Let $ABCD$ be a parallelogram and $ABEF$ be a rectangle with $EF$ lying along the line $CD$. If $AB = 7$ cm and $BE = 6.5$ cm, then the area of the parallelogram is

### Concept & Logic A fundamental theorem in geometry states that a parallelogram and a rectangle resting on the same base and situated between the same parallel lines have equal areas. ### Step-by-Step Solution 1. **Analyze the given figures:** * Parallelogram $ABCD$ and rectangle $ABEF$ share the same base, which is $AB$. * The line $EF$ of the rectangle lies along the line $CD$ of the parallelogram. This means both figures are bounded by the same parallel lines ($AB$ and the line containing $C, D, E, F$). 2. **Determine Area of the Rectangle:** * The area of a rectangle is $\text{Length} \times \text{Breadth}$. * Length ($AB$) = 7 cm * Breadth ($BE$) = 6.5 cm * Area of rectangle $ABEF$ = $7 \times 6.5 = 45.5$ cm² 3. **Apply the theorem:** * Area of parallelogram $ABCD$ = Area of rectangle $ABEF$ = 45.5 cm² ### Exam Strategy & Shortcut If a parallelogram and a rectangle share a base and parallel boundaries, just calculate the rectangle's area. There is no need to find the slant height of the parallelogram. ### Common Pitfall Attempting to use trigonometry or assuming the side $BC$ is equal to $BE$, which overcomplicates a straightforward theorem-based question. ### Final Answer Therefore, the correct answer is **45.5 cm²**.