The Area of a Rhombus is Equal to Half the Product of its Diagonals

Here we will prove that the area of a rhombus is equal to half the product of its diagonals.

Solution:

Given:

PQRS is a rhombus whose diagonals are PR and QS. The diagonals intersect at O.

To prove: ar(rhombus PQRS) = $\frac{1}{2}$ ×PR × QS.

Statement	Reason
1. ar(∆RSQ) = $\frac{1}{2}$ ×Base × Altitude = $\frac{1}{2}$ ×QS × RO.	1. QS ⊥ PR, because diagonals of a rhombus are perpendicular to each other.
2. ar(∆PQS) = $\frac{1}{2}$ ×Base × Altitude = $\frac{1}{2}$ ×QS × PO.	2. As in reason 1.
3. ar(∆RSQ) + ar(∆PQS) = $\frac{1}{2}$ ×QS × (RO + PO).	3. By addition from statements 1 and 2.
4. ar(rhombus PQRS) = $\frac{1}{2}$ ×PR × QS.	4. By addition axiom for area.

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Statement	Reason
1. ar(∆RSQ) = \(\frac{1}{2}\) ×Base × Altitude = \(\frac{1}{2}\) ×QS × RO.	1. QS ⊥ PR, because diagonals of a rhombus are perpendicular to each other.
2. ar(∆PQS) = \(\frac{1}{2}\) ×Base × Altitude = \(\frac{1}{2}\) ×QS × PO.	2. As in reason 1.
3. ar(∆RSQ) + ar(∆PQS) = \(\frac{1}{2}\) ×QS × (RO + PO).	3. By addition from statements 1 and 2.
4. ar(rhombus PQRS) = \(\frac{1}{2}\) ×PR × QS.	4. By addition axiom for area.

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