Originally posted by Evil lord SauronThe midpoint can only be connected to one vertex not multiple (and therefore there is no need to use the plural).
Given an Isosceles triangle, with a middle point on the base. If this point and the opposite vertices are connected to form a median, demonstrate that the median forms two right angles at the base and two congruent angles at the top vertices.
P.S I am willing to play any teens on this site as myself(14)
Originally posted by XanthosNZHe's 14. This is from his geometry class. Be nice.
The midpoint can only be connected to one vertex not multiple (and therefore there is no need to use the plural).
As for the question itself, it is obvious. You could approach by calling the repeated side length x and the repeated angle theta.
Therefore the projection of each similar side is the same ( x*cos theta) and therefore is directly above the midpoint.
Seriously, this question is stupid.
Originally posted by perihelioni'm pretty sure this is what you mean... http://mathworld.wolfram.com/Circle-CircleIntersection.html its in cartesian coords, but i'm sure it can be adapted... 5 min later:
I was just puzzling over a geometry problem today actually...
Whats the area of the intersection of any two circles, given their radii r1 and r2 (pick these so r1 =< r2), and the distance between the centers d.
Theres 3 cases:
1. r1 + r2 =< d, the circles don't overlap at all, intersection area = 0
2. d =< r2 - r1, the smaller circle is completel ...[text shortened]... cle
3. r2 - r1 < d < r2 + r1, partial overlap, thats the interesting case I can't figure out.
Originally posted by Evil lord SauronYour solution is obvious, but even more obvious is SSS congruency. Shared side (the median), two equal sides due to the midpoint, and two sides because of isosceles. Using Side-Side-Side, the two triangles formed must be congruent.
As for my first question, all you really had to do was to consider the two triangles that were formed. They have two sides congruent ( because of the midpoint), the angles and the other ajacent side beacause of the traingle is Isosceles. Thus meaning, using the I theory of congruent triangles (Two sides and the angle between them) the tria ...[text shortened]... lelogram, opposite angles are same distance from the diagonal that joins the other two vertices.
Originally posted by Evil lord SauronI think you might mean that the angle inside the polygon at the intersection of the two altitudes plus angle A is 180 deg.
Thanks, that was my homework!
Now: Consider an acute traingle ABC with its base being BC. The two altitudes of the congruent angles form a polygon with the top vertices. Demonstrate that the obtuse angle formed from the intersection of the two vertices plus angle A is equal to 180 degrees.
(chow on that!)
Originally posted by Evil lord SauronI'll solve this more a cylinder instead. One with dimensions 12cm h and 12cm d (6 cm r), and a density of 1 g/cm^3, because I think that is what you meant.
In an equilateral cilinder made of glass(density 2.5) 12cm tall, a semisphere is dug on the top of the cilinder. Calculate a) the total area of all of the surfaces b) the weight c) the total pressure that the cilinder has if the cavity is filled with water( density 1)
Originally posted by GastelNo no no. You should do this using triple integrals.
I'll solve this more a cylinder instead. One with dimensions 12cm h and 12cm d (6 cm r), and a density of 1 g/cm^3, because I think that is what you meant.
~ = pi
a) A = ~dh + ~r^2 + 2~r^2
b) V = ~r^2h - 4~r^3/6 = ~(36)(12) - 4(36) cm^3
M = (V)(2.5)
c) P = M / ~r^2 = M / ~(36) in g/cm^2 and you will have to convert as necessary.
Start a new thread called "Do my Grade 9 Math homework for me."
Originally posted by XanthosNZNice (And I reserve smilies for only special occasions)
No no no. You should do this using triple integrals.
(This is actually Maple code but it's easy enough to see what it does, it's the easiest way to write this in text)
That'll give you the mass of the glass. Piece of cake.