Statement 1 alone provides insufficient information to answer the question, since it is possible for an isosceles triangle, which has two or three sides of equal length, to have perimeter equal to or not equal to a scalene triangle, which has three sides of different lengths. For example, a triangle with sides of length 10, 10, and 12 has perimeter , the same as a triangle with sides of length 9, 10, and 13, since , but a triangle with sides of length 10, 10, and 13 has perimeter .

Statement 2 alone provides insufficient information to answer the question. Since  and , it follows that the perimeter are equal if and only if ;  we are not told whether this is true or false.

Now assume both statements.  is isosceles, so two of its sides have equal length; however, it cannot hold that  ; if so, then, since  and , it would follow that , which contradicts  being scalene. Therefore, either  or . If , then  is congruent to one other side of , and, consequently, one other side of , contradicting  being scalene. Therefore, ; as stated before, the perimeters are equal if and only if , so the perimeters are not equal.

Statement 1 alone gives insufficient information. By the Triangle Inequality Theorem, the sum of the lengths of the shortest two sides of a triangle must be greater than the length of the longest. Examine these two scenarios:

Case 1: 

This triangle satisfies the triangle inequality, since ; its perimeter is 

Case 2: 

This triangle satisfies the triangle inequality, since ; its perimeter is .

Therefore, Statement 1 alone does not answer whether the perimeter is less than, equal to, or greater than 24.

Assume Statement 2 alone. Again, ; since, by Statement 2, , by substitution, . The perimeter of  is

, and, since , then 

The perimeter of  is greater than 24.

For the sake of simplicity, we will assume the length of each side of the square is 1; this reasoning works independently of the sidelength. The perimeter of the square is, as a result, 4, and the length of each of the diagonals  and  is  times the length of a side, or simply .

The equivalent question becomes whether the perimeter of the triangle is greater than, equal to, or less than 4. The statements can be rewritten as

Statement 1:  - or equivalently,

Statement 2: 

Assume Statement 1 alone. By the Triangle Inequality, the sum of the lengths of two sides of a triangle must exceed the third. Therefore, 

and 

Since from Statement 1, , t

By the Addition Property of Inequality, we can add  to both sides;

The perimeter of the triangle is greater than 4; equivalently,  has greater perimeter than Square .

Assume Statement 2. By similar reasoning, since one side has length , the perimeter is at greater than twice this, or , which is greater than 4, so   has greater perimeter than Square .

Statement 1 is insufficient, as it only gives that two sides are of equal length; it gives no side lengths, nor does it give any measurements that yield the side lengths. 

Assume Statement 2 alone. By the Triangle Inequality, the length of each side must be less than the sum of the lengths of the other two, so

Also, 

Therefore, 

,

and we can find the range of the values of the perimeter 

by adding:

Therefore, the perimeter may or may not be greater than 50.

Assume both statements to be true. An isosceles triangle has two sides of the same length, so either  or . 

If , the perimeter is 

If , the perimeter is 

Either way, the perimeter is less than 50.

For the sake of simplicity, we will assume the length of each side of the square is 1; this reasoning works independently of the side length. The perimeter of the square is, as a result, 4, and the length of each of the diagonals  and  is  times the length of a side, or simply .

The equivalent question becomes whether the perimeter of the triangle is greater than, equal to, or less than 4. The statements can be rewritten as

Statement 1: 

Statement 2: 

We show that these two statements together provide insufficient information. 

By the Triangle Inequality, the sum of the lengths of two sides of a triangle must exceed the third. We can get the range of values of  using this fact:

Also, 

So, 

Add  and  to all three expressions; the expression in the middle is the perimeter of :

Since , for all practical purposes, 

Therefore, we cannot tell whether the perimeter is less than, equal to, or greater than 4. Equivalently, we cannot determine whether the triangle or the square has the greater perimeter.

We demonstrate that both statements together provide insufficient information by examining two cases:

Case 1: 

The perimeter of  is

.

Case 2: .

The perimeter of  is

.

Both cases satisfy the conditions of both statements but different perimeters are yielded. 

Statement 1 is insufficient, as it only gives that two sides are of equal length; it gives no side lengths, nor does it give any measurements that yield the side lengths. 

Assume Statement 2 alone. By the Triangle Inequality, the length of each side must be less than the sum of the lengths of the other two, so

We can find the minimum of the perimeter:

:

The perimeter of  is greater than 50.

Assume Statement 1 alone. We show that this provides insufficient information by examining two scenarios.

Case 1: . By definition,  and , satisfying the condtions of the main body of the problem, and , satisfying the condition of Statement 1. Since the triangles are congruent, all three pairs of corresponding sides have the same length, so the perimeters are equal.

Case 2: Examine this diagram, which superimposes the triangles such that  and  coincide with  and , respectively:

The conditions of the main body and Statement 1 are met, since ,  (their being the same segment and angle, respectively, in the diagram), and  by construction. Note, however, that , so:

.

Making the perimeters different. 

Now assume Statement 2 alone.  is the included side of  and , and   is the included side of  and . The three congruence statements given in the main body and Statement 2 together set the conditions of the Angle-Side-Angle Postulate, so , and the perimeters are indeed the same.

Assume Statement 1 alone. A segment that has as its endpoints the midpoints of two sides of a triangle is a midsegment of the triangle, and its length is half that of the side to which it is parallel. Therefore, the sum of the lengths of the midsegments—that is, the perimeter of the triangle they form, which from Statement 1 is 34, is half the perimeter of the larger triangle, which here is . The perimeter of  is therefore 68.

Assume Statement 2 alone. Here,  itself is the triangle formed by the midsegments. Since the larger triangle has perimeter 136,  has perimeter half this, or 68.

Assume both statements to be true. An isosceles triangle has two sides of equal length, so either  or . By the Triangle Inequality Theorem, the length of each side must be less than the sum of the lengths of the other two; both scenarios are possible, since

and 

.

If , the perimeter of  is

.

If , the perimeter of  is

.

Without further information, it is impossible to determine whether the perimeter is less than or greater than 60.