Hey here we have to use the contradiction. At first take that we have a triangle with integral coordinates then find its area by the determinant method ,where you would get an rational solution. Then find the area by the formula of area of equilateral triangle(i.e.,area=√3÷4×a²). Here you would get an irrational solution. Thats it here you get two different areas for the same triangle.proved by contradiction.
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Rajat Bisht
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2 years, 10 months ago

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@Rajat Bisht
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Waaw awsome solution.........i solved this problem by assuming a triangle with vertices A(a,0) B(-a,0) C(0,b) where a,b are integers......then since triangle is eqilatral so |AB|=|AC| and then by destence formula i got a contradiction
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Aman Sharma
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2 years, 10 months ago

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In an equilateral triangle if two vertices have integers as coordinates then the third vertex won't hve integers as coordinates , they would be irrational nos.
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Archana Kataria
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2 years, 10 months ago

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TopNewestHey here we have to use the contradiction. At first take that we have a triangle with integral coordinates then find its area by the determinant method ,where you would get an rational solution. Then find the area by the formula of area of equilateral triangle(i.e.,area=√3÷4×a²). Here you would get an irrational solution. Thats it here you get two different areas for the same triangle.proved by contradiction. – Rajat Bisht · 2 years, 10 months ago

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– Aman Sharma · 2 years, 10 months ago

Waaw awsome solution.........i solved this problem by assuming a triangle with vertices A(a,0) B(-a,0) C(0,b) where a,b are integers......then since triangle is eqilatral so |AB|=|AC| and then by destence formula i got a contradictionLog in to reply

In an equilateral triangle if two vertices have integers as coordinates then the third vertex won't hve integers as coordinates , they would be irrational nos. – Archana Kataria · 2 years, 10 months ago

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