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 |
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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