To find a triangle that has the same area as Quincy's triangle with vertices A (2,1), B (2,7), and C (7,5), we first need to calculate the area of Quincy's triangle using the formula for the area of a triangle given its vertices:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \]
Substituting the coordinates of points A, B, and C:
- \( (x_1, y_1) = (2, 1) \)
- \( (x_2, y_2) = (2, 7) \)
- \( (x_3, y_3) = (7, 5) \)
Now we can plug these values into the formula:
\[ \text{Area} = \frac{1}{2} \left| 2(7-5) + 2(5-1) + 7(1-7) \right| \]
Calculating each term:
- \( 2(7 - 5) = 2 \times 2 = 4 \)
- \( 2(5 - 1) = 2 \times 4 = 8 \)
- \( 7(1 - 7) = 7 \times -6 = -42 \)
Now combine these results:
\[ \text{Area} = \frac{1}{2} \left| 4 + 8 - 42 \right| = \frac{1}{2} \left| -30 \right| = \frac{1}{2} \times 30 = 15 \]
So the area of Quincy's triangle is 15 square units.
To find another triangle with the same area, you would use the same area formula and ensure that the vertices you choose also yield an area of 15 when plugged into the formula.
For example, vertices D (0,0), E (0,10), and F (5,0) yield:
\[ \text{Area} = \frac{1}{2} \left| 0(10-0) + 0(0-0) + 5(0-10) \right| = \frac{1}{2} \left| 0 + 0 - 50 \right| = \frac{1}{2} \times 50 = 25 \] (This would be incorrect as it yields 25).
Find different combinations until you achieve an area of 15.
If you have specific options for triangles to compare with Quincy's triangle, please provide them, and I can assist in determining which, if any, have the same area.
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