The lengths of three unequal edges of a rectangular solids block are in GP . if the volume of the block is `216 cm^(3)` and the total surface area is `252cm ^(2)` then the length of the longest edge is

Options

A. 12cm
B. 6 cm
C. 18 cm
D. 3 cm

Explanation

To solve the problem step by step, we will use the given information about the rectangular solid block whose edges are in geometric progression (GP).

### Step 1: Define the edges
Let the lengths of the three edges of the rectangular solid be:
- \( L = \frac{A}{R} \)
- \( B = A \)
- \( H = AR \)

Here, \( A \) is the middle term, and \( R \) is the common ratio of the GP.

### Step 2: Write the volume equation
The volume \( V \) of the rectangular solid is given by:
\[ V = L \times B \times H = \frac{A}{R} \times A \times AR \]
This simplifies to:
\[ V = A^3 \]
We know from the problem that the volume is \( 216 \, \text{cm}^3 \):
\[ A^3 = 216 \]
Taking the cube root of both sides, we find:
\[ A = \sqrt[3]{216} = 6 \]

### Step 3: Write the total surface area equation
The total surface area \( S \) of the rectangular solid is given by:
\[ S = 2(LB + BH + HL) \]
Substituting the expressions for \( L \), \( B \), and \( H \):
\[ S = 2\left(\frac{A}{R} \cdot A + A \cdot AR + AR \cdot \frac{A}{R}\right) \]
This simplifies to:
\[ S = 2\left(\frac{A^2}{R} + A^2 + A^2\right) = 2A^2\left(\frac{1}{R} + 1 + R\right) \]
We know from the problem that the total surface area is \( 252 \, \text{cm}^2 \):
\[ 2A^2\left(\frac{1}{R} + 1 + R\right) = 252 \]

### Step 4: Substitute \( A \) into the surface area equation
Substituting \( A = 6 \):
\[ 2(6^2)\left(\frac{1}{R} + 1 + R\right) = 252 \]
Calculating \( 6^2 = 36 \):
\[ 72\left(\frac{1}{R} + 1 + R\right) = 252 \]
Dividing both sides by \( 72 \):
\[ \frac{1}{R} + 1 + R = \frac{252}{72} = \frac{7}{2} \]

### Step 5: Solve for \( R \)
Rearranging gives:
\[ \frac{1}{R} + R = \frac{7}{2} - 1 = \frac{5}{2} \]
Letting \( x = R \), we rewrite this as:
\[ \frac{1}{x} + x = \frac{5}{2} \]
Multiplying through by \( 2x \) to eliminate the fraction:
\[ 2 + 2x^2 = 5x \]
Rearranging gives:
\[ 2x^2 - 5x + 2 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} \]
This gives us two possible values for \( R \):
\[ R = \frac{8}{4} = 2 \quad \text{or} \quad R = \frac{2}{4} = \frac{1}{2} \]

### Step 7: Determine the longest edge
Since \( R \) must be greater than 1 for the edges to be unequal, we take \( R = 2 \).

Now, we can find the lengths of the edges:
- \( L = \frac{A}{R} = \frac{6}{2} = 3 \)
- \( B = A = 6 \)
- \( H = AR = 6 \cdot 2 = 12 \)

Thus, the longest edge is:
\[ H = 12 \, \text{cm} \]

### Final Answer
The length of the longest edge is \( 12 \, \text{cm} \).

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