
Distance between U and T = 1 unit
It is divided into 3 equal parts.
TR=RS=SU=\(\frac{1}{3}\)
R=-1-\(\frac{1}{3}\)=\(\frac{-3}{3}\)-\(\frac{1}{3}\)=\(\frac{-4}{3}\)
S=\(\frac{-3}{3}\)-\(\frac{2}{3}\)=\(\frac{-5}{3}\)
Similarly,
AB = 1 unit
It is divided into 3 equal parts.
P=2+\(\frac{1}{3}\)=\(\frac{6}{3}\)+\(\frac{1}{3}\)=\(\frac{7}{3}\)
Q=2+\(\frac{2}{3}\)=\(\frac{6}{3}\)+\(\frac{2}{3}\)=\(\frac{8}{3}\)


| 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 \(cm^2\) |



| 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 \(cm^2\) |
