Given that ABCD is a rectangle, is the area of triangle ABE>

Given that ABCD is a rectangle, is the area of triangle ABE > 25? (Note: Figure above is not drawn to scale).
\(Area=\frac{1}{2}*AB*BE\)

(1) AB = 6 --> clearly insufficient: BE can be 1 or 100.

(2) AE = 10 --> now, you should know one important property: for a given length of the hypotenuse a right triangle has the largest area when it's isosceles, so for our case area of ABE will be maximized when AB=BE. So, let's try what is the largest area of a right isosceles triangle with hypotenuse equal to 10. Finding legs: \(x^2+x^2=10^2\) (where x=AB=BE) --> \(x=\sqrt{50}\) --> \(area_{max}=\frac{1}{2}\sqrt{50}^2=25\). Since it's the maximum area of ABE then the actual area cannot be more than 25. Sufficient.

Answer: B.
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Hi, and welcome to GMAT Club.

Unfortunately your reasoning is nor correct.

You assume with no ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 10 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 6:8:10. Or in other words: if \(a^2+b^2=10^2\) DOES NOT mean that \(a=6\) and \(b=8\), certainly this is one of the possibilities but definitely not the only one. In fact \(a^2+b^2=10^2\) has infinitely many solutions for \(a\) and \(b\) and only one of them is \(a=6\) and \(b=8\).

For example: \(a=1\) and \(b=\sqrt{99}\) or \(a=2\) and \(b=\sqrt{96}\) or \(a=4\) and \(b=\sqrt{84}\) ...

Hope it's clear.
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Hi,

B states that AE = 10 and triangle ABE is a right triangle. So it makes it a special case "side-based" right triangle where one of the lengths of the sides form ratios of whole numbers, such as 3 : 4 : 5.

Side AE = 10, which means that side AB = 6 and side BE = 8 (ratio 6:8:10 = ratio 3:4:5). Now knowing the sides, you can easily calculate the area which equals 24 < 25.

This is what's so great about the forum. One's faulty assumptions get checked in time. In this case, I had also fallen into the trap of thinking that since hypotenuse is 10 the other sides are 8 and 6. As Bunuel points out, that's clearly the wrong way to think about this.

And knowing the isosceles-right triangle property certainly helps!

F.S 1 is clearly Insufficient.

Another approach for F.S 2 :

We know that \(a^2+c^2 = 10^2 \to a^2+c^2 = 100\)

Also, area of \(\triangle\) ABE - \(\frac{1}{2}*a*c\)

Is\(\frac{1}{2}*a*c>25 \to\) Is \(a*c>50 \to 2*a*c>100?\)

Is \(2*a*c>a^2+c^2 \to\) Is \((a-c)^2<0\). Of-course, the answer is NO.Sufficient.

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Hi Bunnel,
Why cant the reasoning that a right triangle has greatest area when it is isosceles be applied to the first statement as well. Which says AB = 6, hence assuming BE = 6 we would get the area = 1/2*6*6 = 18 < 25

The property says: for a given length of the hypotenuse a right triangle has the largest area when it's isosceles. Thus you cannot apply it to the first statement.
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Bunuel, have you encountered real gmat questions testing this concept: for a given length of the hypotenuse a right triangle has the largest area when it's isosceles ?

Hi Ergenekon,

This is a rarer concept (from the realm of Multi-Shape Geometry), but the GMAT has been known to test it.

The broader issue is more about comparing squares and rectangles though.

For example, compare the areas of this square and rectangles....

10x10
9x11
8x12

Areas:
(10)(10) = 100
(9)(11) = 99
(8)(12) = 96

By increasing one side and decreasing the other by an "equivalent amount", the area decreases.

When it does appear on the GMAT, it's often themed around 'percentage change' in side lengths (re: length is 10% greater, width is 10% less), but the pattern is still the same.

GMAT assassins aren't born, they're made,
Rich

For any right angled triangle the hypotenuse is the largest side and not only isosceles right?

Owing to the convention that the longest side is against the largest angle ?

Posted from my mobile device

Yes. The largest side is always against the largest angle. So, in any right triangle hypotenuse is the largest side.
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Bunuel A right-angled triangle has a maximum area when the triangle is isosceles. Similarly, An isosceles triangle has a maximum area when it is a right-angled triangle. So from (1) we have AB = 6. Let's assume BE = AB = 6 (Making it an isosceles triangle). Thus the are would be 1/2 x 6 x 6 = 18 < 25. Why is this thought process incorrect

The property says: for a given length of the hypotenuse a right triangle has the largest area when it's isosceles. So, you can apply this property when the length of the hypotenuse is known. Thus you cannot apply it to the first statement.
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Hi Hoozan,

In this particular DS question, there are a couple of things that should stand-out:

First, since we're dealing with a DS question, the picture is not necessarily drawn 'to scale' (meaning that side AB of the rectangle might be considerably shorter OR longer than side BC (and by extension, segment BE). Second, with the information in Fact 1, we only have one of the two legs of a right triangle, so we have no way of determining its exact area (nor how big that area can become). As segment BE gets smaller, the area of the triangle becomes smaller - and as BE gets bigger, the area of the triangle becomes bigger.

With the information in Fact 1, while it's certainly possible that triangle ABE is an isosceles, right triangle (with an area of 18), we have NO proof that that is actually the case. Again, as BE gets bigger, the area would increase above 18).

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: Rich.C@empowergmat.com

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