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Okay, so I need to figure out how to solve this problem where AmonD is trying to calculate the area of a quadrilateral with vertices at (1,2), (4,6), (7,2), and (5,1). Hmm, quadrilaterals can be a bit tricky because they come in different types, like squares, rectangles, trapezoids, etc., but this one doesn't look like any of those regular shapes. Maybe it's a general quadrilateral.
First, I remember that for polygons, especially convex ones, there's a formula called the shoelace formula that can calculate the area if you know the coordinates of the vertices. Let me recall how that works. The shoelace formula takes the coordinates of the vertices in order, either clockwise or counter-clockwise, and then multiplies and subtracts specific pairs of coordinates. The formula is something like: for vertices (x1, y1), (x2, y2), ..., (xn, yn), the area is half the absolute value of the sum from i=1 to n of (xiyi+1 - xi+1yi), where xn+1 = x1 and yn+1 = y1.
Let me verify that. Suppose I have a simple square with vertices (0,0), (0,1), (1,1), (1,0). Applying the formula, the sum would be (01 + 01 + 10 + 10) - (00 + 11 + 11 + 00) = (0 + 0 + 0 + 0) - (0 + 1 + 1 + 0) = -2. Taking absolute value and dividing by 2 gives 1, which is correct. So that seems right.
Alright, so applying this to the given quadrilateral. Let me list the coordinates in order. The problem mentions the vertices as (1,2), (4,6), (7,2), and (5,1). I need to make sure they are listed in order, either clockwise or counter-clockwise. Let me plot them mentally or maybe sketch roughly:
(1,2) is somewhere in the lower left, (4,6) is up higher, (7,2) is to the right and down, and (5,1) is a bit to the left and down from (7,2). Connecting these in order: (1,2) -> (4,6) -> (7,2) -> (5,1) -> back to (1,2). Hmm, this should form a quadrilateral. Let me check if the order is correct. If I traverse these points in the given order, does it make a simple quadrilateral without crossing sides? I think so. So I can proceed with the shoelace formula.
Let me set up the coordinates:
x1 = 1, y1 = 2
x2 = 4, y2 = 6
x3 = 7, y3 = 2
x4 = 5, y4 = 1
Then, according to the shoelace formula, the area is 1/2 |(x1y2 + x2y3 + x3y4 + x4y1) - (y1x2 + y2x3 + y3x4 + y4x1)|
Wait, let me make sure. The formula is sum over xiyi+1 - xi+1yi. So for each i, multiply xi by yi+1 and subtract xi+1*yi. Then sum all those and take half the absolute value.
So, breaking it down step by step:
First, list all the vertices in order and repeat the first vertex at the end to close the polygon:
(1,2), (4,6), (7,2), (5,1), (1,2)
Now, compute the sum of xi*yi+1:
(16) + (42) + (71) + (52)
Calculate each term:
1*6 = 6
4*2 = 8
7*1 = 7
5*2 = 10
Sum these: 6 + 8 + 7 + 10 = 31
Next, compute the sum of yi*xi+1:
(24) + (67) + (25) + (11)
Calculate each term:
2*4 = 8
6*7 = 42
2*5 = 10
1*1 = 1
Sum these: 8 + 42 + 10 + 1 = 61
Now subtract the second sum from the first sum: 31 - 61 = -30
Take the absolute value: |-30| = 30
Divide by 2: 30 / 2 = 15
So the area should be 15. Wait, let me check my calculations again because sometimes it's easy to make a mistake in multiplication or addition.
First sum (xi*yi+1):
1*6 = 6
4*2 = 8
7*1 = 7
5*2 = 10
Adding them: 6 + 8 = 14, 14 +7=21, 21+10=31. That's correct.
Second sum (yi*xi+1):
2*4 =8
6*7=42
2*5=10
1*1=1
Adding them:8+42=50, 50+10=60, 60+1=61. Correct.
Difference:31-61=-30. Absolute value 30. Half of that is 15. So area is 15.
But wait, let me visualize again. If the quadrilateral is a bit irregular, maybe 15 is correct. Let me cross-verify using another method. Maybe divide the quadrilateral into two triangles and calculate their areas separately.
To do that, I can pick a diagonal, say from (1,2) to (7,2). Then the quadrilateral is split into two triangles: (1,2), (4,6), (7,2) and (1,2), (7,2), (5,1).
First triangle: points A(1,2), B(4,6), C(7,2)
Area of triangle ABC can be calculated using the formula: |(Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By))/2|
Plugging in:
Ax=1, Ay=2; Bx=4, By=6; Cx=7, Cy=2
Area = |1*(6 - 2) + 4*(2 - 2) + 7*(2 - 6)| / 2
Compute each term:
1*(4) =4
4*(0)=0
7*(-4)= -28
Sum:4 +0 -28= -24
Absolute value:24. Divide by 2:12. So area of first triangle is 12.
Second triangle: points A(1,2), C(7,2), D(5,1)
Using the same formula:
Ax=1, Ay=2; Cx=7, Cy=2; Dx=5, Dy=1
Area = |1*(2 - 1) +7*(1 - 2) +5*(2 - 2)| /2
Compute each term:
1*(1)=1
7*(-1)= -7
5*(0)=0
Sum:1 -7 +0= -6
Absolute value:6. Divide by 2:3. So area of second triangle is 3.
Total area:12 +3=15. Same result as before. So that confirms the area is indeed 15.
Alternatively, maybe using vectors or coordinate geometry another way. Let me try another approach for thoroughness.
Another method is to use the shoelace formula but with coordinates arranged in a table. Let me set up the coordinates in order and apply the formula step by step.
List of vertices:
(1,2)
(4,6)
(7,2)
(5,1)
(1,2) // repeated to close the polygon
Now, write down xi and yi:
i | xi | yi
1 | 1 | 2
2 | 4 | 6
3 | 7 | 2
4 | 5 | 1
5 | 1 | 2
Compute the sum of xi*yi+1:
(16) + (42) + (71) + (52) = 6 +8 +7 +10 =31
Compute the sum of yi*xi+1:
(24) + (67) + (25) + (11) =8 +42 +10 +1=61
Subtract the two sums:31 -61= -30
Take absolute value and divide by 2: |-30|/2=15. Same result.
Alternatively, maybe using vectors. The area of a quadrilateral can also be found by dividing it into two triangles, which we did, and summing their areas, which gave 15. So all methods converge to 15.
Wait, but just to make sure, maybe I can check if the quadrilateral is convex or concave. If it's concave, the shoelace formula still works, right? Because the shoelace formula works for any simple polygon, whether convex or concave, as long as the vertices are ordered correctly without crossing.
Let me check the order of the vertices. If I plot them:
(1,2) is bottom-left, (4,6) is top, (7,2) is bottom-right, (5,1) is slightly left and down from (7,2). Connecting these in order, the sides would be from (1,2) to (4,6): upward slope, then to (7,2): downward slope, then to (5,1): downward slope to the left, then back to (1,2). This should form a convex quadrilateral because all the interior angles are less than 180 degrees. Wait, is that true?
Wait, when moving from (7,2) to (5,1) to (1,2), the turn at (5,1) might create an interior angle. Let me see. The vectors from (7,2) to (5,1) is (-2,-1), and from (5,1) to (1,2) is (-4,1). The angle between these two vectors can be calculated, but maybe it's easier to just assume that the quadrilateral is simple (non-intersecting sides). Since the shoelace formula worked and gave a positive area, and the two triangle method also worked, it's safe to say the area is 15.
Alternatively, maybe use the vector cross product method. For a polygon, the area can be calculated by 1/2 the absolute value of the sum over edges of (xiyi+1 - xi+1yi), which is essentially the shoelace formula again.
Another way: using matrices or determinants. For each triangle, the area is half the determinant of the matrix formed by their coordinates. For example, for triangle ABC, the area is 1/2 |det([[1,2,1],[4,6,1],[7,2,1]])|. Let me compute that.
The determinant for three points (x1,y1), (x2,y2), (x3,y3) is:
x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)
Which is the same as the formula I used earlier. So for triangle ABC:
1*(6 - 2) +4*(2 - 2) +7*(2 -6) =14 +40 +7*(-4)=4 +0 -28= -24. Absolute value 24, half is 12. Same as before.
Similarly for the other triangle. So all methods confirm the area is 15.
I think that's thorough enough. All approaches lead to the same answer, so I'm confident the area is 15.
To calculate the area of the quadrilateral with vertices at (1,2), (4,6), (7,2), and (5,1), we use the shoelace formula:
List the coordinates in order, repeating the first vertex at the end:
[
(1,2),\ (4,6),\ (7,2),\ (5,1),\ (1,2)
]
Compute the sum of (x_i y_{i+1}):
[
1 \cdot 6 + 4 \cdot 2 + 7 \cdot 1 + 5 \cdot 2 = 6 + 8 + 7 + 10 = 31
]
Compute the sum of (y_i x_{i+1}):
[
2 \cdot 4 + 6 \cdot 7 + 2 \cdot 5 + 1 \cdot 1 = 8 + 42 + 10 + 1 = 61
]
Subtract the two sums and take the absolute value:
[
|31 - 61| = |-30| = 30
]
Divide by 2 to get the area:
[
\frac{30}{2} = 15
]
Verification using triangle decomposition:
Split the quadrilateral into two triangles:
Triangle 1 (1,2), (4,6), (7,2): Area = 12
Triangle 2 (1,2), (7,2), (5,1): Area = 3
Total Area: (12 + 3 = 15)
Final Answer:
The area of the quadrilateral is (\boxed{15}).
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发表于 2025-11-15 13:54:41