energetics wrote:

A triangular sign was cut from a rectangular piece of paper. If the paper was cut so as to yield a sign of the greatest possible area, then what is the area of the remaining piece (or pieces) of paper?

(1) The rectangular piece of paper originally had dimensions 16 inches by 24 inches.

(2) The height of the triangle is 16 inches.

General question to experts: To double check... If the GMAT says a figure is a rectangle in DS prompt, is it still possible for it to be a square given new information in the statements? To yield the greatest possible area it would be an equilateral triangle in a square vs. a right triangle in a rectangle. So if (1) said the dimensions are 16 by 16 we are still able to figure it out from just 1) but this fact could be very important in another problem, right?

Given a rectangle we can easily calculate the triangle with max area and that will be the two congruent triangles (each of) its diagonal makes. The area of the triangle in that case is 1/2*breadth*length of rectangle. And the area of the remaining piece of paper is also 1/2*breadth*length of rectangle.From 1) We know the dimensions of the rectangle. Sufficient

Please note that given exact dimensions of any shape be it rectangle, circle or square, or any irregular polygon, we can be sure that max area for a triangle is constant as there is only one max area; even though there may be different ways to make that triangle. This is therefore clearly sufficient.

From 2) We just know the height of the triangle. Had we known the base as well, we could have calculated the area of the remaining piece/ pieces of paper.

Because no matter how you cut the triangle, the max area of the triangle as well as the remaining pieces of paper are same and they are equal to 1/2*breadth*length of rectangle. Insufficient.

Therefore answer is A.Please note that for all types of rectangle, even square the max area of the triangle is 1/2*breadth*length.Area of equilateral triangle made from a square (of side measuring "a" each) is as below:

base= a and height=\(\frac{\sqrt{3}a}{2}\)

Therefore, Area of an equilateral triangle cut from a square is= \(\frac{1}{2}*a*\frac{\sqrt{3}a}{2}\)=\(\frac{\sqrt{3}a^2}{4}\)=\(\frac{1.73a^2}{4}\)

Whereas, if we cut the square in two equal pieces through the diagonal, the area of each of the two triangles made is \(\frac{a^2}{2}\) each, which is clearly greater than \(\frac{1.73a^2}{4}\).

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