(i)\(\frac{2}{5}\)÷ \(\frac{1}{2}\) = \(\frac{2}{5}\) × 2 = \(\frac{4}{5}\)
(ii) \(\frac{4}{9}\) ÷ \(\frac{2}{3}\) = \(\frac{4}{9}\) × \(\frac{3}{2}\) = \(\frac{2}{3}\)
(iii) \(\frac{3}{7}\) ÷ \(\frac{8}{7}\) = \(\frac{3}{7}\)× \(\frac{7}{8}\) =\(\frac{3}{8}\)
(iv) 2\(\frac{1}{3}\) ÷ \(\frac{3}{5}\) = \(\frac{7}{3}\) ÷ \(\frac{3}{5}\) = \(\frac{7}{3}\) ×\(\frac{5}{3}\) = \(\frac{35}{9}\)
(v) 3 \(\frac{1}{2}\) ÷ \(\frac{8}{3}\) = \(\frac{7}{2}\) ÷ \(\frac{8}{3}\) = \(\frac{7}{2}\) × \(\frac{3}{8}\) = \(\frac{21}{16}\)
(vi) \(\frac{2}{5}\) ÷ 1\(\frac{1}{2}\)= \(\frac{2}{5}\) ÷ \(\frac{3}{2}\) = \(\frac{2}{5}\) × \(\frac{2}{3}\) = \(\frac{4}{15}\)
(vii) 3 \(\frac{1}{5}\) ÷ 1 \(\frac{2}{3}\) = \(\frac{16}{5}\) ÷ \(\frac{5}{3}\) = \(\frac{16}{5}\) × \(\frac{3}{5}\) = \(\frac{48}{25}\)
(viii) 2\(\frac{1}{5}\) ÷ 1\(\frac{1}{5}\) = \(\frac{11}{5}\) ÷ \(\frac{6}{5}\) = \(\frac{11}{5}\) × \(\frac{5}{6}\) = \(\frac{11}{6}\)


| 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 \(cm^2\) |
