Subscribe to our YouTube channel for the latest videos, updates, and tips.
Home | About Us | Contact Us | Privacy | Math Blog
Here we will prove that parallelogram on the same base and between the same parallel lines are equal in area.
Given: PQRS and PQMN are two parallelograms on the same base PQ and between same parallel lines PQ and SM.
To prove: ar(parallelogram PQRS) = ar(parallelogram PQMN).
Construction: Produce QP to T.
Proof:
Statement |
Reason |
1. PS = QR. |
1. Opposite sides of the parallelogram PQRS. |
2. PN = QM. |
2. Opposite sides of the parallelogram PQMN. |
3. ∠SPT = ∠RQT. |
3. Opposite sides PS and QR are parallel and TPQ is a transversal. |
4. ∠NPT = ∠MQT. |
4. Opposite sides PN and QM are parallel and TPQ is a transversal. |
5. ∠NPS = ∠MQR. |
5. Subtracting statements 3 and 4. |
6. ∆PSN ≅ ∆RQM |
6. By SAS axiom of congruency. |
7. ar(∆PSN) ≅ ar(∆RQM). |
7. By area axiom for congruent figures. |
8. ar(∆PSN) + ar(quadrilateral PQRN) = ar(∆RQM) + ar(quadrilateral PQRN) |
8. Adding the same area on both sides of the equality in statement 7. |
9. ar(parallelogram PQRS) = ar(parallelogram PQMN). (Proved) |
9. By addition axiom for area. |
From Parallelogram on the Same Base and Between the Same Parallel Lines 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.
Jul 11, 25 02:14 PM
Jul 09, 25 01:29 AM
Jul 08, 25 02:32 PM
Jul 08, 25 02:23 PM
Jul 08, 25 09:55 AM
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.