Find the length of the longest rod that can be placed in a room 16 m long, 12 m broad and `10(2)/(3)m` high.

Options

A. a)23m
B. b)68m
C. c)`22(2)/(3)m`
D. d)`22(1)/(3)m`

Answer

To find the length of the longest rod that can be placed in a room with given dimensions, we need to calculate the diagonal of the room. The formula for the diagonal \(d\) of a rectangular prism (room) is given by:

\[ d = \sqrt{L^2 + B^2 + H^2} \]

where \(L\) is the length, \(B\) is the breadth, and \(H\) is the height of the room.

### Step 1: Identify the dimensions of the room
- Length \(L = 16 \, m\)
- Breadth \(B = 12 \, m\)
- Height \(H = 10 \frac{2}{3} \, m\)

### Step 2: Convert the height into an improper fraction
To convert \(10 \frac{2}{3}\) into an improper fraction:
\[ H = 10 \frac{2}{3} = \frac{10 \times 3 + 2}{3} = \frac{30 + 2}{3} = \frac{32}{3} \, m \]

### Step 3: Calculate the squares of the dimensions
Now we calculate the squares of each dimension:
- \(L^2 = 16^2 = 256\)
- \(B^2 = 12^2 = 144\)
- \(H^2 = \left(\frac{32}{3}\right)^2 = \frac{1024}{9}\)

### Step 4: Sum the squares
Now, we sum the squares:
\[ L^2 + B^2 + H^2 = 256 + 144 + \frac{1024}{9} \]
To add these, we need a common denominator. The common denominator for \(256\) and \(144\) is \(9\):
\[ 256 = \frac{256 \times 9}{9} = \frac{2304}{9} \]
\[ 144 = \frac{144 \times 9}{9} = \frac{1296}{9} \]

Now, adding these:
\[ L^2 + B^2 + H^2 = \frac{2304}{9} + \frac{1296}{9} + \frac{1024}{9} = \frac{2304 + 1296 + 1024}{9} = \frac{4624}{9} \]

### Step 5: Calculate the diagonal
Now we take the square root to find the diagonal:
\[ d = \sqrt{\frac{4624}{9}} = \frac{\sqrt{4624}}{3} \]

Calculating \(\sqrt{4624}\):
\[ \sqrt{4624} = 68 \]
Thus,
\[ d = \frac{68}{3} \, m \]

### Step 6: Convert to mixed number
To convert \(\frac{68}{3}\) into a mixed number:
\[ \frac{68}{3} = 22 \frac{2}{3} \, m \]

### Final Answer
The length of the longest rod that can be placed in the room is:
\[ \boxed{22 \frac{2}{3} \, m} \]