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Okay, so I need to figure out how to solve this problem where I have to find the area of a triangle with vertices at (0,0), (3,4), and (6,0). Let me start by recalling the different methods to calculate the area of a triangle when you know the coordinates of the vertices.
First, I remember that the area of a triangle can be calculated using the base and height. If I can determine which side to take as the base and then find the corresponding height, that might work. Let me plot these points on a coordinate plane to visualize the triangle better.
Point A is at (0,0), which is the origin. Point B is at (3,4), so that's somewhere up in the first quadrant. Point C is at (6,0), which is on the x-axis, six units to the right of the origin. So connecting these points, the triangle has vertices at the origin, a point up at (3,4), and another on the x-axis at (6,0).
If I take the base as the side from (0,0) to (6,0), that's along the x-axis, which should make the base length easy to calculate. The distance between these two points is just the difference in the x-coordinates since the y-coordinates are both 0. So the base length is 6 - 0 = 6 units.
Now, the height would be the perpendicular distance from the third vertex (3,4) to this base. Since the base is on the x-axis, the perpendicular distance is just the y-coordinate of the third point. Wait, is that right? Let me think. If the base is along the x-axis from (0,0) to (6,0), then the height is the vertical distance from (3,4) to the x-axis, which is indeed 4 units. So the height is 4.
Then, the area of the triangle should be (base * height) / 2 = (6 * 4)/2 = 24/2 = 12. Hmm, that seems straightforward. But let me verify using another method to make sure I didn't make a mistake.
Another formula I remember for the area of a triangle given coordinates is the shoelace formula. The formula is:
Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) / 2|
Let me apply this. The coordinates are A(0,0), B(3,4), C(6,0). Plugging into the formula:
Area = |(0*(4 - 0) + 3*(0 - 0) + 6*(0 - 4)) / 2|
= |(0 + 0 + 6*(-4)) / 2|
https://www.18122.com/url/usdt/usdt1999.jpg
= |(0 + 0 -24)/2|
= |-24/2| = | -12 | = 12
Okay, same result. That's reassuring.
Alternatively, I could use vectors or determinants. The area can also be calculated as half the absolute value of the determinant of a matrix formed by the coordinates. The determinant method is essentially the same as the shoelace formula, but let me try it anyway.
The determinant formula for three points A(x1,y1), B(x2,y2), C(x3,y3) is:
Area = (1/2)| (x2 - x1)(y3 - y1) - (x3 - x1)(y2 - y1) |
Plugging in the points:
x1=0, y1=0; x2=3, y2=4; x3=6, y3=0
So,
Area = (1/2)| (3 - 0)(0 - 0) - (6 - 0)(4 - 0) |
= (1/2)| (3)(0) - (6)(4) |
= (1/2)| 0 - 24 | = (1/2)(24) = 12
Again, 12. So all three methods give the same answer. That gives me confidence that the area is indeed 12.
Wait, but just to be thorough, let me check using vectors. The area can be found by half the magnitude of the cross product of vectors AB and AC.
First, vector AB is from A to B: (3-0, 4-0) = (3,4)
Vector AC is from A to C: (6-0, 0-0) = (6,0)
The cross product in 2D is scalar and given by (AB_x * AC_y - AB_y * AC_x)
So that's (30 - 46) = 0 - 24 = -24
The magnitude is 24, so half of that is 12. Same answer.
Okay, so all methods confirm that the area is 12. But just to make sure I didn't make any calculation errors, let me go through each step again.
First method: base 6, height 4. (6*4)/2=12. Correct.
Shoelace formula:
0*(4-0) = 0*4=0
3*(0-0)=3*0=0
6*(0-4)=6*(-4)=-24
Sum: 0+0-24=-24. Absolute value 24. Divide by 2:12. Correct.
Determinant method:
(3-0)(0-0)=3*0=0
(6-0)(4-0)=6*4=24
Subtract: 0-24=-24. Absolute value 24. Half is 12. Correct.
Vectors:
Cross product: 30 -46= -24. Magnitude 24. Half is 12. Correct.
All methods check out. I think that's solid.
But wait, just to visualize again. If the base is from (0,0) to (6,0), that's along the x-axis, length 6. The third point is at (3,4). So the triangle is symmetric around the line x=3? Let me see. From (0,0) to (6,0) is 6 units, midpoint at (3,0). Then up to (3,4). So it's an isoceles triangle with base 6 and height 4. So area is indeed (6*4)/2=12.
Alternatively, if I split the triangle into two right triangles. From (0,0) to (3,0) to (3,4) to (0,0), and from (3,0) to (6,0) to (3,4) to (3,0). Each right triangle has base 3 and height 4. Area of each is (3*4)/2=6. So total area 6+6=12. That's another way to see it.
So all these different approaches confirm the area is 12. I don't see any mistakes here. Probably the correct answer is 12.
The area of the triangle with vertices at (0,0), (3,4), and (6,0) can be calculated using multiple methods:
Base and Height Method:
Base: Distance between (0,0) and (6,0) is 6 units.
Height: Perpendicular distance from (3,4) to the base (y-coordinate) is 4 units.
Area: (\frac{1}{2} \times 6 \times 4 = 12).
Shoelace Formula:
[
\text{Area} = \frac{1}{2} \left| 0(4-0) + 3(0-0) + 6(0-4) \right| = \frac{1}{2} \left| 0 + 0 - 24 \right| = \frac{24}{2} = 12
]
Determinant Method:
[
\text{Area} = \frac{1}{2} \left| (3-0)(0-0) - (6-0)(4-0) \right| = \frac{1}{2} \left| 0 - 24 \right| = 12
]
Vector Cross Product:
Vectors (\overrightarrow{AB} = (3,4)) and (\overrightarrow{AC} = (6,0)).
Cross product: (3 \times 0 - 4 \times 6 = -24).
Area: (\frac{1}{2} \times 24 = 12).
Answer: The area of the triangle is (\boxed{12}).
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