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If we assume these rules

-A rectangle is inside the circle

-All corners of the rectangle touch the edge of the circle

If we say the length is l, the width is w and the diameter is d. What would the equation for d be?

And what would be the equation for a cuboid in a sphere?

It doesn't have to be a square

Sorry- I was mistaken in thinking that an inscribed shape had to be proper- I'll do better than to look at google images next time.

We know that the length from one corner to the opposite must be the diameter. So $$sqrt{l^2+w^2}=d$$ For a cuboid $$sqrt{l^2+w^2+h^2}=d$$

Here is the solution for an inscribed square instead of a rectangle, which may help you find your solution. So, let $s=l=w$. Then, we know the diagonal of the square is equivalent to the diameter of the circle. The diagonal forms a right triangle with the sides; therefore, we have
$$ s^2 +s^2=d_{square}^2$$
where you can solve for $d_{square}$ give by
$$d_{square}=sqrt{2s^2}$$

For a cube, you know the diagonal is equal to the diameter of the sphere. However, the diagonal of one face and the edge of one face form a right triangle such that the hypotenuse is the diagonal of the cube. Therefore, we have
$$d_{square}^2+s^2=d_{cube}^2$$
where you can solve for $d_{cube}$ give by
$$d_{cube}=sqrt{d_{square}^2+s^2}$$

"-All corners of the rectangle touch the edge of the circle"

That is called the rectangle being inscribed in a circle.

"IT IS NOT INSCRIBED!!!"

Yes, it is.

......

Pick the center of the circle and draw four line segments (radii) to the four vertices of the rectangle. (We don't know yet that these radii form two intersecting lines; we want to prove that.) This cuts the rectangle into four triangles. The triangles are all isosceles because all radii of the circle are congruent. The pairs of opposite triangles are congruent as they have three congruent sides. Draw the twelve angles of the four triangles. The base angles of the pairs of adjacent triangles are supplementary because they form a right angle together. Look at the four angles and the center of the circle. Figure each one is $180$ minus the two base angles. Then each two adjacent angles are complimentary. (It's just algebra: $a = 180 -2c$ and $b = 180 -2d$ and $c + d =90$ so $a+b = 360 - 2(c + d) = 360 -2*90 = 180$.) Thus the radii you formed were actually two intersecting lines. So the center of the circle is co-linear with the diagonals of the the rectangle.

So the diagonals of the rectangle are the same as the diameter of the circle.

If the sides of the rectangle are $l,w$ then by the pythagorean theorem the diagonal of the rectangle (which is the diameter of the circle) is $sqrt{l^2 + w^2}$.

To extend to three dimensions a similar argument can prove the diagonal of the rectanguloid is the same as the diameter of the sphere. Such an argument is both too similar and too tedious to go into. (Do the same thing but there are twelve triangles and you have to prove that groups of four of them are co-planar and... well, let's just say "symmetry" and say that if the diagonal is not the diameter we have an asymmetric anomaly; that's actually acceptable argument--- it's just not very convincing if we don't know what we are doing.)

The formula for the diagonal of a rectanguloid can be extended so that if the sides of the rectanguloid or $l,w,h$ the diagonal is $sqrt{l^2 + w^2 + h^2}$

That is worth proving. The diagonal of a base face of the rectanguloid is $d= sqrt{l^2 + w^2}$. This diagonal and the height form a right triangle and the hypotenuse of this right triangle is the diagonal of the rectanguloid. So the diagonal of the rectanguloid is $sqrt {d^2 + h^2} = sqrt{(sqrt{l^2 + w^2}^2 + h^2} = sqrt{l^2 + w^2 + h^2}$.

By induction we can conclude this will hold true for multiple dimensions and an $n$-dimensional rectanguloid with sides $s_1, .... s_n$ inscribed in an an $n$ dimensional sphere, will have a diagonal, diameter of $sqrt{s_1^2 + .... + s_n^2}$.