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?
- 10 cm
- \(\sqrt116\)cm
- \(5\sqrt5 \space cm\)
- \(\sqrt136\space cm\)
The Correct Option is D
Solution and Explanation
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.
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