Let us consider the trapezium \(ABCD\) given below:
Here, \(AB\) \(=\) 13 \(cm\), \(AD\) \(=\) \(BC\) \(=\) 10 \(cm\) and \(CD\) \(=\) 33 \(cm\). Also, \(AB \parallel CD\) and they are 13 \(cm\) apart.
\(4\) arcs are cut with 12.6 \(cm\) as radius with \(A\), \(B\), \(C\) and \(D\) as centre respectively. [Use \(\pi\) \(=\) \(\frac{22}{7}\)]
Answer:
The area of the trapezium \(ABCD\) \(=\) \(cm^2\)
The cumulative area of the shaded regions \(=\) \(cm^2\)