Area of a Parallelogram is Equal to that of a Rectangle Between the Same Parallel Lines
Here we will prove that the area of a parallelogram is equal to that of a rectangle on the same base and of the same altitude, that is between the same parallel lines.
Given: PQRS is a parallelogram and PQ MN is a rectangle on the same base PQ and between the same parallel lines PQ and NR
To prove: ar(Parallelogram PQRS) = ar(Rectangle PQMN)
Proof:
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Statement |
Reason |
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1. PS = QR |
1. Opposite sides of the parallelogram PQRS. |
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2. PN = QM |
2. Opposite sides of the rectangle PQMN. |
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3. ∠PNS = ∠QMR |
3. Both are right angles, PQMN being a rectangle. |
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4. ∆PNS ≅ ∆QMR |
4. By RHS axiom of congruency. |
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5. ar(∆PNS) = ar(∆QMR) |
5. By area axiom for congruent figures. |
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6. ar(∆PNS) + ar(Quadrilateral PQMS) = ar(∆QMR) + ar(Quadrilateral PQMS) |
6. Adding the same area on both sides of the equality in the statement 5. |
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7. ar(Rectangle PQMN) = ar(Parallelogram PQRS). (Proved) |
7. By adding axiom of area. |
Corollaries:
(i) Area of a parallelogram = Base × Height,
because ar(Parallelogram PQRS) = ar(Rectangle PQMN)
= PQ × MQ
= Base × Height.
(ii) Parallelograms with equal base and between the same parallels have the same area.
Here PQRS and MNRS are two parallelograms whose bases PQ and MN are equal, and they are between the same two parallel lines PN and SR. So, the two parallelograms have equal height.
Using ar(Parallelogram) = Base × Height, we find their areas are equal.
(iii) The ratios of the areas of two parallelograms that are between the same parallel lines (that is heights are equal) = Ratio of their bases.
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