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March 1, 2015

March 1, 2015

Posted by **Anonymous** on Wednesday, October 9, 2013 at 7:02pm.

*On my homework, there is a drawing of the wires... It makes sort of like a triangle flag shape.. It is a right triangle. We don't know the BASE of that triangle, which is the distance from point B to C, B being the bottom point and C being the top. So the triangle is a right triangle with the base on the left.. From Point B, there is a wire going down to point A.. The distance from Point C to Point A is 15 meters. The top of the triangle is point D. The leg of the triangle, which is the distance from Point D to C is 12 meters... If you need more information just ask..

- Math -
**Kanukipper**, Monday, November 3, 2014 at 9:02pmFirst you need to determine all of the different segments in terms of x. You have:

AB=Unknown

BC=Unknown

CD=12m

BD=Unknown

All of the Unknown's will be in terms of some X value. For simplicity's sake let's say BC=x (ultimately this is what you want to solve for you can do it the other way and make AB=x). We also know that AC=15m and AB+BC=AC so we can determine AB and BC. So now you have:

AB=15-x

BC=x

CD=12

BD=Unknown

BD is the hypotenuse of a right triangle with BC and CD it's sides. So using the Pythagorean Theorem you can solve for BD in terms of x to get the following:

AB=15-x

BC=x

CD=12

BD=squareroot(x^2+144) or (x^2+144)^(1/2)

Now the important part is to notice the units. BD is 9oz/m and AB is 8oz/m. The total weight however is only 197oz. Not oz/m. So you need to multiply the segment length by it's weight/length in order to determine it's actual weight. Then we know that these two weights equal 197 oz. So your equation will look like this:

8*AB+9*BD=197

So,

8*(15-x)+9*(x^2+144)^(1/2)=197

At this point you can solve for x, which as we originally defined it is the BC length.

You get: x=5 or x=67.47

This is a real world problem, and the entire wire of AC is 15m. So you can't have a segment of it be 67.47m. Therefore the only possible answer is 5m.

P.S. If one value is negative then you have the same issue. There aren't any negative lengths. So only one value will work.

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