![]() ![]() For part (c), it is not easy to see that this triangle is isosceles without the Pythagorean Theorem. ![]() So in these two cases there are alternative explanations and the teacher may wish to emphasize this. Also, in parts (a) and (b), a line of reflective symmetry is not hard to identify. Of the legs are obtained by moving along the grid lines, from one vertex, by the same number of squares vertically and horizontally. For the triangles given in parts (a) and (b) two This method is not, however, always the most efficient. One way to do this is to calculate side lengths using the Pythagorean Theorem. You should be able to continue from here, finding the value of $b$ that maximizes the area (or the square of the area) and finding an expression for $\cos\alpha$ in terms of $b$ then in terms of just constants.This task looks at some triangles in the coordinate plane and how to reason that these triangles are isosceles. The median, $m$, is a constant, so we finally have a formula for the area in terms of only one variable, the side $b$. ![]() The base is obviously $b$, and side $c$ is the hypotenuse of a right triangle with sides $\frac b2$ and $h'$, where $h'$ is the altitude on side $\overline \\ Instead, use the base and height from my second diagram. Your area formula $A=\frac 12ch$ is not useful here, since finding $h$ directly is difficult for you. We can easily solve this for $c$ and we get That does also work, though that does not directly include what your final answer will be, the cosine of the apex angle. In your comments you say you want your two variables that define the isosceles triangle to be $c$, the length of the two equal sides, and $b$, the length of the other side. Then it is easy to find expressions for the vertices coordinates, midpoint of one side, median length, altitude, and triangle area. We can let $c$ be the length of the equal sides and $\alpha$ be the measure of the apex angle. I have placed the isosceles triangle on a Cartesian coordinate system, with the apex vertex at the origin and one of the equal sides on the positive $x$-axis. Here is one way that more directly uses the cosine of the vertex angle. I left that explanation very general, because I can think of many ways to define an isosceles triangle in terms of two variables. Substitute that into your expression for the cosine of the vertex angle, and you are done. Substitute that into the expression for area, and you now have an expression for area in terms of only one variable. Then set the expression for the median length equal to a constant and solve for one of your original objects in terms of the other. Then find expressions for the other relevant properties, namely the length of the median, the area, and the cosine of the vertex angle. Choose two measurable objects that determine the triangle. The shape of an isosceles triangle, up to congruence, has two degrees of freedom. ![]() Since you are having trouble findind the area in terms of median length and cosine of vertex angle, I suggest you do it differently. ![]()
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