion of geometry

kidokid

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Apr 8, 2011
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can you help me with this probblem :D

ok the problem says the ladders below are standing against the wall at the same angle. how high up the wall does the longer ladder go?(all measures are in feet)
the picture has two ladders against a cube; one ladder is 30ft. and the other is 24ft. the one thats 30 ft. How hogh it is is is represented by X and the one that is 24 ft. How high up is 10 ft.There are four multiple choice answers...
a-11.25
b-12.5
c-14
d-26
im trying to find out if my method of solving this problem is correct. Well to start off i created a proportion 30/x=24/10 right and then i multiplied 10 by 30 to get 300 then i divided by 24 to get B-12.5 Is this the correct way of solving this problem :?:
 
If the wall is vertical, the ground is horizontal, and the ladders are parallel, you have proportional right triangles. All the corresponding parts are proportional, to wit,

2430=10x\displaystyle \frac{24}{30}=\frac{10}{x}

This is the same as you have, but I would NOT work it quite the way you did. I didn't like that '300' at all.

45=2430=10x\displaystyle \frac{4}{5}=\frac{24}{30}=\frac{10}{x}

45=10x    25=5x\displaystyle \frac{4}{5}=\frac{10}{x} \implies \frac{2}{5}=\frac{5}{x}

In this last form, I dare you to get a value so large as 300!
 
ok the problem says the ladders below are standing against the wall at the same angle. how high up the wall does the longer ladder go?(all measures are in feet)
the picture has two ladders against a cube; one ladder is 30ft. and the other is 24ft. the one thats 30 ft. How hogh it is is is represented by X and the one that is 24 ft. How high up is 10 ft.There are four multiple choice answers...
a-11.25
b-12.5
c-14
d-26
im trying to find out if my method of solving this problem is correct. Well to start off i created a proportion 30/x=24/10 right and then i multiplied 10 by 30 to get 300 then i divided by 24 to get B-12.5 Is this the correct way of solving this problem

Hello, kidokid,

Tkhunny has shown you a correct method and solution (of course!) :) However, I want to point out that your method was correct, also, though your terminology was not quite accurate. Your starting equation is correct:

30/x=24/10

But you do not simply multiply "10 by 30 to get 300"; you multiply both sides of the equation by 10 to get:

300/x = 24

Next you multiply both sides of the equation by x to get:

300 = 24x

Finally, divide both sides of the equation by 24 to get:

300/24 = x = 12.5

My point is simply that whatever we do to one side of the equation, we must do to the other side of the equation as well. This is the basis for manipulating all algebraic equations.
 
kidokid said:
...trying to find out if my method of solving this problem is correct.
Well to start off i created a proportion 30/x=24/10 right and then i multiplied 10 by 30
to get 300 then i divided by 24 to get B-12.5 Is this >>>\displaystyle > > >the correct way of solving this problem <<<\displaystyle < < <:?:

kidokid,

it is not "the\displaystyle the correct way" of solving this problem; it is a\displaystyle a way.

Because of what the OP has is equiavalent to:

30x = 2410,\displaystyle \frac{30}{x} \ =\ \frac{24}{10},

then stating that 10 is multiplied by 30 to get 300, and then that 300 is divided by 24 begets 12.5
is another way of stating the steps that are worked in the cross-multiplying method.

When a person, such as kidokid, states the above as his way, he is not explicitly stating that what
he is doing to one side, he is to do to the other side.

Cross-multiplying and appropriate dividing to get the answer does not show doing to each side
the same operation, but\displaystyle but it is inherent in the cross-multiplying method. That is,
it implicitly uses the ideas of multiplication and division being done on both sides, respectively.
 
...is another way of stating the steps that are worked in the cross-multiplying method.

When a person, such as kidokid, states the above as his way, he is not explicitly stating that what
he is doing to one side, he is to do to the other side.

Cross-multiplying and appropriate dividing to get the answer does not show doing to each side
the same operation, it is inherent in the cross-multiplying method. That is,
it implicitly uses the ideas of multiplication and division being done on both sides, respectively.

Herein lies the problem. Students often attempt to use "cross multiplying" as a set of "steps" to "obtain a solution" -- with no understanding of what "cross multiplying" really is. As a result, I have seen students attempt to multiply two fractions by "cross multiplying" (or even add two fractions in this way!) as well as other mis-applications. Students must know *why* they are performing an operation -- not just performing a series of "steps" by rote. IMHO.
 
You know I can't resist this one.

"cross-multiply" - profanity and vulgarity. Never do it. :shock:
 
kidokid,

Always try to cross-multiply to solve a problem with a fraction equal
to a fraction wherever convenient.

It's the product of the means equals the product of the extremes.
(And I learned it in my high school geometry class.)

So, ab = cd\displaystyle \frac{a}{b} \ = \ \frac{c}{d} means

bc = ad\displaystyle bc \ = \ ad

And you should know this phrase and its derivation when you use it.
--------------------------------------------------------------------------------

tkhunny said:
Never do it.


kidokid,

in fact, do it as often as possible.

This is a relationship typically taught in elementary algebra,
or in geometry, of which this subforum includes.
 
If you want to learn the concept of "means equal the extremes", I have no objection to that.

"Cross-multiply" - no good.

My views. I welcome others'.
 
Denis said:
I'm lazy....so cross multiply I love!

Well, I'm not lazy, and I expect students to take advantage of it.

A specific site from the Internet page that Denis gave:

http://en.wikipedia.org/wiki/Cross-multiplication

-----------------------------------------------------------------------------------------


Also, kidokid, if you want to compare the sizes of numbers
written as fractions, you can cross-multiply:

Ex.

Which of these numbers is larger than the other, or are they
equal in size:


23\displaystyle \frac{2}{3}versus\displaystyle versus 35\displaystyle \frac{3}{5}

2(5)\displaystyle 2(5)versus\displaystyle versus 3(3)\displaystyle 3(3)

10 > 9\displaystyle 10 \ > \ 9

So, the first number is greater than the second number.

------------------------------------------------------------------------------------

Edit:

Look at the advantages of solving n128 = n12\displaystyle \frac{n - 12}{8} \ = \ \frac{n}{12}

using cross-multiplying over multiplying by the least

denominator on each side.
 
Two points in what is an interesting discussion.

(I) I doubt I am unusually lazy, but I cross-multiply whenever it seems like a useful short-cut. It is like moving a term from one side of an equation to the other and changing the sign. Algebra is mainly a process of simplicifcation, and cross-multiplying simplifies.

(II) Unless the student FULLY understands that cross-multiplication is a short-cut and for what it is a short-cut, the student suddenly has an APPARENT violation of the fundamental rule that what is done to one side of the equation must be done to the other. ABSENT that understanding, a student will almost certainly make mistakes. Moreover, it is extremely easy to foster misunderstanding. For example, saying (a / b) = (c / d) MEANS ad = bc. It is NOT TRUE that those two propositions are logically equaivalent. (a / b) = (c / d) DOES ENTAIL that ad = bc, but ad = bc does NOT entail that (a / b) = (c / d).
 
JeffM said:
For example, saying (a / b) = (c / d) MEANS ad = bc. It is NOT TRUE that those two propositions
are logically equaivalent. (a / b) = (c / d) DOES ENTAIL that ad = bc, but ad = bc does NOT entail that (a / b) = (c / d).

The use of that word is a conditional; it is an if-then.
It is not used for equivalence.


Example:
-----------

Two odd prime numbers added together means\displaystyle means their sum is an even number.

I could have stated "The consequence\displaystyle consequence of adding together two odd
prime numbers is an even integer."

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

There is no equivalence (an if-and-ony-if) intended here using the word "mean."

And, no, an even integer is not necessarily the sum of two odd prime numbers.
 
"Mean" implies a definition, and a definition does entail logical equivalence. I am sorry: "A means B" is not necessarily limited to "If A, then B." English has an UNAMBIGUOUS way to specify a conditional statement. I know you intended "means" to imply a conditional, and people who know what they are doing in algebra undoubtedly read it as such. But many of those who come here for help in algebra do not know what they are doing. My point was that until they understand what they are doing and why when they cross-multiply, they are apt to make mistakes and easily misconstrue what they read or hear.

Yes, "consequence" is better.
 
"Might" imply. So it is ambiguous. Ambiguity is not usually considered a desirable attribute in an explanation.

ONE of the points I raised is that until a kid undertands why cross-multiplication is a valid operation, mistakes will happen, particularly if instructors are not as precise in their language as it is possible for them to be.

I'll let you have the last word (if you want one) as this is getting silly.
 
AS IT STOOD before THIS post, you were having the last word, lookagain:
your edit is at 12:47, which is after Jeff's post.
But I'll take you both out of your misery by taking the last word myself: zzz
 
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