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. |
From The Area of a Rhombus is Equal to Half the Product of its Diagonals to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.