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.

Good luck!

P.S I am willing to play any teens on this site as myself(14)

Originally posted by Evil lord Sauron 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.

Good luck!

P.S I am willing to play any teens on this site as myself(14)

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.

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 completely inside the larger circle, intersection area = area of the smaller circle
3. r2 - r1 < d < r2 + r1, partial overlap, thats the interesting case I can't figure out.

Originally posted by perihelion 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.

i'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:
ah i see now, you don't need the formula for your circle at all, you simply calculate the area of each circular segment individually then sum (ie. work out the area bound by the arc of circle1 and a line connecting the intercept points, then ditto for circle2. http://mathworld.wolfram.com/CircularSegment.html has a good diagram of the area of a segment

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 triangles are congruent, an in particular the angle in question.

Try this: Demonstrate that in a parallelogram, opposite angles are same distance from the diagonal that joins the other two vertices.

Originally posted by Evil lord Sauron 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.

Your 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.

A property of parallelograms is that the diagonals are bisectors of each other. Now the tricky part is that you must realize that opposite triangles formed by the two diagonals are congruent (SSS - easy as pi). Because they are congruent, the altitudes are the same and now we have formed two new triangles, which mean the vertices are equidistant from the diagonal. Also, you can prove that the altitudes intersect the diagonal (for the other two vertices) equidistant from the intersection of the main diagonals. Easy.

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.

Originally posted by Evil lord Sauron 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!)

I think you might mean that the angle inside the polygon at the intersection of the two altitudes plus angle A is 180 deg.

If that is the case, it can be solved by inspection. The altitudes intersect the opposite side of the triangle at 90 deg (by definition) so the sum of their angles in the polygon is 180 deg. Since a 4 sided polygon (regular polygon) always has 360 deg, then the remaining angles (A plus the intersection point) must sum to 180 deg.

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 Evil lord Sauron 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)

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 Gastel 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."

No no no. You should do this using triple integrals.

Something like:

int(int(int(2.5,r=0..6),theta=0..2*pi),z=0..12)-int(int(int(2.5,ro=0..6),phi=0..pi/2),theta=0..2*pi);
(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.

Originally posted by XanthosNZ No no no. You should do this using triple integrals.

Something like:

int(int(int(2.5,r=0..6),theta=0..2*pi),z=0..12)-int(int(int(2.5,ro=0..6),phi=0..pi/2),theta=0..2*pi);
(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.

Nice ðŸ˜‰ (And I reserve smilies for only special occasions)