Triangle $A B C$ has its vertices at $A(2,0), B(6,0)$ and $C(6,3)$. The area of the triangle, in square units, is
(A) 3
(B) 4
(C) 6
(D) 7
(E) 12
First, we sketch the triangle.

Since point $C$ is directly above point $B$ (that is, angle $A B C$ is a right angle), then we can look at triangle $A B C$ as having base $A B$ (of length 4) and height $B C$ (of length 3).

Thus, the area of the triangle is $A=\frac{1}{2} b h=\frac{1}{2}(4)(3)=6$ square units.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-366.jpg?height=382&width=507&top_left_y=1668&top_left_x=1297)

ANswer: (C)