A question on triangle properties here. From Gmatclub book:For a given perimeter equilateral triangle has the largest area.
For a given area equilateral triangle has the smallest perimeter.
and For an isosceles triangle with given length of equal sides right triangle (included angle) has the largest area.
So what is the biggest triangle? If we have to choose between a equilateral and an isosceles with the same perimeter, which one would have the biggest area?
Or these rules are not interrelated? I'm confused here.. Example when the equilateral rules is used: is-the-perimeter-of-triangle-abc-greater-than-87112.html
Why don't we consider a right isosceles here? Why only the equilateral?Example when the isosceles rule is used: the-points-r-t-and-u-lie-on-a-circle-that-has-radius-4-if-168676.html
Then why not consider an equilateral here?..
I hope you get my doubt.
I'm happy to respond.
My friend, I believe you are confusing a few different things.
First of all, for the problem of the largest area with fixed perimeter, it depends very much on the starting criterion. Among all shapes in the plane, no restrictions, the shape with the most area for a fixed perimeter is the circle
. If we limit ourselves to triangles, it's an equilateral triangle
. If we limit ourselves to quadrilaterals, it's a square
. If we limit it to all n-sided polygons for some specific value of n, it would be the regular polygon
(notice that the equilateral triangle and the square are the regular polygons for the cases n = 3 and n = 4 respectively). This is mathematically true and relatively unlikely to be tested on the GMAT. Admittedly, it did play a role in the first question you cited:
Now, your quoted this rule. For an isosceles triangle with given length of equal sides, the right triangle (included angle) has the largest area
Here, notice the nature of the constraint is very different. We are not fixing the entire perimeter, but instead, we are fixing the lengths of the two equal sides, and the third side can vary to be any possible length. This may seem like only a minor change, but it makes this a profoundly different question. In fact, this is not merely true for an isosceles triangle. I will generalize. Suppose segment AB has one fixed length and segment BC has another fixed length, and we construct triangle ABC; triangle ABC will have the maximum possible area when the angle at vertex B equals 90 degrees.
triangles with varying angle.JPG [ 17.43 KiB | Viewed 396 times ]
This is a mathematical rule that is extremely different from the first rule. The nature of the question is quite different, so the nature of the answer is quite different.
Furthermore, in this question:the-points-r-t-and-u-lie-on-a-circle-that-has-radius-4-if-168676.html
The solution involved an isosceles triangle only because it was known at the outset that two side of the triangle were radii, and hence the triangle must be isosceles. This is always the case for a triangle inside a circle that has two radii as two of its three sides. This has absolutely nothing to do with the rule I just discussed.
Does all this make sense?
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