9514 1404 393
Answer:
34 square units.
Step-by-step explanation:
The figure can be decomposed into a trapezoid and a triangle. The point of intersection of the two lines is (6, 4), so a horizontal line at y=4 will create ...
- a triangle of height 2 and base 6
- a trapezoid with bases 6 and 8, and height 4.
Using the relevant area formulas, we find ...
triangle area = 1/2bh = 1/2(6)(2) = 6
trapezoid area = 1/2(b1 +b2)h = 1/2(6+8)(4) = 28
Total shaded area = 6 + 28 = 34 square units.
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The equations of the lines can be written as ...
y = -2x +16
y = -1/3x +6
Equating y, we get
-2x +16 = -1/3x +6
10 = 5/3x . . . . . . . . . add 2x-6
6 = x . . . . . . . . . multiply by 3/5
y = -2(6) +16 = 4
The point of intersection is (6, 4).
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<em>Alternate solution</em>
Once we know the vertices of the shaded area:
- (0, 0)
- (0, 6)
- (6, 4)
- (8, 0)
- (0, 0)
we can form pairwise "determinants" of the form x1y2 -x2y1. Note the first point is repeated at the bottom of the list, and the points are listed in order around the boundary of the area. The points (0, 0) contribute nothing, so we have left them out of the computation below. The area is half the absolute value of the sum of these "determinants".
1/2|(0·4 -6·6) +(6·0)-(8·4)|
= 1/2|-36 -32| = 1/2(68) = 34
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In the attachment, the equations of the lines are written in intercept form, since the problem statement gives the intercepts of the lines.
