Home
Class 12
CHEMISTRY
Total number of Hexagonal faces in a t...

Total number of Hexagonal faces in a truncated octahedron = x
Total number of Hexagonal faces in a truncated tetrahedron = y
Then x + y is:

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to determine the total number of hexagonal faces in both a truncated octahedron and a truncated tetrahedron, and then sum these values. ### Step 1: Determine the number of hexagonal faces in a truncated octahedron. A truncated octahedron is a type of Archimedean solid that has: - 8 regular hexagonal faces - 6 square faces Thus, the total number of hexagonal faces (x) in a truncated octahedron is: \[ x = 8 \] ### Step 2: Determine the number of hexagonal faces in a truncated tetrahedron. A truncated tetrahedron is another type of Archimedean solid that has: - 4 regular triangular faces (which are not relevant for our count of hexagonal faces) - 4 hexagonal faces Thus, the total number of hexagonal faces (y) in a truncated tetrahedron is: \[ y = 4 \] ### Step 3: Calculate the sum of the hexagonal faces. Now we can find the total number of hexagonal faces by adding the values of x and y: \[ x + y = 8 + 4 = 12 \] ### Final Answer: The total number of hexagonal faces in a truncated octahedron and a truncated tetrahedron combined is: \[ x + y = 12 \] ---
Promotional Banner

Similar Questions

Explore conceptually related problems

The number of hexagonal faces that are present in a truncated octahedron is

Please help Sabu decode the jail lock. Chacha Choudhary gave Sabu a formula : f_(1) = ((x)/(z)xx y), f_(2) =((f)/(v) xxu), f_(3) = ((r)/(s)xx w) Sabu can open the lock if he finds the value of 3f_(1) +f_(2) +f_(3) = key where: Number of triangular faces in a truncated tetrahedron = x Number of hexagonal faces in a truncated tetrahedron = x Number of corners in a truncated tetrahedron = z Number of square faces in a truncated octahedron = t Number of hexagonal faces in a truncated octahedron = u Number of corners in a truncated octahedron = u Number of triangular faces in a truchcated cube = w Number of octangonal faces in a truncated cube = r Number of corners in a truncated cube = s What is the KEY ?

Find the number of hexagonal faces that are present in a truncated octahedral.

Total number of bone of human face :-

In a truncated tetrahedron the sum of no.of hexagonal faces and no.of triangular faces is