ABC is an isosceles triangle. What is its perimeter?

1) Statement 1 tells us that one of the sides is = 6, but we do not have information about the other sides. Considering that an isósceles triangle has at least 2 same sized sides we have infinite combinations and posibilities; 6,6,1 ; 6,5,5 ; 6,6,6... NOT SUFFICIENT.

2) Statement 2 Same as the above and the combinations are again infinite; 2,7,7 ; 2,2,1... and so on.. NOT SUFFICIENT.

Combining the statements we are given two sides of the equilateral triangle; 2 and 6. The third side must be equal to one of the mentioned sides.
If we had 6,6 and 2, the equilateral triangle would have a perimeter of 14.
If we had 2,2 and 6 we could NOT form a triangle because the shortest distance between two points is the straight line. So the sum of two of the 3 sides of a triangle must be always greater than the length of the other one and in this case, this is not satified.

So we can assure our triangle has a perimeter of 6+6+2. Ans C.

For solving this question one should know that sum of 2 sides of triangle must be greater than third side.
1) no info on other sides...insuff
2)no info on other sides....insuff

Combining only 2 possibilities
a) 6-6-2 triangle
b) 2-2-6 triangle..but this triangle is just not posssible as 2+2<6( sum of 2 sides of triangle IS NOT greater than third side)

so only 6-6-2 is possible one
and thus we calculate perimeter

sufficient
Ans C

Good day,
Can anybody explain the difference concept. I found out similar question, however the answer is E.
https://gmatclub.com/forum/what-is-the- ... fl=similar

Hi,
Q may be similar but the sides are different...
1) in this Q sides are 6 and 2..
What can be 3rd side?..
6 yes 6,6,2..
2 ?? no because 2,2,6 cannot be sides of a triangle as the sum of two sides 2+2 is < 6, the third side..
Not possible because then it will be just a line and not triangle..
Only one triangle possible..
C

2) now the q u are referring to..
Sides 4,4√2
The triangle can have sides 4,4,4√2 and 4,4√2,4√2..
So E

Hope it clears your doubt
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Great, thanks for quick response

Hi
1 and 2 are not sufficient clearly, so 1+2

somehow i stuck in different point.
Instead of starting with side c< side a+b, i started like lsosceles triangle will have sides in ratio 1:1:root 2.
obvisously we cannot have 6 and 6 as two sides because the third side has to be greater than 6 and we got the third side a 2.
so the sides should be 2,2,6 so perimeter is 10.. ans C

I believe i went seriously wrong somewhere
can someone help me out here.

Hi

First of all, NOT every isosceles triangle has sides in the ratio 1:1:root 2... Only Right Angled Isosceles triangle (with angles 45, 45 and 90 degrees) has this ratio of sides as 1:1:root 2

Next, triangle inequality is an important property of Every triangle - according to which sum of any two sides of a triangle MUST be greater than the third side, and similarly difference of any two sides of a triangle MUST be less than the third side.

On this basis, a triangle with sides 2, 2, 6 is Not possible because 2+2 is not greater than 6. If you try to construct a Triangle with these sides, you wont be able to.
However, a triangle with sides 6, 6, 2 is definitely possible, because in this sum of any two sides is greater than the third side.

So the answer to the question is still C but the perimeter of our triangle will be 6+6+2 = 14.

IMPORTANT RULE: If two sides of a triangle have lengths A and B, then . . .
DIFFERENCE between A and B < length of third side < SUM of A and B

Target question: What is the perimeter of isosceles triangle PQR?

Given: PQR is an isosceles triangle

Statement 1: Side PQ = 6
There are many isosceles triangles that satisfy statement 1. Here are two:
Case a: PQ = 6, PR = 6 and RQ = 5. Note that this is an isosceles triangle AND it meets the above rule. In this case, the perimeter = 6 + 6 + 5 = 17
Case b: Case a: PQ = 6, PR = 6 and RQ = 4. Note that this is an isosceles triangle AND it meets the above rule. In this case, the perimeter = 6 + 6 + 4 = 16
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: Side PR = 2
Applying the same logic we used for statement 1, we can see that statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that one side has length 6
Statement 2 tells us that one side has length 2
Since PQR is an isosceles triangle, the third side (RQ) must have length 6 or 2 (so that we have two sides of equal length).
If side RQ has length 6, then the sides have length 6, 6, and 2. These three lengths satisfy the above rule, in which case, the perimeter = 6 + 6 + 2 = 14
If side RQ has length 2, then the sides have length 6, 2, and 2. These three lengths DO NOT satisfy the above rule. If we make the side with length 6 the third side, and apply the rule, we get 2 - 2 < 6 < 2 + 2. When we simplify this, we get 0 < 6 < 4
So, the three lengths CANNOT be 6, 2 and 2.
So, the perimeter MUST be 14
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer:

RELATED VIDEO

Statement 1 not sufficient as we only know that side PQ = 6 .
Statement 2 alone is not sufficient as we only know that side PR=2
We already know that THE triangle is an isosceles triangle which means 2 sides have the same length and same angle.
therefore together we know 3rd side is either 6 or 2.
Now if we use the triangle properties IE the 3rd side = less than sum of 2 sides and more than difference of 2 sides.
therefore in case the side = 2 then THE property fails because 6-2=4 . Third side should be greater than 4 therore 2 cannot be an option.
So only possibility =6 therefore the triangle is 6 , 6, 2 and the perimeter = 14 . Ans= C

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

When we apply VA(Variable Approach) to geometry, there are 3 variables for a triangle.
Since we have 3 variables(PQ, QR and PR) and 0 equations, E is most likely to be the answer. So, we should consider 1) & 2) first, since we can save time by first checking whether conditions 1) and 2) are sufficient, when taken together.

Conditions 1) & 2)
Since PQR is an isosceles, QR = PQ = 6 or QR = PR = 2.
When we consider a property that the sum of two sides of a triangle is greater than other side, we should have QR + PR > QR. But if QR = PR = 2, this is not true since 2 + 2 < 6.
Thus QR must be equal to PQ = 6.

Therefore, the answer is C.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

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