ABCD is a trapezium, such that AB = CD and AD|BC. AD = 5 cm, BC = 9 cm. If area of ABCD is 35 sq.cm, then CD is :

Options

A. `sqrt29` cm
B. 5 cm
C. 6 cm
D. `sqrt21` cm

Answer

To solve the problem step by step, we will follow the given information about trapezium ABCD.

### Step 1: Understand the properties of trapezium ABCD
We know that in trapezium ABCD:
- AD is parallel to BC (AD || BC)
- AB = CD
- AD = 5 cm
- BC = 9 cm
- Area of trapezium ABCD = 35 sq.cm

### Step 2: Use the formula for the area of a trapezium
The area \( A \) of a trapezium can be calculated using the formula:
\[ A = \frac{1}{2} \times (b_1 + b_2) \times h \]
where \( b_1 \) and \( b_2 \) are the lengths of the parallel sides, and \( h \) is the height.

In our case:
- \( b_1 = AD = 5 \, \text{cm} \)
- \( b_2 = BC = 9 \, \text{cm} \)
- Area \( A = 35 \, \text{sq.cm} \)

### Step 3: Substitute the known values into the area formula
Substituting the values into the area formula:
\[ 35 = \frac{1}{2} \times (5 + 9) \times h \]
\[ 35 = \frac{1}{2} \times 14 \times h \]
\[ 35 = 7h \]

### Step 4: Solve for the height \( h \)
To find \( h \), divide both sides by 7:
\[ h = \frac{35}{7} = 5 \, \text{cm} \]

### Step 5: Determine the lengths of the segments
Since \( AD \) and \( BC \) are parallel and \( AB = CD \), we can denote the length of \( CD \) as \( x \). The difference in lengths of the two parallel sides is \( BC - AD = 9 - 5 = 4 \, \text{cm} \).

### Step 6: Divide the difference equally
Since \( AB = CD \) and the trapezium is symmetric, the difference of 4 cm will be divided equally between \( AB \) and \( CD \):
\[ AB = CD = x \]
Thus, the segments \( BP \) and \( KC \) (where \( P \) and \( K \) are the points where the height meets \( AD \) and \( BC \)) will each be:
\[ BP = KC = \frac{4}{2} = 2 \, \text{cm} \]

### Step 7: Apply the Pythagorean theorem
In triangle \( DKC \):
- \( DK = AD = 5 \, \text{cm} \)
- \( KC = 2 \, \text{cm} \)

Using the Pythagorean theorem:
\[ DC^2 = DK^2 + KC^2 \]
\[ DC^2 = 5^2 + 2^2 \]
\[ DC^2 = 25 + 4 = 29 \]
\[ DC = \sqrt{29} \, \text{cm} \]

### Step 8: Conclusion
Thus, the length of \( CD \) is:
\[ CD = \sqrt{29} \, \text{cm} \approx 5.39 \, \text{cm} \]