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Example 2 - If a triangle and a parallelogram are on same - Paralleograms & triangles with same base & same parallel lines

  1. Chapter 9 Class 9 Areas of parallelograms and Triangles
  2. Serial order wise
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Example 2 (Method 1) If a triangle and a parallelogram are on the same base and between the same parallels, then prove that area of triangle is equal to half the area of parallelogram. Given: A parallelogram ABCD and ∆ ABP on the same base AB and between the same parallels To prove: Area of triangle is equal to half the area of parallelogram. ar (∆ ABP ) = 1/2 ar (ABCD) Construction: Join DP Let DM ⊥ AB & PN ⊥ AB Proof: Example 2 (Method 2) If a triangle and a parallelogram are on the same base and between the same parallels, then prove that area of triangle is equal to half the area of parallelogram. Given: A parallelogram ABCD and ∆ ABP on the same base AB and between the same parallels PC & AB To prove: Area of triangle is equal to half the area of parallelogram. ar (∆ ABP ) = 1/2 ar (ABCD) Proof: In parallelogram ABCD, AB ∥ CD So, PC ∥ AB We draw a line BQ parallel to AP , i.e. , BQ ∥ AP Since BQ ∥ AP & PQ ∥ AB, Both pairs of opposite sides are parallel ∴ ABQP is a parallelogram Parallelograms ABQP & ABCD are on the same base AB and between the same parallel lines AB & PC ∴ Area(ABQP) = Area(ABCD) In parallelogram ABQP, BP is the diagonal So, ∆ ABP ≅ ∆ QBP ⇒ Area(∆ ABP) = Area(∆ QBP) Now, Area(∆ ABP) = Area(∆ QBP) = 1/2 Area(ABQP) ⇒ Area(∆ ABP) = 1/2 Area(ABQP) ⇒ Area(∆ ABP) = 1/2 Area(ABCD) Hence proved

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