Area of Parallelogram Between Parallels
A fundamental theorem in geometry states: parallelograms on the same base and between the same parallel lines have equal area.
This means the shape of the parallelogram does not matter — as long as the base and the distance between the parallel lines (height) remain the same, the area is constant.
This theorem is the foundation for proving many results about areas of triangles and quadrilaterals.
What is Area of Parallelogram Between Parallels?
Theorem: Parallelograms on the same base and between the same parallels are equal in area.
Area = base × height
(height = perpendicular distance between the parallel lines)
Since the base and height are the same for all such parallelograms, their areas are equal regardless of how "slanted" they are.
Derivation and Proof
Proof:
- Let parallelograms ABCD and ABEF have the same base AB and lie between the same parallels AB and CF.
- Since DC ∥ AB and FE ∥ AB, points D, C, F, E lie on the same line (CF).
- In △ADF and △BCE:
- AD = BC (opposite sides of parallelogram ABCD)
- AF = BE (opposite sides of parallelogram ABEF)
- DF = CE (since DC = AB = EF, and D, C are between F, E)
- By SSS: △ADF ≅ △BCE
- Area(ABCD) = Area(ABED) + Area(△BCE) = Area(ABED) + Area(△ADF) = Area(ABEF)
- Therefore Area(ABCD) = Area(ABEF). ∎
Solved Examples
Example 1: Example 1: Equal area parallelograms
Problem: Two parallelograms share base PQ = 10 cm and lie between parallels 6 cm apart. One is a rectangle, the other is slanted. Find both areas.
Solution:
- Area of rectangle = 10 × 6 = 60 cm²
- Area of slanted parallelogram = 10 × 6 = 60 cm² (same base, same height)
Answer: Both areas = 60 cm².
Example 2: Example 2: Finding the area
Problem: ABCD is a parallelogram with AB = 12 cm. The distance between AB and CD is 8 cm. Find the area.
Solution:
- Area = base × height = 12 × 8 = 96 cm²
Answer: Area = 96 cm².
Example 3: Example 3: Finding the height
Problem: A parallelogram has area 150 cm² and base 15 cm. Find the distance between the parallel sides.
Solution:
- Height = Area / Base = 150 / 15 = 10 cm
Answer: Height = 10 cm.
Example 4: Example 4: Triangle between same parallels
Problem: A parallelogram ABCD and triangle ABE share base AB = 14 cm and lie between the same parallels 9 cm apart. Find their areas.
Solution:
- Area of parallelogram = 14 × 9 = 126 cm²
- Area of triangle = ½ × 14 × 9 = 63 cm²
- Triangle area = half the parallelogram area (same base, same parallels)
Answer: Parallelogram = 126 cm², Triangle = 63 cm².
Example 5: Example 5: Comparing areas
Problem: ABCD and PQRS are parallelograms on the same base AB = 8 cm but between different parallels. ABCD has height 5 cm and PQRS has height 7 cm. Compare areas.
Solution:
- Area(ABCD) = 8 × 5 = 40 cm²
- Area(PQRS) = 8 × 7 = 56 cm²
- They are NOT between the same parallels, so areas differ.
Answer: Areas are 40 cm² and 56 cm². The theorem only applies when between the SAME parallels.
Example 6: Example 6: Proof-based problem
Problem: ABCD is a parallelogram. E is any point on BC. Prove that Area(△ABE) + Area(△DEC) = ½ Area(ABCD).
Solution:
- Area(△ABE) = ½ × BE × h (where h is distance between AB and DC)
- Area(△DEC) = ½ × EC × h
- Sum = ½h(BE + EC) = ½h × BC = ½ × BC × h
- But Area(ABCD) = AB × h = BC × h (since AB = CD, but actually base = AB)
- Wait: we need the height from AB. Area(ABCD) = AB × h₁. The triangles use height from parallel sides.
- Since E is on BC, △ADE has same base AD as parallelogram (opposite side) and half the height only when E is midpoint.
- Actually: Area(△ABE) has base BE with height = distance from A to BC.
- The correct approach: Area(△AED) = ½ × AD × h. But area(ABCD) = AD × h. So Area(△AED) = ½ Area(ABCD). Then Area(ABE) + Area(DEC) = Area(ABCD) − Area(AED) = ½ Area(ABCD).
Answer: Proved using the fact that diagonal divides parallelogram into equal triangles and subtracting △AED.
Example 7: Example 7: Application
Problem: A farmer has two parallelogram-shaped fields on the same base 200 m and between the same canal lines 80 m apart. Are the fields equal in area?
Solution:
- Yes. Same base (200 m) and between same parallels (80 m apart).
- Both areas = 200 × 80 = 16,000 m²
Answer: Yes, both fields have area 16,000 m².
Example 8: Example 8: Two bases of parallelogram
Problem: ABCD is a parallelogram. AB = 14 cm, height from AB = 6 cm. AD = 10 cm. Find the height from AD.
Solution:
- Area = AB × h₁ = AD × h₂
- 14 × 6 = 10 × h₂
- h₂ = 84/10 = 8.4 cm
Answer: Height from AD = 8.4 cm.
Real-World Applications
Applications:
- Land measurement: Fields between parallel roads or canals have easily calculable areas.
- Proofs: This theorem is used to prove that a median divides a triangle into two equal areas.
- Architecture: Slanted walls on the same base have the same cross-sectional area.
Key Points to Remember
- Parallelograms on the same base and between the same parallels have equal area.
- Area = base × perpendicular height (distance between parallels).
- A triangle on the same base and between same parallels has half the area of the parallelogram.
- The shape of the parallelogram does not matter — only base and height determine area.
- If the parallels change (different height), areas will be different even with same base.
- This theorem extends to triangles: triangles on same base between same parallels have equal area.
Practice Problems
- Two parallelograms on the same base 18 cm and between parallels 10 cm apart. Find both areas.
- A parallelogram has area 200 cm² and base 25 cm. Find the height.
- ABCD is a parallelogram with AB = 20 cm, height from AB = 7 cm. Find area. Also find height from BC if BC = 10 cm.
- Prove that a triangle on the same base as a parallelogram and between the same parallels has half the area.
- Two fields are parallelograms on base 150 m. One is between parallels 50 m apart, other between parallels 70 m apart. Compare areas.
Frequently Asked Questions
Q1. What does 'between the same parallels' mean?
It means the base lies on one parallel line and the opposite side lies on another parallel line. The perpendicular distance between these two lines is the height.
Q2. Does the slant of the parallelogram affect the area?
No. As long as the base and height (distance between parallels) are the same, the area is the same regardless of slant.
Q3. Does this theorem apply to triangles too?
Yes. Triangles on the same base and between the same parallels have equal area. The area is half that of the corresponding parallelogram.
Q4. What if the parallelograms are on different bases?
The theorem does not apply. They will have equal area only if base × height is the same for both.
