politexnik
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If the triangle is equilateral, then the base is 10 (and the median is not [imath]3\sqrt{17}[/imath]).In an equilateral triangle ABC, AB=BC=10, and the median AM is 3√17. Find the length of the altitude to the base of the triangle. Find the length of the base of the triangle. View attachment 36061
(Note: I am assuming that "isosceles" is meant, instead of "equilateral".)In an equilateral triangle ABC, AB=BC=10, and the median AM is 3√17. Find the length of the altitude to the base of the triangle. Find the length of the base of the triangle. View attachment 36061
If the triangle is equilateral, then the base is 10 (and the median is not [imath]3\sqrt{17}[/imath]).
Assuming you meant isosceles, I can solve it using some trig; what method have you tried?
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Yes, you're right, I mean "isosceles", thanks(Note: I am assuming that "isosceles" is meant, instead of "equilateral".)
What results have you learned recently in class? What new rules were given to you in the section in your textbook that generated this question?
For instance, (this page) has a bunch of rules about medians. Using one of the last "Properties of Medians" that they list, you should be able to create an equation that relates the lengths of the sides of the triangle to the lengths of the three medians (one of which is the height, |BH| = h). Using the fact that AHB is a right triangle, you can create an equation which relates the height, h, and the length of the base, |AC| = b.
When you reply, please include a clear listing of your thoughts and efforts so far. Thank you!
Great! Now, with the various hints and suggestions you've been given, how far have you gotten?Yes, you're right, I mean "isosceles", thanks