Basically the problem above could be expressed as

Any one side of a triangle must be shorter than the sum of the other two sides.

Included, as you pointed out, is not contemplate . It is just between

When we say that one number lies between two numbers, we never include the boundaries of the interval?

\(-1<x<1\)

-1 and 1 NOT included

\(-1 \leq x \leq 1\)

-1 and 1 INCLUDED

More here https://gre.myprepclub.com/forum/gre-quant ... tml#p52039

the triangle is a right-angle triangle.
The ratio between the longest (hypotenuse by default) and shortest side is 5:3
thus, the length can be 5:3, 10:6, 15:9, 20:12, 25:15, 30:18 and so on.
in each case, the length of the 3rd side would be - 4, 8, 12, 16, 20, 24 respectively.
and in each case the area of the triangle would be 6, 24, 54, 96, 150, 226, and only 54, 96 satisfy the condition, which says 50<area<150.
and the corresponding short sides in these cases: 9 and 12 (C and D are the answer)

Hi,

In right triangle LMN, the ratio of the longest side to the shortest side is 5 to 3. If the area of LMN is between 50 and 150, which of the following could be the length of the shortest side?

As it is a right angle triangle with the ratio of the longest side to the shortest side is 5 to 3. Thus the other side has to be 4. Which helps us to find the are of the triangle= 1/2 *3x*4x= 6x

Thus, all multiples of 6 between 50 amd 150 could be the area of the above triangle.

Hence, if the shortest side= 9 i.e. 3x=9; x=3
Thus 4x=12
Making the area of triangle as 1/2*9*12=54

Considering option D; 3x=12; x=4
Thus 4x=16
Area of triangle= 96

Thus, only the above two value of the shortest side satisfies the given condition.

IMO C & D

Hope this helps!

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