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# Triangle-Area Problem

Through an arbitrary point lying inside a triangle,3 straight lines parallel to its sides are drawn. These lines divide the triangle into six parts ,3 of which are triangles. If the areas of these triangles are A,B,C,then what is the area of the initially given triangle?(obviously area is to found in terms of A,B and C)

Note by Biswaroop Roy
3 years, 8 months ago

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The answer that I got was 3(A+B+C) because think about it. The way this question is posed, it must hold true for all triangles. If we let the triangle be an equilateral triangle, and let the intersection of these three lines be the center (remember, this is to make things easier), then we produce 3 rhombi and 3 triangles (like a hazardous sign). If we split the rhombi into equilateral triangles, you'll notice that we have 9 equal equilateral triangles, 3 of which are A, B, and C. This means that there are 3 times their sum. If you want to get down and dirty with the math, use Heron's formula and try messing around with that. Hope this helps! · 3 years, 8 months ago

The issue with assuming a specific case, is that your answer only works for that specific case, and you do not know what happens in the other cases. For example, we could get the same value using functions like $$\frac{ A^2 + B^2 + C^2 } { A+B+C}$$ or $$\frac{ 9ABC} { AB + BC + CA }$$.

You could do some further checks by taking other points, or looking at other triangles. For example, if you consider the point to be at a vertex, then $$3(A+B+C)$$ is clearly the wrong answer. Staff · 3 years, 8 months ago

Answer is not 3(A+B+C). Originally the question had 4 options And obviously an option was 3(A+B+C) Needless to say I thought the answer was that.Surprising the answer turned out to be (sq.root(A)+sq.root(B)+sq.root(C))^2. I still haven't got this answer.And I am still working on it. You might notice that this answer also holds true for symmetric figures like equilateral triangle. · 3 years, 8 months ago

Hint: Similar Triangles. Use this to find the areas of the 'medium triangles' and hence the parallelograms. Staff · 3 years, 8 months ago

Thanks. · 3 years, 7 months ago

Hmm... Nevermind then... · 3 years, 8 months ago

Some one please try this one. I am working on it for quite some time now,and I would be glad if some one solves it. · 3 years, 8 months ago