If the area of rectangle $ABCD$ with integral sides is $48$ sq.cm. If $E$ be any point on AD. How to find the sum of maximum and minimum area (sq.cm) of $\triangle BCE$?
How to find the sum of maximum and minimum area of a triangle within a rectangle?
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geometry
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0Are $A$, $B$, $C$, $D$ in counterclockwise (or clockwise) order? Then the question is very easy. – 2011-10-21
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0Hint: draw a diagram. If necessary remember how to compute the area of a triangle and a rectangle. Note the answer will be different if AD is a diagonal, rather than a side as is indicated by the way that the question is posed. – 2011-10-21
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0@André Nicolas:I am not sure. – 2011-10-21
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1@André: I’d be willing to make a small bet that $\overline{AD}$ is a side of the rectangle. That’s a cute little question. – 2011-10-21
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1It is usual for the vertices to be named in order. If you draw the diagram as Mark Bennet suggests it will help. There is also a big hint in the fact that you are not given the dimensions of $ABCD$. Is it $6 \times 8, 1 \times 48$ or what? – 2011-10-21
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0@FoolForMath: If $AD$ is a diagonal of the rectangle, then you can choose $E$ so that area of triangle is as close to $0$ as you want, so there is no minimum unless you allow "degenerate" triangles of area $0$. If instead $AD$ is a side, then the answer is instantaneous. – 2011-10-21
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0@AndréNicolas: if you allow the degenerate answer $0$, the diagonal version is only twice the work of the one where AD is a side, and I would think Brian M. Scott has the right call on that. – 2011-10-21
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0@Ross Millikan,`@AndréNicolas`,`@Mark Bennet`,`@Brian M. Scott`:So,for the usual clockwise or anticlockwise notations,the answer seems to me is $48$ sq.cm.Since the areas of two or more triangle lying between the same parallel lines will be equal,so here we would have equal maximum and minimum area for $\triangle BCE$. – 2011-10-21
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0@FoolForMath: I agree. – 2011-10-21
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0@FoolForMath: Yes, that is the reason. Why don't you write it up as an answer? – 2011-10-21
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0@André Nicolas:Done! :) – 2011-10-21
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0@FoolForMath: I upvoted. However, there should be editing. The statement about lying between the same parallel lines is not true if the "bases" have different sizes. Might as well look carefully at your overall wording, and take the opportunity to fix the couple of instances of x,y to $x$, $y$ or $x, y$, whichever looks good. – 2011-10-21
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0@André Nicolas:Aha yes I forgot to add the "same base" constraint.and please feel free to edit the answer further based on your understanding,I would never doubt it:) – 2011-10-21
1 Answers
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Assuming the usual clockwise or anticlockwise notations.
Since we know that the areas of two or more triangle having the same base and lying between the same parallel lines will be equal, so here we would have equal maximum and minimum area for $\triangle BCE$.
Again, the area of a triangle is given by $\frac{1}{2}\times x \times y$ where $x$,$y$ are the sides of rectangle. Here the constraint is $x \times y = 48 $ sq.cm. So choosing any value of $x,y$ satisfying this constraint would give the same area for any $\triangle BCE$.
Hence, in this case, the maximum area + minimum area = $48$ sq.cm.
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3I think you can even post it as a non-CW answer. The site explicitly permits (and encourages?) that. – 2011-10-21