In a trapezium, the two non-parallel sides are equal in length, each being of 5 units. The parallel sides are at a distance of 3 units apart. If the smaller side of the parallel sides is of length 2 units, then the sum of the diagonals of the trapezium is

Options

A. `10sqrt5` `units`
B. `6sqrt5` `units`
C. `5sqrt5` `units`
D. `3sqrt5` `units`

Answer

To solve the problem step by step, we will first visualize the trapezium and then apply the Pythagorean theorem to find the lengths of the diagonals.

### Step 1: Visualize the Trapezium
Let's denote the trapezium as ABCD, where AB and CD are the parallel sides, with AB being the shorter side (2 units) and CD being the longer side. The non-parallel sides AD and BC are both equal to 5 units. The distance between the parallel sides (height) is given as 3 units.

### Step 2: Set Up the Coordinates
We can place the trapezium in a coordinate system:
- Let point A be at (0, 0).
- Let point B be at (2, 0) (since AB is 2 units long).
- Let point D be at (x, 3) and point C be at (x + c, 3), where c is the length of the longer parallel side CD.

### Step 3: Calculate the Length of CD
To find the length of CD, we need to determine the value of x. Since AD and BC are equal (5 units), we can use the Pythagorean theorem.

### Step 4: Apply Pythagorean Theorem
For triangle AOD (where O is the foot of the perpendicular from D to AB):
- The height (AD) is 3 units.
- The base (AO) is x units.
- The hypotenuse (AD) is 5 units.

Using the Pythagorean theorem:
\[ AD^2 = AO^2 + OD^2 \]
\[ 5^2 = x^2 + 3^2 \]
\[ 25 = x^2 + 9 \]
\[ x^2 = 16 \]
\[ x = 4 \]

Now, point D is at (4, 3).

### Step 5: Calculate Length of CD
Since we know that AB = 2 units and the distance between the two parallel sides is 3 units, we can find the coordinates of point C:
- Let CD = c, then point C will be at (4 + c, 3).

### Step 6: Find the Length of CD
Using the distance formula for BC:
\[ BC^2 = (4 + c - 2)^2 + (3 - 0)^2 \]
\[ 5^2 = (2 + c)^2 + 3^2 \]
\[ 25 = (2 + c)^2 + 9 \]
\[ 16 = (2 + c)^2 \]
Taking the square root:
\[ 4 = 2 + c \]
Thus, \( c = 2 \).

### Step 7: Calculate the Length of the Diagonals
Now we can calculate the lengths of the diagonals AC and BD.

For diagonal AC:
- A = (0, 0) and C = (6, 3)
Using the distance formula:
\[ AC = \sqrt{(6 - 0)^2 + (3 - 0)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \]

For diagonal BD:
- B = (2, 0) and D = (4, 3)
Using the distance formula:
\[ BD = \sqrt{(4 - 2)^2 + (3 - 0)^2} = \sqrt{4 + 9} = \sqrt{13} \]

### Step 8: Sum of the Diagonals
Finally, we find the sum of the diagonals:
\[ AC + BD = 3\sqrt{5} + \sqrt{13} \]

### Final Answer
The sum of the diagonals of the trapezium is \( 3\sqrt{5} + \sqrt{13} \).