# Area of a Triangle is Half that of a Parallelogram on the Same Base and between the Same Parallels

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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