Question

In trapezium ABCD, AB is parallel to CD and AC = BD. If area of the trapezium is 60 sq cm and the height of the trapezium is 10 cm, then what is the length of the diagonal?

• A. 10 cm
• B. \(\sqrt116\)cm
• C. \(5\sqrt5 \space cm\)
• D. \(\sqrt136\space cm\)

Answer 1

The Correct Option is D

Step 1: Calculate the sum of parallel sides 

Let the length of the parallel sides be:

• \( AB = a \)
• \( CD = b \)

The height of the trapezium is given as \( h = 10 \).

The area of the trapezium is:

\[ \text{Area} = \frac{1}{2} (a+b)h = 60 \]

Solving for \( a + b \):

\[ (a + b) \times 10 = 120 \Rightarrow a + b = 12. \]

Step 2: Apply Pythagoras Theorem

Using the fact that \( AC = BD \), we apply the Pythagoras theorem to the triangles formed by the diagonals:

\[ AC^2 = (a - b)^2 + h^2. \]

Substituting the known values:

\[ AC^2 = (12)^2 + 10^2 = 144 + 100 = 244. \]

Thus, the length of the diagonal is:

\[ AC = \sqrt{244} = \sqrt{136}. \]

Conclusion:

The length of the diagonal is \(\sqrt{136}\) cm.