Parallelograms on the same base and between the same parallel lines have equal area. A triangle is half the parallelogram on the same base and height. Triangles on the same base and between the same parallel lines have equal area. The line joining mid-points of two sides of a triangle is parallel to the third side and half as long. Joining mid-points of any quadrilateral forms a parallelogram (Varignon) of half the area. These theorems are used to prove equal areas of triangles/parallelograms inside figures.
1. Main Theorems
TheoremStatementProof IdeaTheorem 1Parallelograms on same base & between same parallel lines have equal areaCongruent triangles by AASTheorem 2Triangle = ½ parallelogram on same base & same parallelsDiagonal divides parallelogram into 2 equal trianglesTheorem 3Triangles on same base & between same parallel lines have equal areaBoth = ½ same parallelogramMid-point Theorem (Extra important)Line joining mid-points of two sides → parallel to third side & half lengthVector or similar triangles
2. Area Formulas You Must Know
ShapeArea FormulaParallelogrambase × heightTriangle½ × base × heightTrapezium½ × (sum of parallel sides) × heightRhombusbase × height OR ½ × d₁ × d₂
3. Most Important Proof Questions (90% SEE)
- Parallelograms on same base & same height → equal area
- Triangle = ½ parallelogram on same base & height
- Triangles on same base & same height → equal area
- Mid-points E,F of AB,AC → EF || BC and EF = ½ BC
- Varignon Theorem: Joining mid-points of quadrilateral → parallelogram
- Area of parallelogram formed by mid-points = ½ original parallelogram
4. 5 Most Common SEE Questions + Short Solutions
- ABCD parallelogram, E,F mid-points of AB,CD → prove EFGH parallelogram & area = ½ ABCD → EF || AD, FG || AB → EFGH parallelogram → area ½
- Triangles with same base & same height → equal area → Both have same base and same perpendicular distance → area equal
- Parallelogram ABCD, P on diagonal AC → prove area ΔABP = area ΔCDP → Same height from B and D to AC → equal area
- Trapezium ABCD, AB||DC, E mid-point of AD, F mid-point of BC → prove EF || AB || DC & EF = ½(AB+DC) → Mid-line theorem in triangle
- ABCD parallelogram, M mid-point of AB, N mid-point of AD → prove area AMND = ¼ ABCD → MN joins mid-points → MN = ½ CD → area ¼
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