Data Sufficiency

a ) Is the perimeter of triangle T greater than the perimeter of
square S?

(1) The length of the longest side of T is twice the length
of a side of S.
(2) T is isosceles.


Can you use this logic. The length of the other two sides of a triangle thats not the largest must add up to be greater than the largest sides. Therefore the perimeter is greater than 2 times the largest sides?

Expanding this problem a little bit more. What if we are given the shortest side or the medium sides or both? Can we make any conclusion about the largest side?


b ) Is the perimeter of triangle T greater than the perimeter of
triangle S?

(1) The length of the longest side of T is twice the length of the longest
side of S.
(2) T is isosceles.

yellowho wrote:a ) Is the perimeter of triangle T greater than the perimeter of
square S?

(1) The length of the longest side of T is twice the length
of a side of S.
(2) T is isosceles.


Can you use this logic. The length of the other two sides of a triangle thats not the largest must add up to be greater than the largest sides. Therefore the perimeter is greater than 2 times the largest sides?

Yes, we should use that logic.
P(T) > p(S) ?

T has sides a, b and c (also, a>=b>=c)
S has side length of x

Statement 1:
The length of the longest side of T is twice the length of a side of S
=>a = 2x

P(T) = a+b+c > 4x
P(S) = 4x
Clearly, P(T) > P(S) ---- Sufficient

Statement 2:
T is isosceles
=> a=b

Not enough information about square sides - Insufficient

So, A

Expanding this problem a little bit more. What if we are given the shortest side or the medium sides or both? Can we make any conclusion about the largest side?

Even in this case the result will be same, since if the smallest (or medium sized side) has length twice the length of the square side (2x here), the larger side would be definitely greater (>2x) hence the perimeter would be quite larger than the perimeter of square.

yellowho wrote:
b ) Is the perimeter of triangle T greater than the perimeter of
triangle S?

(1) The length of the longest side of T is twice the length of the longest
side of S.
(2) T is isosceles.

P(T) > P(S) ?
T has sides (a>=b>=c) and S has sides (x>=y>=z)

Statement 1:
The length of the longest side of T is twice the length of the longest side of S.

=> a = 2x
So, P(T) = a+b+c > 4x
P(S) = x+y+z <= 3x (as y and z has to be smaller or equal to x)

Clearly, P(S) > P (T) --- Sufficient

Statement 2:
T is isosceles.
Not much information given to compare the perimeters of the the triangles T and S -- Insufficient

So, A

hi all
in this problem i think we must employ the rool: inequality of triangle, which says that the sum of any two legs must be greater than the third leg, and the difference between any two legs must be less that the third leg
say legs if triangle are a,b,c with longest leg c. according to the above rool a+b>c.and a-b<c but we don`t need later in the problem
and side of square-s, we need to find whether
a+b+c>4s
(1) c=2s, is a+b+2s>4s, a+b>2s. s=c/2, is a+b>2*(c/2)-here cancel 2, and left with is a+b>c. for sure yes (according to the inequlity of triangle)
(2) no useful information about the sides insuff