#1**0 **

marthasimonsMay 30, 2023

#1**0 **

First, we can find the length of AB. Since AOB is a right angle, we can use the Pythagorean Theorem to get:

AB^2 = OA^2 + OB^2 = 1^2 + 1^2 = 2 AB = sqrt(2)

Similarly, we can find the length of AC and BC:

AC^2 = OA^2 + OC^2 = 1^2 + 1^2 = 2 AC = sqrt(2)

BC^2 = OB^2 + OC^2 = 1^2 + 1^2 = 2 BC = sqrt(2)

Now, let's consider the cube inscribed in the tetrahedron. Let D be the vertex of the cube opposite from O, and let E, F, and G be the vertices of the cube on faces AOB, AOC, and BOC, respectively. Since AOB, AOC, and BOC are all right angles, we know that the line segments DE, DF, and DG are perpendicular to faces AOB, AOC, and BOC, respectively.

Let x be the side length of the cube. Then, we have:

DE^2 = (AB - x)^2 + (OB - x)^2 = (sqrt(2) - x)^2 + (1 - x)^2 DF^2 = (AC - x)^2 + (OA - x)^2 = (sqrt(2) - x)^2 + (1 - x)^2 DG^2 = (BC - x)^2 + (OC - x)^2 = (sqrt(2) - x)^2 + (1 - x)^2

Note that these expressions are all equal, since DE, DF, and DG are simply three different line segments perpendicular to three different faces of the cube. Thus, we have:

(sqrt(2) - x)^2 + (1 - x)^2 = DG^2 = (sqrt(2) - x)^2 + (1 - x)^2

Expanding and simplifying, we get:

2(sqrt(2) - x)^2 + 2(1 - x)^2 = 2 2(3 - 2sqrt(2)x + x^2) + 2(1 - 2x + x^2) = 2 6 - 4sqrt(2)x + 2x^2 + 2 - 4x + 2x^2 = 2 4x^2 - 4sqrt(2)x + 1 = 0

Solving for x using the quadratic formula, we get:

x = (2sqrt(2) ± sqrt(8 - 4))/8 x = (2sqrt(2) ± sqrt(4))/8 x = (2sqrt(2) ± 2)/8

Since x must be positive, we take the positive root:

x = (2sqrt(2) + 2)/8 x = sqrt(2)/4 + 1/4 x = (1/2)sqrt(2)/2 + 1/4 x = (1/2)(sqrt(2)/sqrt(2*2)) + 1/4 x = (1/2)(1/√2) + 1/4 x = (1/2√2) + (1/4) x = 1/√8 + 1/4 x = (√2/4) + 1/4 x = (1/2) + 1/4 x = 3/4 P.s. do my hw https://writemyessay.ca/do-my-homework-for-me

Therefore, the side length of

marthasimonsMay 1, 2023