Experimental Verification
Therefore, the area of a triangle is equal to half of the area of the parallelogram on the same base and between the same parallel lines.
Summary
Therefore, the area of a triangle is equal to half of the area of the parallelogram on the same base and between the same parallel lines.
Things to Remember
Therefore, the area of the triangle is equal to half of the area of the parallelogram on the same base and between the same parallel lines.
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Experimental Verification
Theoretical proof:
Given: The triangle ABC and the parallelogram BCDE are on the same base BC and between the same parallel lines BC and AD.
TO prove: Area ofΔABC = \(\frac{1}{2}\) BCDE
Construction: Draw AM⊥CB and EN⊥BC then AM is the height ofΔABC and EN be the height of BCDE.
| S.N. | Statement | S.N. | Reasons |
| 1. | Area of BCDE = base *height = BC * EN | 1. | Area of BCDE = base * height |
| 2. | Area of ABC = \(\frac{1}{2}\) * BC * AM | 2. | Area ofΔ =\(\frac{1}{2}\) base * height |
| 3. | Area of ABC = \(\frac{1}{2}\) * BC * EN | 3. | Perpenicular distance between the same parallel lines are equal, i.e. AM = EN |
| 4. | Area of ABC = \(\frac{1}{2}\) of BCDE | 4. | From the statement (1) BC * EN = Area of BCDE |
Lesson
Geometry
Subject
Compulsory Mathematics
Grade
Grade 10
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