Need solution

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

Therefore, the side length of the cube is 1/3, as required.

What is tetrahedron?

Four triangular faces make up the three-dimensional shape of a tetrahedron. The foundation of the pyramid is one of the triangles, with the other three triangles joining it.

Let D be the vertex of the cube on face ABC. 

Since the opposite vertex of the cube is at O, we have OD = 1. 

Let the side length of the cube be x.

Consider triangle AOB.

AB² = AO² + OB² = 1 + 1 = 2

Similarly, find that BC² = AC² = 2.

Since ABC is a right triangle with angles A, B, and C being 90° -

sin A = BC / AB = √2 / 2

sin B = AC / AB = √2 / 2

sin C = BC / AC = 1

Consider tetrahedron ABCO. Since AOB, AOC, and BOC are right angles -

∠AOCB = π - ∠AOC - ∠BOC = π/2

∠AOBC = π - ∠AOB - ∠BOC = π/2

∠ABCO = π - ∠AOC - ∠AOB = π/2

So triangles AOC, AOB, and BOC are all right triangles with hypotenuse 1 and angles A, B, and C, respectively. 

Using the sine rule -

sin AOC = AO / OC = 1

sin AOB = sin BOC = BO / OC = 1

Therefore, the areas of triangles AOC, AOB, and BOC are -

Area(AOC) = (1/2) × AO × OC × sin AOC = (1/2) × 1 × 1 × 1 = 1/2

Area(AOB) = Area(BOC) = (1/2) × BO × OC × sin AOB = (1/2) × 1 × 1 × 1 = 1/2

Now, consider triangle AOD.

sin AOD = sin(180° - AOB - AOC) = sin(BOC) = √2 / 2

Using the sine rule -

AD / sin AOD = OD / sin OAD

AD / (√2 / 2) = 1 / x

AD = (√2 / 2) * (1 / x)

The area of triangle AOD is -

Area(AOD) = (1/2) × AD × OD × sin AOD = (1/2) × (√2 / 2) × (1 / x) × 1 × (√2 / 2) = 1 / (2x²)

Now, consider the tetrahedron ABCO. 

The volume of the tetrahedron is -

V = (1/3) × Area(ABC) × OD = (1/3) × (√3 / 4) × 1 = √3 / 12

The volume of the cube is -

V = x³

Since the cube is inscribed in the tetrahedron -

√3 / 12 = x³

So, now there is -

x = 1/3

Therefore, the side length of the cube is 1/3, as required