
The sum of the measures of all interior angles of a triangle is \(180\degree\).
\(\angle PQR + \angle PRQ + \angle QPR = 180\degree\)
\(25\degree + 65\degree + \angle QPR = 180\degree\)
\(90\degree + \angle QPR = 180\degree\)
\(\angle QPR = 180\degree - 90\degree = 90\degree\)
Therefore, \(Δ\) \(PQR\) is right-angled at point \(P\).
Hence, \((PR)^2 + (PQ)^2= (QR)^2\)
Thus, (ii) is true.


| 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 \(cm^2\) |


In the case of right-angled triangles, identify the right angles.



| 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 \(cm^2\) |
