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Right Triangle with GREATEST AREA

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by Stockmoose16 » Wed Oct 22, 2008 12:08 pm
I was doing an MGMAT challenge question, and I came across the following statement:

The right triangle with the largest area will be an isosceles right triangle (where both the base and height are of equal length).

This doesn't make sense. A 30-60-90 right triangle would have an area of (X^2 * sqrt 3)/2, which is clearly greater than an isoceles right triangle (45-45-90), which would have an area of just (x^2)/2

Can anyone explain?
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by gmat009 » Wed Oct 22, 2008 12:15 pm
30-60-90 right triangle

sides will be x:(sqrt 3)x:2x

A1= 1/2 *((sqrt 3)x)*x

45-45-90

sides will be x:x:((sqrt 2)x)

A2= 1/2 *x*x

A1>A2
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by Stockmoose16 » Wed Oct 22, 2008 12:20 pm
gmat009 wrote:30-60-90 right triangle

sides will be x:(sqrt 3)x:2x

A1= 1/2 *((sqrt 3)x)*x

45-45-90

sides will be x:x:((sqrt 2)x)

A2= 1/2 *x*x

A1>A2
That's the same thing I showed above.

For A1: 1/2 * (sq rt 3 * X) *X = (X^2 * sqrt 3)/ 2, which would certainly be GREATER THAN A2.

According to the MGMAT question, ISOCELES TRIANGLES have the greatest area of all right triangles. You just proved that 30-60-90 triangles have the greatest area of all right triangles.
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Re: Right Triangle with GREATEST AREA

by Stockmoose16 » Wed Oct 22, 2008 12:36 pm
Stockmoose16 wrote:I was doing an MGMAT challenge question, and I came across the following statement:

Given a hypotenuse, The right triangle with the largest area will be an isosceles right triangle (where both the base and height are of equal length).

This doesn't make sense. A 30-60-90 right triangle would have an area of (X^2 * sqrt 3)/2, which is clearly greater than an isoceles right triangle (45-45-90), which would have an area of just (x^2)/2

Can anyone explain?
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by stop@800 » Wed Oct 22, 2008 10:19 pm
we are comparing in the incorrects :)


As we are changing base and height, we need ot keep hyp same


the 30 60 90 will be (order P B H)
1 sqrt(3) 2


the 45 90 45 will be
1 1 sqrt(2)
multiply all by sqrt(2)
so we have

sqrt(2) sqrt(2) 2



now compare the areas
1 sqrt(3) 2
it will be
sqrt(3) / 2 = 1.73/2


and for
sqrt(2) sqrt(2) 2

it will be
sqrt(2) * sqrt(2) /2
= 1


so definitely 45 45 90 is greater

Hope this helps

The modified question also contains given a hyp
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