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Math Help - Function of x (P and A of a Rectangle)

  1. #1
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    Function of x (P and A of a Rectangle)

    Hello,

    Here's a problem which I'm unable to solve. I'd be grateful if you helped me through:

    A rectangle with altitude x is inscribed in a triangle ABC with base b and altitude h. Express the perimeter P and area S of the rectangle as functions of x.


    Answers are P=2b+2(1-b/h)x and S=b(1-x/h)x

    Thanks.
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  2. #2
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    My first thought was that there was not enough information given- we are told the height of the triangle but not where the vertex is. "Tilting" the triangle so that the vertexs is way over to one side shrinks things down doesn't it? It turns out that, no that is not true.

    Here's what I did: set up a coordinate system so that one vertex of the base is at (0, 0) and the other is at (b, 0). We don't know the x-coordinate of the third vertex so call that "a". The y-coordinate of that third coordinate is "h"- the third vertex is at (a, h).

    Now, it is easy to see that the equation of the line through (0, 0) and (a, h) is y= (h/a)x (when x=0, y= (h/a)0= 0 and when x= a, y= (h/a)a= h since the "a"s cancel). It takes only a little more calculation to see that the line through (a, h) and (b, 0) has equation y= (h/(a-b))(x- a) (when x= a, y= (h/(a-b))(a-b)= h since the "a-b" terms cancel and when x= b, y= (h/(a-b))(b- b)= (h/(a-b))(0)= 0).

    Now there is one ambiguity about this problem- it doesn't say anything about the orientation of the rectangle. A figure is "inscribed" in a polygon as long as its vertices lie on the polygon. A rectangle might be inscribed in a triangle as long as one side of the rectangle is on any side of the triangle. I am going to assume that one side of the rectangle lies along the base of the triangle (with length b).

    Also, because I am using "x" in the equations of the lines, I am going to call the height of the rectangle "u" instead.

    Okay, the upper vertex of the "left" side of the rectangle lies on the line y= (h/a)x. When y= u, u= (h/a)x so x= (a/h)u. The upper vertex of the "right" side of the rectangle lies on the line y= (h/(a-b))(x- b) so when y= u, u= (h/(a-b))(x- b), x- b= ((a-b)/h)u and x= b+ ((a-b)/h)u. The length of the base of the rectangle is the difference of those two x values: b+ ((a-b)/h)u- (a/h)u= b+ (a/h)u- (b/h)u- (a/h)u= b- (b/h)u.

    Now we have that the base of the rectangle is b- (b/h)u= b(1- u/h) and the height is u.

    Replacing "u" with the original "x" of the problem, the base is b(1- x/h) and height is x.

    The perimeter is, of course, twice the base plus twice the height and the area is base times height. Do those calculations and you should see that you get the answers given.
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