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Re: The area of a triangular region with perimeter 8 [#permalink]
2
jezzsk8 wrote:
How do you determine the maximum value of the base is 4 for the triangle rule to be valid? Like is it always p/2 and if so why?



IMPORTANT RULE: If two sides of a triangle have lengths A and B, then . . .
(A-B) < third side < (A+B)

SInce the perimeter is 8, the triangle rule to satisfy if we consider integer value is 3.

Here the base is 2 and the other two sides is 3, here it satisfy the triangle equation.

Now area = 1/2 *2 * 3 (since the maximum value of altitude < 3)

=3

But the sides can also be fraction and still the sides can be add upto = 8.

So I considered the maximum value of the base to be 4 and still it is < 8.

Hence B >A .

If you are unclear let me know.
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Re: The area of a triangular region with perimeter 8 [#permalink]
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Carcass wrote:
Quantity A
Quantity B
The area of a triangular region with perimeter 8
8


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.



Answer is B

As the area of triangle is maximum when base and height has 90 degree angle.Now three different types of triangles are possible.
1)45-45-90
2)30-60-90
3)x-y-90

case 1= sides are ; 1x-1x-under root 2
sum=3.4x=8
x=2.35
area:1/2 *2.35*2.35 =2.76


Case 2: sides are; 1x-under root 3x-2x
sum of sides =x+under root 3x+2x= 8
4.7x=8
x=1.7
area =1/2 *1.7*2.89=2.46

similarly case 3
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Re: The area of a triangular region with perimeter 8 [#permalink]
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