Elementary Geometry Word Problems Here!

In the diagram to the right, point $P$ lies on the line ${\displaystyle y=\frac{3}{2}x}$ and is one vertex of square $PQRS$. Point $R$ has coordinates $(5,0)$.
a. Find the coordinates of point $P$.
b. If point $R$ lies on line $l$, and line $I$ divides quadrilateral $PQRS$ into two regions of equal area, find the equation for line $l$.

Convert from radians to degrees:
$-<!--\; -\; -->\frac{3\mathrm{\pi <!--\; \pi \; -->}}{8}$

Convert from radians to degrees:
$\frac{5\mathrm{\pi <!--\; \pi \; -->}}{12}$

Convert from radians to degrees:
$\frac{4\mathrm{\pi <!--\; \pi \; -->}}{3}$

Convert from degrees to radians:
$-<!--\; -\; -->{315}^{\circ <!--\; \circ \; -->}$

Let $X=\{(x,y)\in <!--\; \in \; -->{\mathbb{R}}^2\mid <!--\; \mid \; -->y=0\text{ or }x\ge <!--\; \ge \; -->0\}$ and $f:X\to <!--\; \to \; -->\mathbb{R}$ defined by $f(x,y)=x$ . I want to find $C\subset <!--\; \subset \; -->X$ closed such that $f(C)\subset <!--\; \subset \; -->\mathbb{R}$ isn't closed. How to prove that f is a closed map?

Is there a form of non-Euclidean geometry in which perpendicular lines never cross, or cross twice or something?

Convert from degrees to radians:
${210}^{\circ <!--\; \circ \; -->}$

ABCD is a parallelogram, P is any point on AC. Through P, MN is drawn parallel to BA cutting BC in M and AD in N. SR is drawn parallel to BC cutting BA in S and CD in R. Show that [ASN]+[AMR]=[ABD] (where $[.]$ denotes the area of the rectilinear figure).

My attempt : used base division method to find the area of ASN but I get extra variables which is tough...

I have a circle with a line that is drawn through it at two points. I have the coordinate for the center of the circle, the two coordinates where it intersects the circle and the radius as well. I want to put the two rectangles around the two pieces from the circle. My question is how can I get the coordinates of the rectangle?

When using the definition and properties of the inner product, we get the parallelogram law:
$$\begin{gathered} \Vert <!--\; \Vert \; -->x+y{\Vert <!--\; \Vert \; -->}^2=⟨<!--\; ⟨\; -->x+y,x+y⟩<!--\; ⟩\; -->=⟨<!--\; ⟨\; -->x,x⟩<!--\; ⟩\; -->+⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->+⟨<!--\; ⟨\; -->y,x⟩<!--\; ⟩\; -->+⟨<!--\; ⟨\; -->y,y⟩<!--\; ⟩\; -->, \\ \Vert <!--\; \Vert \; -->x-<!--\; -\; -->y{\Vert <!--\; \Vert \; -->}^2=⟨<!--\; ⟨\; -->x-<!--\; -\; -->y,x-<!--\; -\; -->y⟩<!--\; ⟩\; -->=⟨<!--\; ⟨\; -->x,x⟩<!--\; ⟩\; -->-<!--\; -\; -->⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->-<!--\; -\; -->⟨<!--\; ⟨\; -->y,x⟩<!--\; ⟩\; -->+⟨<!--\; ⟨\; -->y,y⟩<!--\; ⟩\; --> \end{gathered}$$
Adding these two expressions:
$\Vert <!--\; \Vert \; -->x+y{\Vert <!--\; \Vert \; -->}^2+\Vert <!--\; \Vert \; -->x-<!--\; -\; -->y{\Vert <!--\; \Vert \; -->}^2=2⟨<!--\; ⟨\; -->x,x⟩<!--\; ⟩\; -->+2⟨<!--\; ⟨\; -->y,y⟩<!--\; ⟩\; -->=2\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}^2+2\Vert <!--\; \Vert \; -->y{\Vert <!--\; \Vert \; -->}^2$
as required. But the above does not simplify to Pythagoras' theorem $\Vert <!--\; \Vert \; -->x+y{\Vert <!--\; \Vert \; -->}^2=\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}^2+\Vert <!--\; \Vert \; -->y{\Vert <!--\; \Vert \; -->}^2$if $x$ and $y$ are orthogonal.

How can we get Pythagoras from the parallelogram law?

Convert from degrees to radians:
$-<!--\; -\; -->{500}^{\circ <!--\; \circ \; -->}$

Convert from radians to degrees:
$-<!--\; -\; -->\frac{7\mathrm{\pi <!--\; \pi \; -->}}{2}$

Convert from degrees to radians:
${300}^{\circ <!--\; \circ \; -->}$

I came across this question while I was taking one of the pratice Mu Alpha Theta tests for my school and I wasn't sure how to solve it. It reads: In $\mathrm{\Delta <!--\; \Delta \; -->}USA$, $\mathrm{\angle <!--\; \angle \; -->}S$ is bisected by $\stackrel{\to <!--\; \to \; -->}{SY}$, with $Y$ on side $\stackrel{¯<!--\; ¯\; -->}{UA}$. If all sides have integer values, $\stackrel{¯<!--\; ¯\; -->}{US}=18$ and $\stackrel{¯<!--\; ¯\; -->}{UY}=12$; find the smallest possible perimeter of $\mathrm{\Delta <!--\; \Delta \; -->}USA$.

The answer key with the solutions skipped to the part with the triangle inequalities, so that is why I am not sure what to do. I think there is some ratio when it comes to splitting a triangle with an angle bisector but I am not sure what it is. My question is how would I solve this question, and for future reference what is that ratio for an angle bisector in a triangle.

If E is the midpoint of the segment BC, $L_1\parallel <!--\; \parallel \; -->L_2$ , $AD=4cm,DE=5cm,EA=7cm$ , find the area of the quadrilateral ABCD.

Given two noncoplanar lines p and q, and a point A, does there always exist a line that passes through p, q and A?

Given a family of parallelograms such that the corresponding edges of all members are parallel and any two members of this family intersect. Can we conclude that any three members of this family intersect?
Intuitively I thought the answer should be positive. But I am having difficulty to write a rigorous proof (or find a counter example). Can anyone provide help me with this? Thanks

$\frac{1}{a+b}+\frac{1}{a+c}=\frac{3}{a+b+c}$ in a triangle
Find the angle $\mathrm{\alpha <!--\; \alpha \; -->}$ of a triangle with sides a, b and c for which the equality

$\begin{array}{|cccccccccc|}\hline \text{Time}& \text{9am}& \text{10:30 am}& \text{Noon}& \text{1:30 pm}& \text{3 pm}& \text{4:30 pm}& \text{6 pm}& \text{7:30 pm}& \text{9 pm}\\ t& 0& 1.5& 3& 4.5& 6& 7.5& 9& 10.5& 12\\ P\left(t\right)& 200& 728& 1193& 1329& 1583& 1291& 804& 256& 0\\ \hline\end{array}$
Use a midpoint Riemann sum with four intervals of equal size to estimate the total number of people seeking care during the 12-hour period.