What is the perimeter of the isoceles right triangle ABC
S1 AB = 9
S2 BC = 4
i dont understand why the answer is C since we dont know which sides are equal
a data from gmat club
This topic has expert replies
-
- Legendary Member
- Posts: 1119
- Joined: Fri May 07, 2010 8:50 am
- Thanked: 29 times
- Followed by:3 members
-
- Legendary Member
- Posts: 1085
- Joined: Fri Apr 15, 2011 2:33 pm
- Thanked: 158 times
- Followed by:21 members
my answer is e
st(1) AB=9 and AB can be either hypotenuse or non-hypotenuse side, Not Sufficient
st(2) BC=4 the same applies as for st(1), Not Sufficient
combined st(1&2) the hypotenuse should be longer than non-hypotenuse side, hence hypotenuse is 9 and non-hypotenuse sides are 4, which is not possible as 2*(4^2) will return 32 and not 81 (or 9^2) Not Sufficient.
If this question would be rephrased to only right triangle (not isosceles right triangle) then the answer would be c
st(1) AB=9 and AB can be either hypotenuse or non-hypotenuse side, Not Sufficient
st(2) BC=4 the same applies as for st(1), Not Sufficient
combined st(1&2) the hypotenuse should be longer than non-hypotenuse side, hence hypotenuse is 9 and non-hypotenuse sides are 4, which is not possible as 2*(4^2) will return 32 and not 81 (or 9^2) Not Sufficient.
If this question would be rephrased to only right triangle (not isosceles right triangle) then the answer would be c
diebeatsthegmat wrote:What is the perimeter of the isoceles right triangle ABC
S1 AB = 9
S2 BC = 4
i dont understand why the answer is C since we dont know which sides are equal
Success doesn't come overnight!
GMAT/MBA Expert
- Ian Stewart
- GMAT Instructor
- Posts: 2621
- Joined: Mon Jun 02, 2008 3:17 am
- Location: Montreal
- Thanked: 1090 times
- Followed by:355 members
- GMAT Score:780
If the sides of a triangle are a, b and c, it is always true that the sum of any two sides of the triangle must exceed the third side - that is, that
a + b > c
a + c > b
b + c > a
So with both statements together, it's certainly possible for the three sides to be 9, 9 and 4. It is not, however, possible for the sides to be 4, 4 and 9, since 4+4 is not greater than 9. So the answer is C.
a + b > c
a + c > b
b + c > a
So with both statements together, it's certainly possible for the three sides to be 9, 9 and 4. It is not, however, possible for the sides to be 4, 4 and 9, since 4+4 is not greater than 9. So the answer is C.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
ianstewartgmat.com
ianstewartgmat.com
-
- Master | Next Rank: 500 Posts
- Posts: 184
- Joined: Tue Sep 07, 2010 9:43 am
- Thanked: 6 times
- Followed by:1 members
Hi Ian,
Yes, it is possible to form a triangle using both the statements, but the answer should be E in this case as the question states that the given sides form an isosceles right triangle and it is not possible to form an isosceles right triangle with 9,9 and 4. Please correct me if I am worng.
Yes, it is possible to form a triangle using both the statements, but the answer should be E in this case as the question states that the given sides form an isosceles right triangle and it is not possible to form an isosceles right triangle with 9,9 and 4. Please correct me if I am worng.
-
- Legendary Member
- Posts: 1448
- Joined: Tue May 17, 2011 9:55 am
- Location: India
- Thanked: 375 times
- Followed by:53 members
Hi,pemdas wrote: If this question would be rephrased to only right triangle (not isosceles right triangle) then the answer would be c
No, Even if it is rephrased to right triangle, the answer is E, because if 4,9 are two sides, third side can be either sqrt(9^2 - 4^2) or sqrt(9^2+4^2).
The answer would be C if the triangle given is isosceles(not isosceles right triangle) because third side cannot be 4(as 4+4<9). It can only be 9.
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
-
- Legendary Member
- Posts: 1119
- Joined: Fri May 07, 2010 8:50 am
- Thanked: 29 times
- Followed by:3 members
ahh yeah... i should pay more attention while learning... thank youIan Stewart wrote:If the sides of a triangle are a, b and c, it is always true that the sum of any two sides of the triangle must exceed the third side - that is, that
a + b > c
a + c > b
b + c > a
So with both statements together, it's certainly possible for the three sides to be 9, 9 and 4. It is not, however, possible for the sides to be 4, 4 and 9, since 4+4 is not greater than 9. So the answer is C.
-
- Legendary Member
- Posts: 1448
- Joined: Tue May 17, 2011 9:55 am
- Location: India
- Thanked: 375 times
- Followed by:53 members
Hey,diebeatsthegmat wrote: ahh yeah... i should pay more attention while learning... thank you
Please check the source from which this question is posted. Are you sure it is isosceles right triangle or just isosceles triangle?
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
GMAT/MBA Expert
- Ian Stewart
- GMAT Instructor
- Posts: 2621
- Joined: Mon Jun 02, 2008 3:17 am
- Location: Montreal
- Thanked: 1090 times
- Followed by:355 members
- GMAT Score:780
Good point - I've seen so many questions like this one before, I just glossed over the word 'right' in the question. If the question asks about an 'isosceles right triangle', the question doesn't make any sense, since it's impossible to construct an isosceles right triangle with the dimensions given. So I'm sure the question means to ask just about "isosceles triangle ABC", and not about "isosceles right triangle ABC".rahulvsd wrote:Hi Ian,
Yes, it is possible to form a triangle using both the statements, but the answer should be E in this case as the question states that the given sides form an isosceles right triangle and it is not possible to form an isosceles right triangle with 9,9 and 4. Please correct me if I am worng.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
ianstewartgmat.com
ianstewartgmat.com