In a quadrilateral ABCD, `angleB=angleC=60^@ and angleD=90^@`. Find the length of CD, if `AD=root(3)(3) and BC=root(6)(3) cm`.

Options

A. 1. `3(root(2)(3)-1) cm `
B. 2. `6(sqrt3-1)cm `
C. 3. `2(root(3)(3)-1) cm `
D. 4. `sqrt3(root(2)(3)-1) cm`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have a quadrilateral ABCD with the following angles:
- \( \angle B = 60^\circ \)
- \( \angle C = 60^\circ \)
- \( \angle D = 90^\circ \)

We are also given the lengths:
- \( AD = 3\sqrt{3} \) cm
- \( BC = 6\sqrt{3} \) cm

### Step 2: Draw the figure
Draw quadrilateral ABCD with the given angles and label the sides accordingly.

### Step 3: Draw parallel lines
Draw a line parallel to AC from point A to point E, and another parallel line from point D to point F. This will help us in applying trigonometric ratios.

### Step 4: Analyze triangle EFC
In triangle EFC, since \( EF \) is parallel to \( AD \), we can use the sine ratio:
\[ \sin 60^\circ = \frac{EF}{EC} \]
Given that \( EF = AD = 3\sqrt{3} \), we can substitute:
\[ \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{EC} \]

### Step 5: Solve for EC
Cross-multiplying gives:
\[ EC = \frac{3\sqrt{3}}{\frac{\sqrt{3}}{2}} = 6 \text{ cm} \]

### Step 6: Find BE
Now, we know \( BC = 6\sqrt{3} \) cm, so:
\[ BE = BC - EC = 6\sqrt{3} - 6 \]

### Step 7: Find AE
Since \( \angle ABE = 60^\circ \) and \( \angle AEF = 90^\circ \), triangle ABE is also 60-60-60, making it an equilateral triangle. Thus:
\[ AE = BE = 6\sqrt{3} - 6 \]

### Step 8: Find FD
Since \( FD \) is parallel to \( AE \), we have:
\[ FD = AE = 6\sqrt{3} - 6 \]

### Step 9: Analyze triangle EFC again
In triangle EFC, we can use the tangent ratio:
\[ \tan 60^\circ = \frac{EF}{FC} \]
Substituting the values:
\[ \sqrt{3} = \frac{3\sqrt{3}}{FC} \]

### Step 10: Solve for FC
Cross-multiplying gives:
\[ FC = \frac{3\sqrt{3}}{\sqrt{3}} = 3 \text{ cm} \]

### Step 11: Find CD
Finally, we can find \( CD \):
\[ CD = BE + FC = (6\sqrt{3} - 6) + 3 \]
\[ CD = 6\sqrt{3} - 3 \]

### Step 12: Factor out the common term
Factoring out the common term gives:
\[ CD = 3(2\sqrt{3} - 1) \text{ cm} \]

### Final Answer
Thus, the length of \( CD \) is \( 3(2\sqrt{3} - 1) \) cm.

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