Isosceles triangle |
We have to admit that we really aren't certain what the OQ wanted to know, but Sarah decided that the questioner wanted to know how to determine the length of the two equal sides of an isosceles triangle of known base and height. Or at least we think that's what she figured. Celebi started out with a sensible statement...
"By drawing a straight line down the center of an isosceles triangle, it can be divided into two congruent right triangles, and the Pythagorean theorem can easily be used to solve for the length of an unknown side"...
...and moved on. Unfortunately, things deteriorated as the post dragged on. Sarah's process for solving this problem is long and torturous, involving drawing the triangle to scale and in a specific orientation. She then instructs her reader to "Draw a straight line down the middle of the triangle from the vertex to the base."Since the question was probably submitted by a geometry student, it might have been easier to say "construct a perpendicular bisector," but Sarah's content editor would probably have called that "jargon" since no one takes geometry (or any math, for that matter) in J-school. In any case, Celebi wants her reader to |
- Write the values of the lengths of the known sides of the triangle next to the sides they match. These values may come from a specific math problem or from measurements for a certain project. Write "3 in." next to the line drawn in Step 2 and "4 in." on either side of this line at the base of the triangle.
- Determine which side is of unknown length and use the Pythagorean theorem to solve for it using a calculator. The unknown side is the hypotenuse of each of the two triangles.
- Label the hypotenuse "C" and either of the legs of the triangle "A" and the other one "B."
Buried in the "Tips" section of the post is the main reason our researcher flagged Celebi's post as a Dumbass of the Day candidate. That's where she defines the hypotenuse as "the line that connects the base and height of a right triangle." Dumbass, indeed...
copyright © 2016-2022 scmrak
MM - GEOMETRY
No comments:
Post a Comment