Expert Help with High School Geometry

Hyperbolic Angle Measure of Polygons
Is there a way to determine the angle measure of a regular polygon in hyperbolic space? I know that this depends on the length of the sides. A an example, I know that for an equilateral triangle with side length a and angle A, then $\sec <!--\; \; -->A=1+\frac{2e^a}{e^{2a}+1}$.
Is there a similar formula for higher regular polygons?

Finding volume of solid revolution about x-axis
I'm asked to find the volume of a container of height 13.5 The container is made by rotating $y=1.5x^2$ where $0\le <!--\; \le \; -->x\le <!--\; \le \; -->3$ about the axis $x=-<!--\; -\; -->0.5$ (the bottom is flat).
I tried finding the right volume a number of ways, but I believe my most solid attempt so far is this:
I invert the function to get $y=\sqrt{\frac{2}{3}x}$.
From this I tried to calculate $\mathrm{\pi <!--\; \pi \; -->}\ast <!--\; \ast \; -->{\int <!--\; \int \; -->}_0^3\sqrt{\frac{2}{3}x}dx$ which gave me $3\mathrm{\pi <!--\; \pi \; -->}$ which I thought was correct, but it wasnt. Plotting this, I realized the height becomes 3 when I integrate the inverted function from 0 to 3, so I thought I should maybe integrate from 0 to 13.5. In that case I get $\frac{243\mathrm{\pi <!--\; \pi \; -->}}{4}$ which is also incorrect. I'm sure I've misinterpreted what it means to rotate the function about the axis $x=-<!--\; -\; -->0.5$

Using trigonometric ratios to express area of regular polygons
I am very confused. My book just asked me to use trigonometric ratios to express the area of a regular polygon with 9 sides and lengths of 8.
So far I have learned how to use the sin cos and tan in right triangles and have no idea how this applies to all polygons. Can someone please explain this to me

Is it always possible to draw an ellipse given any two points and the center?
I have 2 points and I want to draw an ellipse that passes through those points.
I also have the center of the ellipse.
My question is: is it always possible to draw such an ellipse?
$\frac{(x-<!--\; -\; -->c_x{)}^2}{a^2}+\frac{(y-<!--\; -\; -->c_y{)}^2}{b^2}=1$
When trying to automatically render ellipses for information visualization, I tried calculating the radiuses a and b by solving the two-equation system. However, sometimes $a^2$ or $b^2$ would be negative and, therefore, a or b would return me NaN.
Edit: A test case where this is failing is with:
$P_1=(610,320)$
$P_2=(596,887)$
$C=(289,648)$

Probability that a geometric random variable is even
Toss an unfair coin until we get HEAD. Suppose the total number of tosses is a random variable X, and $Pr(HEAD)=p$. What is the probability that X is even? Denote this event as A.

A company has five employees on its health insurance plan. Each year, each employee independently has an .80 probability of no hospital admissions. If an employee requires one or more hospital admissions, the number of admissions is modeled by a geometric distribution with a mean of 1.50. The numbers of hospital admissions of different employees are mutually independent. Each hospital admission costs 20,000. Calculate the probability that the company’s total hospital costs in a year are less than 50,000.

How do you find the perimeter of a triangle whose coordinates are (10,2) and (7,-3)?

Find the volume of a solid which is bounded by the paraboloid $4z=x^2+y^2$, the cone $z^2=x^2+y^2$ and the cylinder $x^2+y^2=2x$
I approached this problem by trying to find the volume bounded by the paraboloid and the cylinder and then subtracting it from the volume bounded by the cone and the cylinder. But I am getting the wrong answer. I converted the all the bounds into cylindrical co-ordinates.
For finding the volume bounded by the cone and the cylinder,
Bounds of integration: $z=0$ to $z=r,r=0$ to $r=2\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}),\mathrm{\theta <!--\; \theta \; -->}=0$ to $\mathrm{\theta <!--\; \theta \; -->}=2\mathrm{\pi <!--\; \pi \; -->}$.
For finding the volume bounded by the paraboloid and the cylinder,
Bounds of integration: $z=0$ to $z=\frac{r^2}{4},r=0$ to $r=1,\mathrm{\theta <!--\; \theta \; -->}=0$ to $\mathrm{\theta <!--\; \theta \; -->}=2\mathrm{\pi <!--\; \pi \; -->}$.

Constructible polygons
I know certain polygons can be constructed while others cannot. Here is Gauss' Theorem on Constructions:
$\cos <!--\; \; -->(2\pi /n)$ is constructible iff $n=2^r{p}_1{p}_2···p_k$, where each $p_i$ is a Fermat prime.
Can this is be used to determine the constructibility of a regular $p^2$ polygon? If so how and what would be the $\cos <!--\; \; -->(2\mathrm{\pi <!--\; \pi \; -->}/n)$ here?

Given a probability density, how can I sample from the induced distribution?
Let f be an integratable function such that $\int <!--\; \int \; -->f(x)dx=1$. If we want to take random samples from this, using whatever programming language one pleases, we should compute $F(t)={\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{t}f(x)dx$, invert this function and feed it numbers drawn from a uniform distribution on [0,1].
However I now want to sample coordinates $(u,v)\in <!--\; \in \; -->{\mathbb{R}}^2\mathrm{\setminus <!--\; \setminus \; -->}\{0\}$ such that the probability of (u,v) lying in a set E is given by
${\int <!--\; \int \; -->}_E(u^2+v^2{)}^{-<!--\; -\; -->\frac{3}{2}}dudv$
I don't see how I can now try to find
$F(s,t)={\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^s{\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^t(u^2+v^2{)}^{-<!--\; -\; -->\frac{3}{2}}dudv$
as I get into troubles close to the zero. What other way is there to obtain a sample following this distribution?
A note for the context: If we consider all lines in the Euclidean plane with Cartesian coordinates, not passing through the origin, they can be represented via $ux+vx=1$. If we impose the condition that the probability densitiy should be invariant under Euclidean transformations, then we arrive at the above distribution.

What is the distance between (1,5) and (2,12)?

How can I find the volume of a solid of revolution
Find the volume of the solid obtained by rotating the region bounded by $y=x^2$ and $x=y^2$. Rotating about $y=1$.
I got an intercept of those functions which was (1,1). I tried to use washer method then I got
$\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_0^1[(x{)}^2-<!--\; -\; -->(\sqrt{x}{)}^2]dx$ and I took integral of the functions but my volume was not right number. I think my way to solve was not right. Could you post correct way to find the volume?

The number of Bernoulli trials required to produce exactly 1 success and at least 1 failure.
Let X be the number of Bernoulli (p) trials required to produce exactly 1 success and at least 1 failure. Find the distribution of X.
How can I answer this question if I don't know how many trials have taken place?
For context, the section this comes from was about Poisson distribution.
$(\genfrac{}{}{0ex}{}{n}{k})p^k(1-<!--\; -\; -->p{)}^{(n-<!--\; -\; -->k)}\to <!--\; \to \; -->\frac{e^{-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}}{\mathrm{\mu <!--\; \mu \; -->}}^k}{k!}$
$\text{as }n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}\text{ and }p\to <!--\; \to \; -->0\text{ with }np=\mathrm{\mu <!--\; \mu \; -->}$
However I read in a different book that you can use the Geometric probability model for Bernoulli trials: Geom(p)
$P(X=x)=q^{X-<!--\; -\; -->1}P$
where p is prob. of success, X is number of trials, and q is prob. of failure

Lines $CF,GE$ are parallel to diagonal $DB$.

Are the cyclic quadrilaterals $CBGDC,FBGDF$ (with same sides and angles) congruent? If not, how are they be described?

Distance between (2,-14) and (-9,5)?

Finding a volume integral in an ellipsoid
I am trying to find the volume integral of $\mathrm{\rho <!--\; \rho \; -->}={\mathrm{\rho <!--\; \rho \; -->}}_0\left(\frac{R^2-<!--\; -\; -->r^2}{R^2}\right)$ inside an ellipsoid given by $\frac{x^2}{(3R{)}^2}+\frac{y^2}{(4R{)}^2}+\frac{z^2}{(5R{)}^2}=1$
I've tried using jacobian to move from an ellipsoid to an unit ball by these mapping relationships $x=3Ru$, $y=4Rv$, $z=5Rw$.
But the resulting integral is still heavy
$\int <!--\; \int \; -->{\mathrm{\rho <!--\; \rho \; -->}}_0(1-<!--\; -\; -->(9u^2+16w^2+25w^2))60dudvdw$
Does anyone have any insight to a more elegant way.

Show that
$T(X)=(R,V)=(X_{(n)}-<!--\; -\; -->X_{(1)},\frac{X_{(n)}+X_{(1)}}{2})$
is a minimal sufficient statistic for $\mathrm{\theta <!--\; \theta \; -->}$.

Why is probability on Lie groups nice?
The question may sound weird. I am a probabilist who has heard of works connecting probability with (compact) Lie groups. What is the motivation? Is it generalisation just for the sake of generalisation?
Applebaum’s book is the classic but I am firstly interested to know why doing probability on Lie groups is a worthwhile effort. I know little lie theory from a differential geometric point of view. Is it enough for reading about connections between Lie groups and probability?

Finding volume of a shape using double integral
I'm trying to find the volume of a given shape:
$V=\{\begin{array}{l}\sqrt{x}+\sqrt{y}+\sqrt{z}\le <!--\; \le \; -->1\\ x\ge <!--\; \ge \; -->0,y\ge <!--\; \ge \; -->0,z\ge <!--\; \ge \; -->0\end{array}$
using double integral. Unfortunately I don't know how to start, namely:
$z=(1-<!--\; -\; -->\sqrt{y}-<!--\; -\; -->\sqrt{x}{)}^2$
and now what should I do? Wolfram can't even plot this function, I'm unable to imagine how it looks like...
Would it be simpler with a triple integral?

Let ABC be an isoscles triangle with $AB=AC$.The bisector of $\mathrm{\angle <!--\; \angle \; -->}B$ meets AC at D.Given that $BC=BD+AD$,we need to figure out $\mathrm{\angle <!--\; \angle \; -->}A$.