integer values of 'x'

msbinu wrote:Rules of triangle are
1. length of any side cannot be greater that the sum of other 2
2. Length cannot be shorter than difference of other sides

Applying rule 2
Length of third cannot be shorter than 2
. Let the length be 3 . its > 2 and <= 22 ( rule 1 )
length 4 : > 2 and < = 22
.
.
Length 10 > 2 and <=22

Length 11 >2 but now > 22 .

Hence side can be any thing from 3 to 10 .
Hence total 7 possibilities are there for the third side

@msbinu

The two rules required here are not as given by you, dear pal. You too revise this please, and mind it's an acute angled triangle, above all. The correct answer to this question will come out to be [spoiler](B) 9, which is the OA too, not (A) 7[/spoiler].

Rule 1: For an acute angled triangle, the square of the LONGEST side MUST BE LESS than the sum of squares of the other two sides.

Rule 2: For any triangle, sum of any two sides must be greater than the third side.

Thanks

phillybeat wrote:

sanju09 wrote:If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible?
(A) 7
(B) 9
(C) 11
(D) 12
(E) 13


[spoiler]Source: https://gmat-math.blocked/2010/02/4gmat[/spoiler]

is it A

Can someone please clarify the rules for an acute angled triange ( in the discussion ) !!!

RIGHT ANGLE
1 ANGLE = 90, C^2 + B^2 = A^2

ABTUSE ANGLE
1 ANGLE > 90, A^2 + B^2 < C^2

ACUTE ANGLE
3 ANGLES < 90, A^2 + B^2 > C^2

I HOPE THIS RULES WILL HELP YOU TO SOLVE THE PROBLEM AND THE RIGHT ANSWER IS "E"

10^2 + 12^2 > C^2
244 > c^2
224 > 3^2 TO 244 > 15^2

(15^2 = 225)
(16^2 = 256, WHICH IS > 244) SO 3 TO 15 GIVES "13" INTEGERS