Here we will prove that the area of a triangle is half that of a parallelogram on the same base and between the same parallels.
Given: PQRS is a parallelogram and PQM is a triangle with the same base PQ, and are between the same parallel lines PQ and SR.
To prove: ar(∆PQM) = \(\frac{1}{2}\) × ar(Parallelogram PQRS).
Construction: Draw MN ∥ SP which cuts PQ at N.
Proof:
Statement |
Reason |
1. SM ∥ PN |
1. SR ∥ PQ being opposite sides of the parallelogram PQRS. |
2. SP ∥ MN |
2. By construction |
3. PNMS is a parallelogram |
3. By definition of parallelogram because of statements 1 and 2. |
4. ar(∆PNM) = ar(∆PSM) |
4. PM is a diagonal of the parallelogram PNMS. |
5. 2ar(∆PNM) = ar(∆PSM) + ar(∆PNM) |
5. Adding the same area on both sides of equality in statement 4. |
6. 2ar(∆PNM) = ar(parallelogram PNMS) |
6. By addition axiom of area. |
7. MN ∥ RQ |
7. A line parallel to one of the two parallel lines, is also parallel to the other line. |
8. MNQR is a parallelogram. |
8. Similar to statement 3. |
9. 2ar(∆MNQ) = ar(parallelogram MNQR) |
9. Similar to statement 6. |
10. 2{ar(∆PNM) + ar(∆MNQ)} = ar(parallelogram PNMS) + ar(parallelogram MNQR) |
10. Adding statements 6 and 9. |
11. 2ar(∆PQM) = ar(parallelogram PQRS), that is ar(∆PQM) = \(\frac{1}{2}\) × ar(parallelogram PQRS). (Proved) |
11. By addition axiom of area. |
Corollaries:
(i) Are of a triangle = \(\frac{1}{2}\) × base × altitude
(ii) If a triangle and a parallelogram have equal bases and are between the same parallels then ar(triangle) = \(\frac{1}{2}\) × ar(parallelogram)
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