(i) \(\frac{7}{24}\)-\(\frac{17}{36}\)
L.C.M of 24 and 36 is 72.
\(\frac{7}{24}\)-\(\frac{17}{36}\) = \(\frac{7\times3}{24\times3}-\frac{17\times2}{36\times2}\) = \(\frac{21}{72}\)= -\(\frac{13}{72}\)
(ii) \(\frac{5}{63}\)-(\(-\frac{6}{21}\))
= \(\frac{5}{63}+\frac{2}{7}\)
= L.C.M of 63 and 7 is 63.
\(\frac{5}{63}+\frac{2}{7}\)= \(\frac{5}{63}+\frac{18}{63}\) =\(\frac{5+18}{63}\) = \(\frac{23}{63}\)


| 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 \(cm^2\) |
