What is the mean proportional of
`(15+ sqrt200) and (27- sqrt 648)?`

Options

A. 4
B. `14sqrt7`
C. `3sqrt5`
D. `5sqrt3`

Answer

To find the mean proportional of \( (15 + \sqrt{200}) \) and \( (27 - \sqrt{648}) \), we will follow these steps:

### Step 1: Identify the values of A and B
Let:
- \( A = 15 + \sqrt{200} \)
- \( B = 27 - \sqrt{648} \)

### Step 2: Simplify \( \sqrt{200} \) and \( \sqrt{648} \)
First, we simplify \( \sqrt{200} \):
\[ \sqrt{200} = \sqrt{100 \times 2} = \sqrt{100} \times \sqrt{2} = 10\sqrt{2} \]

Next, we simplify \( \sqrt{648} \):
\[ \sqrt{648} = \sqrt{324 \times 2} = \sqrt{324} \times \sqrt{2} = 18\sqrt{2} \]

### Step 3: Substitute the simplified values back into A and B
Now we substitute these values back:
\[ A = 15 + 10\sqrt{2} \]
\[ B = 27 - 18\sqrt{2} \]

### Step 4: Calculate the mean proportional
The mean proportional \( M \) of \( A \) and \( B \) is given by:
\[ M = \sqrt{A \cdot B} \]
Thus, we need to calculate \( A \cdot B \):
\[ A \cdot B = (15 + 10\sqrt{2})(27 - 18\sqrt{2}) \]

### Step 5: Expand the product
Using the distributive property (FOIL method):
\[ A \cdot B = 15 \cdot 27 + 15 \cdot (-18\sqrt{2}) + 10\sqrt{2} \cdot 27 + 10\sqrt{2} \cdot (-18\sqrt{2}) \]
Calculating each term:
1. \( 15 \cdot 27 = 405 \)
2. \( 15 \cdot (-18\sqrt{2}) = -270\sqrt{2} \)
3. \( 10\sqrt{2} \cdot 27 = 270\sqrt{2} \)
4. \( 10\sqrt{2} \cdot (-18\sqrt{2}) = -180 \)

Now combine these results:
\[ A \cdot B = 405 - 270\sqrt{2} + 270\sqrt{2} - 180 \]
The \( -270\sqrt{2} \) and \( +270\sqrt{2} \) cancel each other out:
\[ A \cdot B = 405 - 180 = 225 \]

### Step 6: Take the square root
Now we find the mean proportional:
\[ M = \sqrt{225} = 15 \]

### Final Answer
The mean proportional of \( (15 + \sqrt{200}) \) and \( (27 - \sqrt{648}) \) is \( 15 \).

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