We are told PGN is obtuse, so it has one angle larger than 90 degrees. However, we don't know what that angle is. To find the perimeter we need all three sides. 

I) Relates the two shorter sides.

II) Relates the longest side to one of the short sides. 

However, we cannot find any of our side lengths, so we cannot find the perimeter.

The perimeter of \(\displaystyle \bigtriangleup ABC\) is equal to the sum of the lengths of the sides; that is, \(\displaystyle p = AB + BC + AC\). 

From Statement 1 alone, we get 

\(\displaystyle AB + AC - BC = 21\)

we can add \(\displaystyle 2 \cdot BC\) to both sides to get

\(\displaystyle AB + AC - BC + 2 \cdot BC = 21 + 2 \cdot BC\)

\(\displaystyle AB + AC + BC = 21 + 2 \cdot BC\)

\(\displaystyle p= 21 + 2 \cdot BC\)

However, without any further information, we cannot determine the actual perimeter. 

A similar argument shows that Statement 2 alone gives insufficient information as well.

However, suppose we were to multiply both sides of the equation in Statement 1 by 2, then add both sides of Statement 2:

\(\displaystyle 2AB + 2AC - 2BC = 42\)

\(\displaystyle \underline{AB + AC +2 BC = 45}\)

\(\displaystyle 3AB + 3AC\)             \(\displaystyle = 87\)

Divide both sides by 3:

\(\displaystyle AB + AC = 29\)

Since 

\(\displaystyle AB + AC - BC = 21\),

we can substitute 29 for \(\displaystyle AB + AC\) and find \(\displaystyle BC\):

\(\displaystyle 29 - BC = 21\)

\(\displaystyle BC = 8\)

While we cannot find \(\displaystyle AB\) or \(\displaystyle AC\) individually, this is not necessary; in the perimeter formula, we can substitute 29 for \(\displaystyle AB + AC\) and 8 for \(\displaystyle BC\):

\(\displaystyle p = (AB + BC )+ AC = 29 + 8 = 37\).

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 \(\displaystyle 10+10+12 = 32\), the same as a triangle with sides of length 9, 10, and 13, since \(\displaystyle 9+10+13=32\), but a triangle with sides of length 10, 10, and 13 has perimeter \(\displaystyle 10+10+13 = 33\).

Statement 2 alone provides insufficient information to answer the question. Since \(\displaystyle AB = DE\) and \(\displaystyle BC = EF\), it follows that the perimeter are equal if and only if \(\displaystyle AC = DF\);  we are not told whether this is true or false.

Now assume both statements. \(\displaystyle \bigtriangleup ABC\) is isosceles, so two of its sides have equal length; however, it cannot hold that  \(\displaystyle \overline{AB} \cong \overline{BC}\); if so, then, since \(\displaystyle \overline{AB} \cong \overline{DE}\) and \(\displaystyle \overline{BC}\cong \overline{EF}\), it would follow that \(\displaystyle \overline{DE}\cong \overline{EF}\), which contradicts \(\displaystyle \bigtriangleup DEF\) being scalene. Therefore, either \(\displaystyle \overline{AC} \cong \overline{AB}\) or \(\displaystyle \overline{AC} \cong \overline{BC}\). If \(\displaystyle \overline{DF} \cong \overline{AC}\), then \(\displaystyle \overline{DF}\) is congruent to one other side of \(\displaystyle \bigtriangleup ABC\), and, consequently, one other side of \(\displaystyle \bigtriangleup DEF\), contradicting \(\displaystyle \bigtriangleup DEF\) being scalene. Therefore, \(\displaystyle AC \ne DF\); as stated before, the perimeters are equal if and only if \(\displaystyle AC = DF\), 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: \(\displaystyle AB = 11, BC = CD = 6\)

This triangle satisfies the triangle inequality, since \(\displaystyle 6+6= 12 >11\); its perimeter is \(\displaystyle 11+6+6 = 23 < 24\)

Case 2: \(\displaystyle AB = 11, BC = CD = 7\)

This triangle satisfies the triangle inequality, since \(\displaystyle 7+7=14 >11\); its perimeter is \(\displaystyle 11+7+7 = 25>24\).

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

Assume Statement 2 alone. Again, \(\displaystyle AB + BC > AC\); since, by Statement 2, \(\displaystyle AC = 12\), by substitution, \(\displaystyle AB + BC > 12\). The perimeter of \(\displaystyle \bigtriangleup ABC\) is

\(\displaystyle p = AB + BC + AC\), and, since \(\displaystyle AB + BC > 12\), then 

\(\displaystyle p = (AB + BC )+ AC> 12+12= 24\)

The perimeter of \(\displaystyle \bigtriangleup ABC\) 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 \(\displaystyle \overline{SU}\) and \(\displaystyle \overline{QR}\) is \(\displaystyle \sqrt{2}\) times the length of a side, or simply \(\displaystyle \sqrt{2}\).

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: \(\displaystyle AB = 1+1\) - or equivalently, \(\displaystyle AB = 2\)

Statement 2: \(\displaystyle BC = 2 \sqrt{2}\)

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

\(\displaystyle AC+BC > AB\)

and 

\(\displaystyle AC+BC > 2\)

Since from Statement 1, \(\displaystyle AB = 2\), t

By the Addition Property of Inequality, we can add \(\displaystyle AB\) to both sides;

\(\displaystyle p =( AC+BC )+AB > 2+2 = 4\)

The perimeter of the triangle is greater than 4; equivalently, \(\displaystyle \bigtriangleup ABC\) has greater perimeter than Square \(\displaystyle SQUR\).

Assume Statement 2. By similar reasoning, since one side has length \(\displaystyle 2 \sqrt{2}\), the perimeter is at greater than twice this, or \(\displaystyle 4 \sqrt{2}\), which is greater than 4, so  \(\displaystyle \bigtriangleup ABC\) has greater perimeter than Square \(\displaystyle SQUR\).

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

\(\displaystyle AC < AB + BC\)

\(\displaystyle AC < 12 + 18\)

\(\displaystyle AC < 30\)

Also, 

\(\displaystyle BC < AB + AC\)

\(\displaystyle 18< 12 + AC\)

\(\displaystyle 6 < AC\)

Therefore, 

\(\displaystyle 6< AC < 30\),

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

\(\displaystyle p =AB + BC + AC\)

by adding:

\(\displaystyle AB + BC + 6< AB + BC + AC < AB + BC + 30\)

\(\displaystyle 12+18 + 6< p < 12+18 + 30\)

\(\displaystyle 3 6< p < 60\)

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 \(\displaystyle AC =AB = 12\) or \(\displaystyle AC=BC = 18\). 

If \(\displaystyle AC = 12\), the perimeter is 

\(\displaystyle p =AB + BC + AC\)

\(\displaystyle p =12 + 18+ 12 = 42\)

If \(\displaystyle AC = 18\), the perimeter is 

\(\displaystyle p =AB + BC + AC\)

\(\displaystyle p =12 + 18+ 18 = 48\)

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 \(\displaystyle \overline{SU}\) and \(\displaystyle \overline{QR}\) is \(\displaystyle \sqrt{2}\) times the length of a side, or simply \(\displaystyle \sqrt{2}\).

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: \(\displaystyle AB = \sqrt{2}\)

Statement 2: \(\displaystyle BC= 1\)

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 \(\displaystyle AC\) using this fact:

\(\displaystyle AC < BC + AB\)

\(\displaystyle AC < 1+ \sqrt{2}\)

Also, 

\(\displaystyle AB < AC + BC\)

\(\displaystyle AB -BC < AC\)

\(\displaystyle \sqrt{2} -1 < AC\)

So, 

\(\displaystyle -1 + \sqrt{2} < AC < 1+ \sqrt{2}\)

Add \(\displaystyle AB\) and \(\displaystyle BC\) to all three expressions; the expression in the middle is the perimeter of \(\displaystyle \bigtriangleup ABC\):

\(\displaystyle -1 + \sqrt{2} + AB + BC < AC + AB + BC< 1+ \sqrt{2} + AB + BC\)

\(\displaystyle -1 + \sqrt{2} + \sqrt{2 }+ 1< p< 1+ \sqrt{2} + \sqrt{2 }+ 1\)

\(\displaystyle 2 \sqrt{2} < p< 2+ 2\sqrt{2}\)

Since \(\displaystyle \sqrt{2} \approx 1.4\), for all practical purposes, 

\(\displaystyle 2 (1.4) < p< 2+ 2 (1.4)\)

\(\displaystyle 2.8 < p< 2+2 .8\)

\(\displaystyle 2.8 < p< 4 .8\)

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: \(\displaystyle AB = BC = AC = 12\)

\(\displaystyle AB + BC =12+12= 24\)

\(\displaystyle BC + AC = 12+12= 24\)

The perimeter of \(\displaystyle \bigtriangleup ABC\) is

\(\displaystyle AB + BC + AC = 12+12+12=36\).

Case 2: \(\displaystyle AB = AC = 10, BC = 14\).

\(\displaystyle AB + BC = 10+14= 24\)

\(\displaystyle BC + AC = 14+10= 24\)

The perimeter of \(\displaystyle \bigtriangleup ABC\) is

\(\displaystyle AB + BC + AC = 10+14+10=34\).

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

\(\displaystyle AB + AC > BC\)

\(\displaystyle AB + AC > 28\)

We can find the minimum of the perimeter:

\(\displaystyle p =\left (AB+ AC \right ) + BC > 28+28 = 56 > 50\):

The perimeter of \(\displaystyle \bigtriangleup ABC\) is greater than 50.

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

Case 1: \(\displaystyle \bigtriangleup ABC \cong \bigtriangleup DEF\). By definition, \(\displaystyle \overline{AB} \cong \overline{DE}\) and \(\displaystyle \angle A \cong \angle D\), satisfying the condtions of the main body of the problem, and \(\displaystyle \overline{BC} \cong \overline{EF}\), 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 \(\displaystyle A\) and \(\displaystyle B\) coincide with \(\displaystyle D \,\) and \(\displaystyle E\), respectively:

The conditions of the main body and Statement 1 are met, since \(\displaystyle \overline{AB} \cong \overline{DE}\), \(\displaystyle \angle A \cong \angle D\) (their being the same segment and angle, respectively, in the diagram), and \(\displaystyle \overline{BC} \cong \overline{EF}\) by construction. Note, however, that \(\displaystyle DF = DC + CF = AC + CF\), so:

\(\displaystyle DF > AC\).

\(\displaystyle DE+ DF+EF > AB + AC + BC\)

Making the perimeters different. 

Now assume Statement 2 alone. \(\displaystyle \overline{AB}\) is the included side of \(\displaystyle \angle A\) and \(\displaystyle \angle B\), and  \(\displaystyle \overline{DE}\) is the included side of \(\displaystyle \angle D\) and \(\displaystyle \angle E\). The three congruence statements given in the main body and Statement 2 together set the conditions of the Angle-Side-Angle Postulate, so \(\displaystyle \bigtriangleup ABC \cong \bigtriangleup DEF\), and the perimeters are indeed the same.