(i) \(\frac{2}{7}\) × \(\frac{3}{4}\)= \(\frac{3}{14}\) or \(\frac{3}{5}\) × \(\frac{5}{8}\) =\(\frac{3}{8}\)
Converting these fractions into like fractions,
\(\frac{3}{14}\) = \(\frac{3\times4}{14\times 4}\) = \(\frac{12}{56}\)
\(\frac{3}{8}\)= \(\frac{3\times 7}{8\times 7}\) =\(\frac{21}{56}\)
Since \(\frac{21}{56}\)> \(\frac{12}{56}\),
∴ \(\frac{3}{8}\)> \(\frac{3}{14}\)
Therefore, \(\frac{3}{5}\) of \(\frac{5}{8}\) is greater.
(ii) \(\frac{1}{2}\) × \(\frac{6}{7}\) = \(\frac{3}{7}\)
\(\frac{2}{3}\) × \(\frac{3}{7}\) = \(\frac{2}{7}\)
Since 3 > 2,
∴ \(\frac{3}{7}\)> \(\frac{2}{7}\)
Therefore, \(\frac{1}{2}\)of \(\frac{6}{7}\) is greater.


| 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 \(cm^2\) |

Find: (a) \(\frac{1}{2}\) of (i) 2 \(\frac{3}{4}\) (ii) 4 \(\frac{2}{9}\)
(b) \(\frac{5}{8}\) of (i) 3 \(\frac{5}{6}\) (ii) 9 \(\frac{2}{3}\)


| 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 \(cm^2\) |
