Here we will prove that triangles with equal areas on the same base have equal corresponding altitudes (or are between the same parallels).
Given: PQR and SQR are two triangles on the same base QR, and ar(∆PQR) = ar(∆SQC). Also, PN and SM are their corresponding altitudes.
To prove: PN = SM (or PS ∥ QR).
Construction: Join PS.
Proof:
Statement |
Reason |
1. \(\frac{1}{2}\) × QR × PN = \(\frac{1}{2}\) × QR × SM. |
1. Are of a triangle = \(\frac{1}{2}\) × base × altitude, and ar(∆PQR) = ar(∆SQR). |
2. PN = SM. |
2. Cancelling \(\frac{1}{2}\) × QR from statement 1. |
3. PN ∥ SM. |
3. PN ⊥ QR and SM ⊥ QR. |
4. PNMS is a rectangle. |
4. PMNS is a parallelogram by statements 2 and 3, and two angles are right angles. |
5. PN = SM (or PS ∥ QR). (Proved) |
5. By statement 4, PNMS is a rectangle. |
Corollary: Parallelograms with equal area on the same base have equal corresponding altitudes (or are between the same parallels).
Here, ar(parallelogram PQRS) = ar(parallelogram PQMN)
Therefore, ar(∆PRQ) = ar(∆PNQ)
Therefore, RN ∥ PQ. But RS ∥ PQ, NM ∥ PQ.
Therefore, RN ∥ RS and RN ∥ NM
Having common point (R or N), all the lines are coincident.
Therefore, the parallelogram have equal altitudes.
From Triangles with Equal Areas on the Same Base have Equal Corresponding Altitudes 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.
May 19, 24 03:36 PM
May 19, 24 03:19 PM
May 19, 24 02:23 PM
May 19, 24 01:26 PM
May 19, 24 10:42 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.