what is the area of the closed figure?

A closed figure is bound by the followning system:

x = 0

y =10

y = -x if -6 <(or equal to) x <(or equal to) 0

2x+3y = 6 if -12 <(or equal to) x <(or equal to) -6

What is the area of the closed figure?

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  • 1 decade ago
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    The closed figure ends up being made up of three separate regular figures. We need calculate the area of each and add them up. To do that, we need to find the "breakpoints" that form the vertices of the various subfigures.

    First, we have x = 0 (the vertical Y axis), and y = 10 (a line parallel to the horizontal X axis). Finally, we have two separate conditions that describe everything we need to know (all case in which x <= 0).

    That pair of conditions looks to differ, so let's just assume they do. (Not a problem with this simple set of figures. We might not want to assume that if we had to work through 43 subfigures.) So, let's first mark out the rectangle formed by the Y axis on the right (x = 0), the y = 10 line parallel to the Y axis, and the perpendiculars to each of those lines from the point where x = -6. That point is subject to the both of those two conditions so we will pick the one valid from the origin to that point:

    y = -x substitute

    y = -(-6) simplify

    y = 6

    So the point is (-6, 6). The rectangle, then, is height = 10 - 6 = 4 and length = 0 - (-6) = 6. Its area (height * length) is: 4 * 6 = 24. One down, two to go.

    Let's consider the triangle we made in finding the endpoint of the line from the origin to (-6, 6). That line is its hypotenuse since the intersection of x =0 and y = 6 is a right angle. So one of the legs is an altitude and one a base (of course, call one the altitude, the other is the base, and the opposite naming pair is true, but it doesn't matter for calculating area!). The area of a triangle is given by altitude * base * 1/2 so if we can measure those two legs, we can calculate the area. Well, one leg is length = 0 - (-6) = 6 and the other length is length = 6 - 0 = 6. So the area is: 1/2 * 6 * 6 = 18.

    Add that to the 24 for the rectangle: 24 + 18 = 42. So far, so good. Now just the last triangle to find.

    At x = -6, the other condition takes over. So, the starting point is (x = -6, 2x + 3y = 6). Let's find y by using the second condition's equation and x = -6:

    2x + 3y = 6 substitute x = -6

    2*(-6) + 3y = 6 add 12 to each side

    -12 + 12 + 3y = 6 + 12 simplify; divide each side by 3

    3y/3 = 18

    y = 6

    So the point is (-6, 6) which is nice in a way. NOT that it would matter, but if y had come to some other value here, like, say, y = -2, then the figure would have taken on a disturbing (but correct) aspect. (That kind of thing leads students to worry they made a bad mistake and waste a lot of time trying to find out just what.)

    This gives us one leg of a right triangle (the perpendicular from (-6, 6) up to the y = 10 line). We just need to know where the line from this point going to x = -12 crosses that y = 10 line. Well, we know the condition's value: 2x + 3y = 6 anywhere from x = -6 to -12. Let's find what x equals when y = 10. Hopefully it will be less than or equal to -12 so we can stop with just these three figures. Hopefully also because we do not know how to evaluate the figure when x < -12 ! So:

    2x + 3y = 6 substitute y = 10

    2x + 3*10 = 6 simplify; add -30 to each side

    2x = 6 - 30 simplify; divide each side by 2

    2x/2 = -24/2 simplify

    x = -12

    Very nice, we stay within the range we know how to evaluate! So, the line forming the hypotenuse of this right triangle hits the y = 10 line exactly at (-12, 10) forming the last vertex we need.

    The first length was length = 10 - 6 = 4. And the second length we need in this triangle is length = -6 - (-12) = 6. The area of this triangle then, is 1/2 * 6 * 4 = 12.

    Adding that to the area of the other two subfigures we get: 42 + 12 = 54. And that is the total area of the complete closed figure.

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