prove that two triangles having the same base lying on the same side of the common base and having equal areas lie between the same parallels, one of which contains the same base.
Answers
Proof:
If two triangles have same base length and same area, then their altitudes corresponding to the equal bases will also be equal. Let me show you.
Let the equal area of the two triangles be A and the length of the equal base be b. Let the altitude corresponding to the base of each triangle be h₁ and h₂. Such that.
⇒ A = 1/2 × b × h₁ = 1/2 × b × h₂
⇒ h₁ = h₂ (When we cut 1/2 and b from both sides)
So we proved that the corresponding altitudes to the bases are equal.
The altitudes are equal, which means the perpendicular distance between the third vertex of the each triangle and the equal base are equal. So that when we draw a line passing through these third vertices, it will be parallel to the line in which the base belongs, because any two parallel lines are equidistant from each other. So the triangles will include in two parallel lines.
Hence proved!!!
Plz ask me if you have any doubt because I was not able to explain it clearly at all. Hope this helps you.
Thank you...
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Answer:
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