What is the area of the triangle BCD? (Diagram shows BCD on the base of a figure that looks like a Trapezoid DEBC, with C on the top right, B on the bottom right, E on the top left, and D on the bottom left. There is a vertical line from C to the base at A, where \(CA=6\) and \(BA=10\). There is also a right angle at E, where \(DE=6\) and \(DB=13\). The figure seems to have a few typos in the labels in the explanation part).

Explanation: Triangle BAC তে Angle BCA একটি সমকোণ। \(CA=6\) এবং \(BA=10\). Pythagoras এর উপপাদ্য অনুযায়ী, \(AC^{2}+BC^{2}=AB^{2}\) \(\Rightarrow 6^{2}+BC^{2}=10^{2}\) \(\Rightarrow BC^{2}=100-36=64\) \(\Rightarrow BC=8\). Again, Triangle DEB তে Angle DEB একটি সমকোণ। \(DE=5\) এবং \(DB=13\). (Assuming \(DE=5\) from the calculation, not 6 as stated in the text and \(DB=13\)). Pythagoras এর উপপাদ্য অনুযায়ী, \(DE^{2}+EB^{2}=DB^{2}\) \(\Rightarrow 5^{2}+EB^{2}=13^{2}\) \(\Rightarrow EB^{2}=169-25=144\) \(\Rightarrow EB=12\). Trapezium DEBC তে BC এবং DE parallel এবং তাদের মধ্যে লম্ব দূরত্ব EB. (Assuming DE and BC are parallel). Area of trapezium DEBC \(= \frac{1}{2} \times (BC+ED) \times EB = \frac{1}{2} \times (8+5) \times 12 = 78\). Area of triangle DEB \(= \frac{1}{2} \times EB \times DE = \frac{1}{2} \times 12 \times 5 = 30\). So, area of triangle DCB \(= \text{Area of Trapezium} - \text{Area of Triangle DEB} = 78 - 30 = 48\).