A rhombus has an area equal to one-fifth the sum of the areas of the squares built on its four sides. The ratio of the long diagonal to the short diagonal is:

Options

A. `2+sqrt(3)`
B. `2-sqrt(3)`
C. `1/(2)`
D. 2

Answer

To solve the problem, let's break it down step by step:

### Step 1: Understand the Problem
We have a rhombus whose area is equal to one-fifth the sum of the areas of the squares built on its four sides. We need to find the ratio of the long diagonal to the short diagonal.

### Step 2: Define the Sides of the Rhombus
Let the length of each side of the rhombus be \( A \). Since all sides of a rhombus are equal, we can say:
- Side 1 = \( A \)
- Side 2 = \( A \)
- Side 3 = \( A \)
- Side 4 = \( A \)

### Step 3: Calculate the Area of the Squares
The area of one square built on a side of length \( A \) is:
\[ \text{Area of one square} = A^2 \]
Since there are four squares, the total area of the squares is:
\[ \text{Total area of squares} = 4 \times A^2 = 4A^2 \]

### Step 4: Area of the Rhombus
According to the problem, the area of the rhombus is one-fifth of the total area of the squares:
\[ \text{Area of rhombus} = \frac{1}{5} \times 4A^2 = \frac{4A^2}{5} \]

### Step 5: Relate the Area of the Rhombus to its Diagonals
The area of a rhombus can also be expressed in terms of its diagonals \( D_1 \) and \( D_2 \):
\[ \text{Area of rhombus} = \frac{1}{2} \times D_1 \times D_2 \]
Setting the two expressions for the area equal gives:
\[ \frac{4A^2}{5} = \frac{1}{2} \times D_1 \times D_2 \]

### Step 6: Rearranging the Equation
Multiplying both sides by 2 to eliminate the fraction:
\[ \frac{8A^2}{5} = D_1 \times D_2 \]

### Step 7: Apply the Pythagorean Theorem
In a rhombus, the diagonals bisect each other at right angles. Using the Pythagorean theorem:
\[ \left(\frac{D_1}{2}\right)^2 + \left(\frac{D_2}{2}\right)^2 = A^2 \]
This simplifies to:
\[ \frac{D_1^2}{4} + \frac{D_2^2}{4} = A^2 \]
Multiplying through by 4 gives:
\[ D_1^2 + D_2^2 = 4A^2 \]

### Step 8: Set Up the Quadratic Equation
Now we have two equations:
1. \( D_1 \times D_2 = \frac{8A^2}{5} \)
2. \( D_1^2 + D_2^2 = 4A^2 \)

Let \( x = \frac{D_1}{D_2} \). Then \( D_1 = xD_2 \). Substitute this into the equations:
- From \( D_1 \times D_2 = \frac{8A^2}{5} \):
\[ xD_2^2 = \frac{8A^2}{5} \]
- From \( D_1^2 + D_2^2 = 4A^2 \):
\[ x^2D_2^2 + D_2^2 = 4A^2 \]
Factoring gives:
\[ (x^2 + 1)D_2^2 = 4A^2 \]

### Step 9: Solve for \( D_2^2 \)
Now we can express \( D_2^2 \):
\[ D_2^2 = \frac{4A^2}{x^2 + 1} \]

### Step 10: Substitute Back to Find \( x \)
Substituting \( D_2^2 \) back into the equation for \( D_1 \times D_2 \):
\[ x \cdot \sqrt{\frac{4A^2}{x^2 + 1}} \cdot \sqrt{\frac{4A^2}{x^2 + 1}} = \frac{8A^2}{5} \]
This simplifies to:
\[ \frac{8A^2x}{x^2 + 1} = \frac{8A^2}{5} \]
Cancelling \( 8A^2 \) gives:
\[ \frac{x}{x^2 + 1} = \frac{1}{5} \]

### Step 11: Cross Multiply and Rearrange
Cross-multiplying gives:
\[ 5x = x^2 + 1 \]
Rearranging gives:
\[ x^2 - 5x + 1 = 0 \]

### Step 12: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{5 \pm \sqrt{25 - 4}}{2} = \frac{5 \pm \sqrt{21}}{2} \]

### Step 13: Determine the Ratio
Since we assumed \( D_1 > D_2 \), we take the positive root:
\[ \frac{D_1}{D_2} = \frac{5 + \sqrt{21}}{2} \]

### Conclusion
The ratio of the long diagonal to the short diagonal is:
\[ D_1 : D_2 = \frac{5 + \sqrt{21}}{2} : 1 \]