Statement (1) informs us that the triangle is an equilateral triangle, with all sides equal to 6. Therefore, the height would divide the triangle into two 30-60-90 triangles, which have side lengths in a ratio of \(\displaystyle 1:\sqrt3:2\). For this triangle the hypotenuse would be 6 and the base would be 3. The height would therefore be \(\displaystyle 3\sqrt3\). SUFFICIENT

Statement (2) also lets you deduce that the triangle is an equilateral triangle. Since the triangle is either isoscles or equilateral (at least 2 sides are equal), that means that two angles are equal. Therefore, if one angle is  \(\displaystyle 60^{\circ}\), the other two also must be  \(\displaystyle 60^{\circ}\). See above. SUFFICIENT

Since we don't know what type the triangle is, we would not only need information about the lengths of the side but also about the characteristics of the triangle. 

Statement 1 gives us the length of a side. However, we can't do anything, since we don't know the length of DC, which would allow us the know BD with the Pythagorean Theorem.

Statement 2 also only gives us information about one side of the triangle. Alone it doesn't allow us to calculate any other length.

Even taken together these statements are insufficient since, we don't know any pair of lengths to use in the Pythagorean Theorem. Even though the triangle looks like a isosceles triangle, it doesn't mean that it is.

To find the length DC, we need to know AD and AC or any of those two provided that ABC is isosceles or equilateral.

Statement 1 tells us the area of the triangle with information about a part of side AC. Since we don't know properties of the triangle, these other lengths can vest many values, just AC can be 12, 24 or 48. Therefore we don't have enough information.

Statement 2 gives us information about angles of the triangle. From what we are told we can see that the triangle is isosceles. Indeed, we know that \(\displaystyle \measuredangle {ADB}=90^{\circ}\) since BD is the height. Therefore \(\displaystyle \measuredangle {BDC}=27^{\circ}\). Now, that we know that the triangle is isosceles, we know that AC must be 12, since D is the midpoint of AC. Therefore DC must be 6.

Hence, both statements taken together are sufficient.

Statement 1 alone proves that \(\displaystyle \Delta ABC \sim \Delta DEF\) by the Angle-Angle Postulate, but does not prove anything about the sides, which would be needed to answer this question. Statement 2 alone only gives a relationship between one side of each triangle; without any further information, this is insufficient.

The two statements together, however, present sufficient evidence. Statement 1 proves that the triangles are similar. Statement 2 gives the ratio of one side in \(\displaystyle \Delta ABC\) to the corresponding side in \(\displaystyle \Delta DEF\), so, as the triangles are similar, this ratio is shared by all three pairs of corresponding sides. Since

\(\displaystyle \frac{AB }{DE}= 1.1\),

1.1 is this common ratio, and the ratio of the area of \(\displaystyle \Delta ABC\) to \(\displaystyle \Delta DEF\) is \(\displaystyle 1.1^{2} = 1.21\), making \(\displaystyle \Delta ABC\) the larger triangle.

Statement 1 alone gives a proportion between two pairs of corresponding sides of the triangles. This is not enough to prove the triangles similar without the third side proportionality (SSS Simiilarity statement) or the congruence of the included angles (SAS Similarity statement).

Statement 2 gives two congruencies between corresponding angles, which by the Angle-Angle statement is enough to prove the triangles similar.

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 \(\displaystyle SQUR\).