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)





9th Grade Math

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