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# Equal Area Triangles on the Same Base I

Alignments to Content Standards: G-GPE.B.5

1. On graph paper, sketch a line segment with end points $A=(0,2)$ and $B=(0,6)$. Plot all points $C=(x,y)$ such that the triangle ABC has an area of 6 square units.
2. What two basic geometric theorems are needed to solve this problem?

## IM Commentary

This task is an relatively easy application of the formula for the area of a triangle and use of parallel lines. A more sophisticated version of the same set of ideas is given in G-GPE Equal Area Triangles on the Same Base II Despite its simplicity, students may be surprised by the answer, as their intuition might tell them that only points $C$ reasonably close to $A$ and $B$ could provide a triangle with area equal to 6.

Depending on the sophistication of the students, the activity could be done on graph paper. The problem also could be use to addres Standard for Mathematical Practice 6 by focusing on the careful use of definitions and terminology. In particular, the terms "the base" and "the height" are often used in regard to triangle, which hides the fact that any of the three sides can be used as a base. This is crucial when applying the triangle area formula in this task. Similarly, student's aforementioned lack of intuition as to the locus of such points possibly stems from an incorrect intuition that altitudes of triangles are necessarily interior to the triangle.

1. Viewing $AB$ as the base $b$ of the triangle, we recognize that by the formula for the area of a triangle, the triangle in question would have area given by $6=\frac{bh}{2}$. Since the length of $AB$ is 4, the height $h$ of the triangle (again, relative to the base $AB$) must thus be 3. A triangle with base $AB$ has height 3 exactly the third vertex, $C$, is on one of those lines that are distance 3 from the $y$-axis. In coordinates, these are the lines with equations $x=3$ or $x=-3$.
2. We need to know a) the formula for the area of a triangle and b) realize that any point a line parallel to a line through $AB$ is the same distance from that line.