CPhill

It's called Graphing Calculator 3D

(1/4) of the perimeter  = a side length

So

148 * (1/4)  =  37

The side length forms the hypotenuse of a right triangle

And (1/2) of the known diagonal length = 12    forms one leg of this triangle

So...using the Pythagorean Theorem, the other leg = 

sqr ( 37^2 - 12^2 )   = sqrt ( 1369 - 144) = sqrt ( 1225) = 35

And twice this is the length of the other diagonal =  70

Ah....I made a mistake....let me do a corrected post...thanks for pointing out my error !!

They are trying to trick you, mathtoo....it's a "disguised" conic

For instance...look at this graph : https://www.desmos.com/calculator/kdzjzqgkr4

It appears that for the function     y / x^2 = 3  the point (0,0) couldn't be on the graph

But....rearranging the equation as   y = 3x^2    shows that it is a parabola with a vertex at (0,0)

Well   the "width" or "length" is really dependent upon your point  of view

Whatever....the floor dimensions are 20 x 18

And the height = 8 ft

And it is a rectangular prism....

Thanks, EP

That looks pretty circular to me.....!!!!!

Another tricky one, RP

A perpendicular line  to   x  =   a     will be     y =  b

So..... a line perpendicular to x = 8   will   be   y = the y coordinate of the given point

So....we have

y = -2

See the graph here to get a feel for this :  https://www.desmos.com/calculator/2av3sxef7g

Multiply both sides by x^2y^2    and we get

y^2 + x^2  =    x + 4   rearrange as

x^2 - 1x + y^2  = 4      complete the square on x

x^2 - 1x + 1/4   + y^2 = 4 + 1/4     factor the first three terms

(x - 1/2)^2 + (y - 0)^2 = 17/4

This is a circle centered at     (1/2, 0 )  and a radius of  sqrt (17) / 2

We first want to make the denominator a real number

We can do this by multiplying the the numerator/denominator by the conjugate of 2i - 1

This is 2i + 1

Next...we see that to  make the numerator a real number....we need to have  (16 + p) i = 

0i

I see that I can do this if   p = - 16

Does that make sense ???

OK.....good job   !!!