geometry
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Source: Beat The GMAT — Problem Solving |
The way I see it is:
Biggest is a circle
28 / pi = c. 9 (circumference) --> Area therefore c. 62
Second is an equilateral triangle - other shapes of triangles should be smaller
28 / 3 = 9 1/3 --> Area c. 60.5
Third is square
28 / 4 = 7 --> Area is 49
Rectangle is smallest
Range from (using integers) 1x13 = 13 to 6x8 = 48
Biggest is a circle
28 / pi = c. 9 (circumference) --> Area therefore c. 62
Second is an equilateral triangle - other shapes of triangles should be smaller
28 / 3 = 9 1/3 --> Area c. 60.5
Third is square
28 / 4 = 7 --> Area is 49
Rectangle is smallest
Range from (using integers) 1x13 = 13 to 6x8 = 48
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2010gmat
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u re right,..circle will have the max area...
but am not sure about the figure with smallest area...stuck between rectangle and triangle...
u re also right that equilateral triangle will have max area...and thats a rule in general...among quadrilaterals square will have max area...
but am not sure about the figure with smallest area...stuck between rectangle and triangle...
u re also right that equilateral triangle will have max area...and thats a rule in general...among quadrilaterals square will have max area...
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We can construct both rectangles AND triangles such that their areas approach 0.2010gmat wrote:u re right,..circle will have the max area...
but am not sure about the figure with smallest area...stuck between rectangle and triangle...
u re also right that equilateral triangle will have max area...and thats a rule in general...among quadrilaterals square will have max area...
For a rectangle, just let the width get very very very close to zero, and the area will approach zero.
For a triangle, just let the height get very very very close to zero, and the area will approach zero.


















