
Draw a line segment \(AD\) perpendicular to \(BC\).
It is an altitude for this triangle.
It can be observed that the length of \(BD\) and \(DC\) is also same.
Therefore, \(AD\) is also a median of this triangle.


| So No | Base | Height | Area of parallelogram |
|---|---|---|---|
| a. | 20 cm | - | 246 \(cm^2\) |
| b. | - | 15 cm | 154.5 \(cm^2\) |
| c. | - | 8.4 cm | 48.72 \(cm^2\) |
| d. | 15.6 cm | - | 16.38 \(cm^2\) |
| Base | Height | Area of triangle |
|---|---|---|
| 15 cm | - | 87 \(cm^2\) |
| - | 31.4 mm | 1256 \(mm^2\) |
| 22 cm | - | 170.5 \(cm^2\) |



| So No | Base | Height | Area of parallelogram |
|---|---|---|---|
| a. | 20 cm | - | 246 \(cm^2\) |
| b. | - | 15 cm | 154.5 \(cm^2\) |
| c. | - | 8.4 cm | 48.72 \(cm^2\) |
| d. | 15.6 cm | - | 16.38 \(cm^2\) |
| Base | Height | Area of triangle |
|---|---|---|
| 15 cm | - | 87 \(cm^2\) |
| - | 31.4 mm | 1256 \(mm^2\) |
| 22 cm | - | 170.5 \(cm^2\) |
