Parallelograms on Same Base and Between Same Parallels
Statement / Question: Prove that parallelograms standing on the same base and between the same parallel lines are equal in area.
Given
▱ABCD & ▱ABEF on base AB between FC ∥ AB.
To Prove
Area of ▱ABCD = Area of ▱ABEF
Construction
Draw MN ⊥ base AB.
Theoretical Proof
| Statements | Reasons |
|---|---|
| 1. Area of ▱ABCD = AB × MN | 1. Area of parallelogram = Base × Height |
| 2. Area of ▱ABEF = AB × MN | 2. Area of parallelogram = Base × Height |
| 3. Area of ▱ABCD = Area of ▱ABEF | 3. From Statements (1) and (2) |
Parallelograms Standing on Equal Bases
Statement / Question: Prove that parallelograms standing on equal bases and between the same parallel lines are equal in area.
Given
▱ABCD & ▱PQRS on equal bases AB = PQ between DS ∥ AQ.
To Prove
Area of ▱ABCD = Area of ▱PQRS
Construction
Draw MN ⊥ AB & XY ⊥ PQ.
Theoretical Proof
| Statements | Reasons |
|---|---|
| 1. MN = XY and AB = PQ | 1. Given AB = PQ, and heights between parallel lines are equal (MN = XY) |
| 2. Area of ▱ABCD = AB × MN | 2. Area of parallelogram = Base × Height |
| 3. Area of ▱PQRS = PQ × XY | 3. Area of parallelogram = Base × Height |
| 4. Area of ▱ABCD = Area of ▱PQRS | 4. From Statements (1), (2), and (3) |
Parallelogram & Rectangle on Same Base
Statement / Question: Prove that a parallelogram and a rectangle standing on the same base and between the same parallel lines are equal in area.
Given
▱ABCD & ▭ABEF on base AB between FC ∥ AB.
To Prove
Area of ▱ABCD = Area of ▭ABEF
Construction
In ▭ABEF, AF ⊥ AB (perpendicular height).
Theoretical Proof
| Statements | Reasons |
|---|---|
| 1. Area of ▱ABCD = AB × AF | 1. Area of parallelogram = Base × Height |
| 2. Area of rectangle ▭ABEF = AB × AF | 2. Area of rectangle = Length × Breadth |
| 3. Area of ▱ABCD = Area of rectangle ▭ABEF | 3. From Statements (1) and (2) |
Parallelogram & Square on Same Base
Statement / Question: Prove that a parallelogram and a square standing on the same base and between the same parallel lines are equal in area.
Given
▱ABCD & □ABEF on base AB between FC ∥ AB.
To Prove
Area of ▱ABCD = Area of square □ABEF
Construction
In square ABEF, AF ⊥ AB (perpendicular height).
Theoretical Proof
| Statements | Reasons |
|---|---|
| 1. Area of ▱ABCD = AB × AF = AB2 | 1. Area of parallelogram = Base × Height (AF = AB in square ABEF) |
| 2. Area of square □ABEF = AB2 | 2. Area of square = Side2 |
| 3. Area of ▱ABCD = Area of square □ABEF | 3. From Statements (1) and (2) |