
The sum of all interior angles of a triangle is \(180°\). By using this property, these problems can be solved as follows.
(i) \(x + 50° + 60°\)
= \(180° x + 110° = 180° x\)
= \(180° -110° = 70°\)
(ii) \(x + 90° +30°\)
= \(180° x + 120°\) =\(180° x\)
= \(180° - 120°\)
=\(60°\)
(iii) \(x + 30° + 110° =180° x + 140° = 180°\)
\(x = 180° - 140° = 40°\)
(iv) \(50° + x + x = 180°\)
\(50° + 2x = 180°\)
\(2x = 180° - 50° = 130°\)
\(x= \frac{130\degree}{2}=65\degree\)
(v) \(x + x + x = 180°\)
\(3x = 180°\)
\(x=\frac{180}{3}=60\degree\)
(vi) \(x + 2x + 90° = 180°\)
\(3x = 180° - 90° = 90 º\)
\(x = \frac{90\degree}{3}=30\degree\)


| 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 \(cm^2\) |
