Triangles on the Same Base and between the Same Parallels are Equal in Area
Here we will prove that triangles on the same base and between the same parallels are equal in area.
Given: PQR and SQR are two triangles on the same base QR and are between the same parallel lines QR and MN, i.e., P and S are on MN.
To prove: ar(∆PQR) = ar(∆SQR).
Construction: Draw QM RP cutting MN at M.
Proof:
|
Statement |
Reason |
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1. QRPM is a parallelogram. |
1. MP ∥ QR and QM ∥ RP by construction. |
|
2. ar(∆PQR) = \(\frac{1}{2}\) × ar(parallelogram QRPM). ar(∆SPQ) = \(\frac{1}{2}\) × ar(parallelogram QRPM). |
2. Area of a triangle = \(\frac{1}{2}\) × area of a parallelogram, on the same base, and between the same parallels. |
|
3. ar(∆PQR) = ar(∆SQR). (Proved) |
3. From statements in 2. |
Corollaries:
(i) Triangles with equal bases and between the same parallels are equal in area.
(ii) If two triangles have equal bases, ratio of their areas = ratio of their altitudes.
(iii) If two triangles have equal altitudes, ratio of their areas = ratio of their bases.
(iv) A median of a triangle divides the triangle in two triangles of equal area.
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