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bruh

Okay now im even more confused -_-

Do one shape at a time

You should keep it as ONE shape because a quadrilateral is very simple.

To find AREA      I'd go two triangles .... BOTH with base 7     one with h = 2   one with h = 2

There is a formula for all triangles.

https://www.mathwarehouse.com/geometry/triangles/area/herons-formula-triangle-area.php

Careful....    Pythagorean Theorem only works for RIGT triangles !

So which way is easier and faster lol should I split it in half into 2 triangles?

Nope

The area of a triangle is the base times height times one half. Use pythagoras theorem to find the hypotunse of a triangle and add that to the base and height of the triangle to find the perimeter. The area of a square is one of the sides squared (ex: s^2), hence it is why it is called square. The perimeter of a square is one of the sides times 4.

Try these math sites:

https://www.symbolab.com/ (there is some chemistry there)

https://www.mathway.com/Algebra

https://www.desmos.com/calculator

...or break it in to two triangles     area of triangle = 1/2 base x height

So would it be AB, BC, AD, and DC or am i doing this wrong

Exactly!!! Kudos !

holy trapezoid batman!

Almost.

Break it into the obvious rectangle, and then two triangles on each side result.

Use the fabled distance formula we keep talking about to get the lengths of the various sides you need 

to compute the areas.

Perimeter is 12 and the area is 6.

You'll have to work through much of the algebra here yourself.

$\text{you have two ellipses of the same size, one centered at } (k,0), \text{ the other at }(0,0)$

$\text{Looking at the geometry it must be that }B \text{ is the leftmost x intercept of }\dfrac{(x-k)^2}{9}+y^2=1\\ \text{and }D \text{ is the rightmost x intercept of }\dfrac{x^2}{9}+y^2=1$

$B = (k-3,0)\\ D=(3,0)$

$\text{The }x \text{ coordinates of the intersections are given by}\\ (x-k)^2=x^2\\ -2kx+k^2 = 0\\ x = \dfrac{k}{2}\\ y = \pm \sqrt{1-\dfrac{k^2}{36}}$

$\text{the quadrilateral formed by joining these points is a rhombus, so we solve for }k\\ \text{that makes the corner angle 90 degrees}\\AD = A-D = \left(\dfrac k 2,\sqrt{1-\dfrac{k^2}{36}}\right)-(3,0) = \left(\dfrac k 2 - 3,\sqrt{1-\dfrac{k^2}{36}}\right)\\ CD=C-D = \left(\dfrac k 2,-\sqrt{1-\dfrac{k^2}{36}}\right)-(3,0) = \left(\dfrac k 2 - 3,-\sqrt{1-\dfrac{k^2}{36}}\right)\\ \text{the angle between }AD \text{ and }CD \text{ being 90 degrees} \Rightarrow AD \cdot CD = 0$

$AD \cdot CD = \dfrac{5 k^2}{18}-3 k+8=0\\$

$k = \dfrac{24}{5} \text{ or the size zero square when }k=6$

P=12

A=6

it becomes positive??

You are getting there....but what happens when you square a NEGATIVE number   (1-4)^2 = (-3)^2       Hmmmmm....did you forget?

The BASE is 4   obviously ....right?

Distance between two points ( I KNOW you have seen this in recent days)

d^2 = (x1-x2)^2 + (y1-y2)^2     (this is just a Pythag Theorem re-written)

So using points    4,1    and  2,4           d^2 = (4-2)^2 + (1-4)^2   = 13     right?  

      so d = sqrt 13    the length of ONE leg....there are TWO of these legs 

               so  2 sqrt 13    PLUS the base (4) = perimeter = 11.21 units           

  Cool now?   You can do this.......

I really hope this is right. I worked it out and got 44.8266 rounded to 44.83. Did I finally get the answer? I am so very sorry for bothering you!

Im so confused!! how did u get that for the perimeter?

How many moles of ammonia are there?    20

2 moles of ammonia come from ONE mole of N2       How many moles of N2 is that?   What did you get?

THEN use    PV = n RT to find   V for the nitrogen....  Let me know waht you find....G'Luck !

$p_1=(x_1,y_1)\\p_2=(x_2,y_2)\\ d(p_1,p_2) = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The perimeter is not the HEIGHT !         Height for this triangle is   3

Perimeter = 4 (the base) + TWO times sqrt ((4-2)^2 + (1-4)^2) =  4   +   2 * sqrt 13= 11.21 units