Parallelograms on Same Base and Between Same Parallels
This topic establishes a powerful result: parallelograms on the same base and between the same parallel lines have equal area, regardless of their shape or how slanted they are.
This theorem is a cornerstone of NCERT Class 9 Chapter: Areas of Parallelograms and Triangles. It leads to several important corollaries about triangles, including the result that a median divides a triangle into two equal areas.
The key insight is that area depends only on the base and the perpendicular distance between the parallels (the height), not on the position of the sides.
What is Parallelograms on Same Base and Between Same Parallels?
Definition: Two parallelograms are said to be on the same base and between the same parallels if:
- They share a common base (one side of each parallelogram is the same line segment).
- Their opposite sides (parallel to the base) lie on the same line (i.e., both are enclosed between the same pair of parallel lines).
Theorem: Parallelograms on the same base and between the same parallels are equal in area.
Important:
- The parallelograms need NOT be congruent (same shape) — only their areas are equal.
- The base is common, and the height (perpendicular distance between the parallels) is the same.
- Area of parallelogram = base × height. Since both the base and height are the same, the areas are equal.
Parallelograms on Same Base and Between Same Parallels Formula
Key Formulas and Results:
1. Area of a parallelogram:
Area = base × height
2. Main theorem:
- If parallelograms ABCD and ABEF share base AB and lie between parallels AB and DE, then:
- Area(ABCD) = Area(ABEF)
3. Parallelogram and rectangle on same base:
- A rectangle is a special parallelogram.
- A parallelogram and a rectangle on the same base and between the same parallels have equal area.
4. Triangle on same base (related result):
- A triangle and a parallelogram on the same base and between the same parallels:
- Area of triangle = (1/2) × Area of parallelogram
5. Two triangles on same base and between same parallels:
- They have equal area.
6. Median divides triangle into equal areas:
- A median of a triangle divides it into two triangles of equal area.
Derivation and Proof
Proof of the Theorem: Parallelograms on the same base and between the same parallels have equal area.
Given: Parallelograms ABCD and ABEF on the same base AB, lying between parallels AB and FC (where D, E lie on line FC).
To prove: Area(ABCD) = Area(ABEF)
Proof:
- In triangles ADF and BCE:
- ∠DAF = ∠CBE (corresponding angles, AD ∥ BC and AF ∥ BE, with transversal FC)
- ∠AFD = ∠BEC (corresponding angles, AF ∥ BE, transversal FC)
- AD = BC (opposite sides of parallelogram ABCD are equal)
- By AAS congruence: triangle ADF ≅ triangle BCE.
- Therefore: Area(ADF) = Area(BCE).
Now:
- Area(ABCD) = Area(ABED) + Area(BCE)
- Area(ABEF) = Area(ABED) + Area(ADF)
- Since Area(ADF) = Area(BCE):
- Area(ABCD) = Area(ABEF). Proved.
Alternative explanation:
- Both parallelograms have the same base AB.
- Both have the same height (perpendicular distance between the two parallel lines).
- Area = base × height is the same for both.
Types and Properties
Types of problems involving this theorem:
1. Equal areas of parallelograms
- Given two parallelograms on the same base and between the same parallels, establish or use the fact that their areas are equal.
2. Parallelogram and rectangle comparison
- A rectangle and a parallelogram with the same base and height have equal area.
- This is a special case of the theorem.
3. Triangle area = half of parallelogram area
- A triangle and a parallelogram on the same base and between the same parallels: area of triangle = (1/2) × area of parallelogram.
- Triangles on the same base and between the same parallels have equal area.
5. Median divides triangle into equal areas
- Application of the equal-area triangles result.
6. Proof-based problems
- Prove that certain regions have equal area using the theorem and its corollaries.
Solved Examples
Example 1: Example 1: Equal area parallelograms
Problem: Parallelograms ABCD and ABEF are on the same base AB = 10 cm and between the same parallels. The height between the parallels is 6 cm. Find their areas.
Solution:
Area of ABCD:
- Area = base × height = 10 × 6 = 60 cm²
Area of ABEF:
- By the theorem, Area(ABEF) = Area(ABCD) = 60 cm²
Answer: Both parallelograms have area = 60 cm².
Example 2: Example 2: Parallelogram and rectangle
Problem: A rectangle ABCE and a parallelogram ABCD are on the same base AB = 12 cm and between the same parallels. The height is 5 cm. Show their areas are equal.
Solution:
Area of rectangle ABCE:
- Area = length × breadth = 12 × 5 = 60 cm²
Area of parallelogram ABCD:
- Area = base × height = 12 × 5 = 60 cm²
Answer: Both areas = 60 cm². A rectangle is a parallelogram, so the theorem applies directly.
Example 3: Example 3: Triangle and parallelogram
Problem: A triangle ABP and a parallelogram ABCD share base AB and lie between the same parallels AB and CD. If the area of the parallelogram is 80 cm², find the area of the triangle.
Solution:
By the corollary:
- Area of triangle = (1/2) × Area of parallelogram
- Area of triangle = (1/2) × 80 = 40 cm²
Answer: Area of triangle ABP = 40 cm².
Example 4: Example 4: Equal area triangles
Problem: Triangles ABC and ABD are on the same base AB and between the same parallels AB and CD. If the area of triangle ABC is 36 cm², find the area of triangle ABD.
Solution:
By the theorem:
- Triangles on the same base and between the same parallels have equal area.
- Area(ABD) = Area(ABC) = 36 cm²
Answer: Area of triangle ABD = 36 cm².
Example 5: Example 5: Median divides triangle into equal areas
Problem: In triangle ABC, D is the mid-point of BC. Prove that Area(ABD) = Area(ACD).
Solution:
Given: D is the mid-point of BC, so BD = DC.
Proof:
- Triangles ABD and ACD have the same base lengths (BD = DC).
- They share the same height: the perpendicular from A to line BC.
- Area(ABD) = (1/2) × BD × h
- Area(ACD) = (1/2) × DC × h
- Since BD = DC: Area(ABD) = Area(ACD).
Answer: The median AD divides triangle ABC into two triangles of equal area.
Example 6: Example 6: Finding height between parallels
Problem: Two parallelograms on the same base of 15 cm have an area of 90 cm² each. Find the distance between the parallel lines.
Solution:
Using Area = base × height:
- 90 = 15 × h
- h = 90/15 = 6 cm
Answer: The distance between the parallels = 6 cm.
Example 7: Example 7: Proving areas using the theorem
Problem: ABCD is a parallelogram. E is any point on AB. Prove that Area(ΔDEC) = (1/2) × Area(ABCD).
Solution:
Proof:
- Triangle DEC and parallelogram ABCD share the same base DC.
- E lies on AB, which is parallel to DC (since ABCD is a parallelogram).
- Therefore, triangle DEC and parallelogram ABCD are between the same parallels AB and DC.
- Area(ΔDEC) = (1/2) × Area(ABCD).
Proved.
Example 8: Example 8: Comparing areas of different regions
Problem: In parallelogram ABCD, E and F are points on DC. Prove that Area(ΔAEB) = Area(ΔAFB).
Solution:
Proof:
- Both triangles AEB and AFB have the same base AB.
- Both E and F lie on DC, which is parallel to AB.
- Therefore, both triangles are between the same parallels AB and DC.
- By the theorem: triangles on the same base and between the same parallels have equal area.
- Area(ΔAEB) = Area(ΔAFB). Proved.
Example 9: Example 9: Area ratio problem
Problem: ABCD is a parallelogram with area 120 cm². M is the mid-point of AB. Find Area(ΔDMC).
Solution:
Method:
- ΔDMC has base DC (= AB, opposite sides of parallelogram).
- M lies on AB, which is parallel to DC.
- ΔDMC and parallelogram ABCD share base DC and are between parallels AB and DC.
- Area(ΔDMC) = (1/2) × Area(ABCD) = (1/2) × 120 = 60 cm²
Note: The position of M on AB does not matter; any triangle with base DC and vertex on AB has the same area.
Answer: Area(ΔDMC) = 60 cm².
Example 10: Example 10: Diagonals divide parallelogram into equal triangles
Problem: Prove that the diagonals of a parallelogram divide it into four triangles of equal area.
Solution:
Given: Parallelogram ABCD. Diagonals AC and BD intersect at O.
Proof:
- Diagonal AC divides ABCD into triangles ABC and ACD of equal area (same base and height).
- O is the mid-point of both diagonals (diagonals of a parallelogram bisect each other).
- In ΔABC: BO is a median (O is mid-point of AC). So Area(ΔAOB) = Area(ΔBOC).
- In ΔACD: DO is a median (O is mid-point of AC). So Area(ΔAOD) = Area(ΔDOC).
- Area(ΔABC) = Area(ΔACD) ⇒ Area(ΔAOB) + Area(ΔBOC) = Area(ΔAOD) + Area(ΔDOC)
- Since each half is equal: all four triangles — AOB, BOC, COD, DOA — have equal area.
Each triangle has area = (1/4) × Area of parallelogram.
Real-World Applications
Applications:
- Land measurement: Comparing areas of differently-shaped plots that share the same base and are between the same boundaries.
- Architecture: Understanding that slanted and upright parallelogram-shaped panels between the same supports have equal area for material estimation.
- Proof techniques: The theorem is used extensively in geometric proofs to establish area equalities without computing actual areas.
- Coordinate geometry: Verifying area relationships between parallelograms and triangles on coordinate planes.
- Art and design: Tiling and tessellation patterns rely on equal-area parallelograms.
- Physics: Vector parallelograms in force diagrams — equal area implies equal magnitude of cross product.
Key Points to Remember
- Parallelograms on the same base and between the same parallels have equal area.
- The parallelograms need not be congruent — only their areas are equal.
- Area = base × height. Same base and same height guarantee equal area.
- A triangle on the same base and between the same parallels has half the area of the parallelogram.
- Triangles on the same base and between the same parallels have equal area.
- A median divides a triangle into two triangles of equal area.
- The diagonals of a parallelogram divide it into four triangles of equal area.
- A rectangle is a special case of a parallelogram; the theorem applies to rectangles as well.
- The proof uses AAS congruence of the extra triangles formed at the ends.
- This theorem forms the foundation for many area-related proofs in geometry.
Practice Problems
- Parallelograms PQRS and PQTV are on the same base PQ = 14 cm and between the same parallels. If the height is 8 cm, find the area of each.
- A triangle and a parallelogram share a base of 10 cm and are between the same parallels 7 cm apart. Find both areas.
- In parallelogram ABCD, E is any point on CD. Prove that Area(ΔABE) = (1/2) × Area(ABCD).
- ABCD is a parallelogram with area 72 cm². Its diagonals intersect at O. Find the area of each of the four triangles formed.
- In triangle ABC, D is the mid-point of BC. If Area(ABC) = 48 cm², find Area(ABD).
- Two triangles are on the same base of 12 cm and between parallels 5 cm apart. Find the area of each triangle.
- Prove that triangles on the same base and between the same parallels have equal area.
- ABCD is a parallelogram. P is a point inside ABCD. Prove that Area(ΔAPB) + Area(ΔCPD) = Area(ΔAPD) + Area(ΔBPC).
Frequently Asked Questions
Q1. What does 'same base and between same parallels' mean?
Two figures are on the same base if they share one side. They are between the same parallels if the shared base is one parallel line and the opposite sides/vertices of both figures lie on the other parallel line.
Q2. Are the parallelograms congruent?
No. The parallelograms on the same base and between the same parallels have equal area but are generally NOT congruent. They can have different shapes (different angles and side lengths).
Q3. Why do they have equal area?
Area of a parallelogram = base × height. Both parallelograms share the same base and the same height (the perpendicular distance between the parallels). So their areas must be equal.
Q4. What is the relationship between a triangle and a parallelogram on the same base?
If a triangle and a parallelogram share the same base and are between the same parallels, the area of the triangle is exactly half the area of the parallelogram.
Q5. Do two triangles on the same base and between the same parallels have equal area?
Yes. Two triangles on the same base and between the same parallels have equal area. This follows from the fact that Area = (1/2) × base × height, and both base and height are the same.
Q6. How does a median divide a triangle?
A median divides a triangle into two triangles of equal area. The two smaller triangles share the same height (from the vertex to the base) and have equal bases (the median bisects the opposite side).
Q7. Is this theorem in the CBSE Class 9 syllabus?
Yes. This theorem and its corollaries are part of CBSE Class 9 Mathematics, Chapter: Areas of Parallelograms and Triangles.
Q8. How do diagonals divide a parallelogram?
The diagonals of a parallelogram divide it into four triangles of equal area. Each triangle has area equal to one-quarter of the parallelogram's area.
