1) What is the length of DE in triangle CDE if angle C is 30°, angle D is 45°, and CE is 5 square root of 2? 2) Find

Sure! Here are the detailed step-by-step solutions to the given problems:

1) To find the length of DE in triangle CDE, we can use the Law of Sines, which states that the ratio of the length of a side to the sine of its opposite angle is constant for all three sides of a triangle. Let"s denote the length of DE as x.

Using the Law of Sines, we have:
\[\frac{{DE}}{{\sin(45°)}} = \frac{{CE}}{{\sin(30°)}}\]

Since \(\sin(45°) = \sin(30°) = \frac{{\sqrt{2}}}{{2}}\), we can simplify the equation to:
\[DE = CE = 5\sqrt{2}\]

Hence, the length of DE is \(5\sqrt{2}\).

2) To find the third side of a triangle when two sides are given, we can use the Law of Cosines, which relates the lengths of the sides to the cosine of one of the angles. Let"s denote the third side as x.

Using the Law of Cosines, we have:
\[x^2 = 5^2 + 7^2 - 2 \cdot 5 \cdot 7 \cdot \cos(60°)\]

Simplifying the equation:
\[x^2 = 25 + 49 - 70 \cdot \frac{1}{2} = 74\]

Taking the square root of both sides, we get:
\[x = \sqrt{74}\]

Hence, the length of the third side is \(\sqrt{74}\) cm.

3) To determine the type of triangle ABC using its coordinates, we can calculate the lengths of its sides and analyze the relationships between them. Let"s calculate the lengths of AB, BC, and AC using the distance formula.

The distance formula is given by:
\[d = \sqrt{{(x_2 - x_1)^2 + (y_2 - y_1)^2}}\]

Calculating the lengths:
AB = \(\sqrt{{(0 - 3)^2 + (6 - 9)^2}} = \sqrt{{9 + 9}} = \sqrt{18}\)
BC = \(\sqrt{{(4 - 0)^2 + (2 - 6)^2}} = \sqrt{{16 + 16}} = \sqrt{32}\)
AC = \(\sqrt{{(4 - 3)^2 + (2 - 9)^2}} = \sqrt{{1 + 49}} = \sqrt{50}\)

By analyzing the relationships between the sides, we can conclude that:
- If all three sides are equal, it is an equilateral triangle. (\(AB = BC = AC\))
- If any two sides are equal, it is an isosceles triangle. (\(AB = BC\) or \(BC = AC\) or \(AC = AB\))
- If none of the sides are equal, it is a scalene triangle.

Based on the calculated lengths, we can conclude that triangle ABC is a scalene triangle.

4) To find the area of the rhombus AVCD, we need to know the length of its diagonals. Let"s denote the length of AK as x.

Since AK is the angle bisector of angle SAV and VAD equals 60°, we can use the properties of the rhombus to find the length of AK. In a rhombus, the diagonals bisect each other at 90° angles, and the diagonals are perpendicular bisectors of each other.

Let"s denote the length of SK as y. Since AK bisects angle SAV, we can form a right triangle SKA with angle SAK as 30° and angle AKS as 90°.

Using trigonometry, we have:
\(\tan(30°) = \frac{SK}{AK}\)
\(\sqrt{3} = \frac{y}{x}\)
\(y = x \sqrt{3}\)

Since the diagonals of a rhombus are perpendicular bisectors of each other, we can use the Pythagorean theorem in triangle SAV to find the length of SV. Let"s denote the length of SV as z.

Using the Pythagorean theorem, we have:
\(SV^2 = SA^2 + AV^2\)
\(SV^2 = (2x)^2 + (2y)^2\)
\(SV^2 = 4x^2 + 4y^2\)
\(SV^2 = 4x^2 + 4(x \sqrt{3})^2\)
\(SV^2 = 4x^2 + 12x^2\)
\(SV^2 = 16x^2\)
\(SV = 4x\)

Since VK is also a diagonal of the rhombus, we can conclude that VK = SV = 4x.

Now, to find the area of the rhombus AVCD, we can use the formula:
Area = \(\frac{{d_1 \cdot d_2}}{2}\), where \(d_1\) and \(d_2\) are the lengths of the diagonals.

Plugging in the values, we have:
Area = \(\frac{{AK \cdot VK}}{2} = \frac{{x \cdot 4x}}{2} = 2x^2\)

Hence, the area of the rhombus AVCD is \(2x^2\).