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Question 1210148: Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of triangle ABC?
A is at (0,0), B is at (1,0), and C is at (1,1).
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let the coordinates of the vertices be $A=(0,0)$, $B=(1,0)$, and $C=(1,1)$.
We need to find the area of triangle $ABC$.
The coordinates of the vertices are given as $A=(0,0)$, $B=(1,0)$, and $C=(1,1)$.
We can use the formula for the area of a triangle given the coordinates of its vertices:
Area = $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$
where $(x_1, y_1) = (0, 0)$, $(x_2, y_2) = (1, 0)$, and $(x_3, y_3) = (1, 1)$.
Area = $\frac{1}{2} |0(0-1) + 1(1-0) + 1(0-0)|$
Area = $\frac{1}{2} |0 + 1 + 0|$
Area = $\frac{1}{2} |1|$
Area = $\frac{1}{2}$
Alternatively, we can notice that the triangle $ABC$ is a right triangle with base $AB$ and height $BC$.
The length of $AB$ is $1-0 = 1$.
The length of $BC$ is $1-0 = 1$.
The area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$.
Area = $\frac{1}{2} \times 1 \times 1 = \frac{1}{2}$.
Final Answer: The final answer is $\boxed{\frac{1}{2}}$
Answer by ikleyn(52798)  (Show Source):
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