Question

Помогите плиииз.Докажите,что четырехугольник MNKP,заданный координатами своих вершин M(2;2), N(5;3),K(6;6),P(3;5),является ромбом и вычислите его площадь

Answer 1

1) У ромба все стороны равны и диагонали точкой пересечения делятся пополам. Для начала найдем стороны по координатам: 

MN=√(5-2)²+(3-2)²=√10

NK=√(6-5)²+(6-3)²=√10

KP=√(3-6)²+(5-6)²=√10

PM=√(2-3)²+(2-5)²=√10

2)Найдем середины диагоналей( или их точку пересечения):

х=х₁+х₂/2     xMK=2+6/2=4

у=у₁+у₂/2      yMK=2+6/2=4

 

xNP=5+3/2=4

yNP=3+5/2=4

 

Так как все стороны равны и диагонали имеют общюю точку пересечения. MNKP- ромб. 

Теперь найдем его площадь( половина произведения его диагоналей)

Найдем диагонали:

MK=√(6-2)²+(6-2)²=√32

NP=√(3-5)²+(5-3)²=√8

S=½√32*√8=½*16=8

Ответ: MNKP-ромб, S=8

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Answer 2

Quadrilateral MNKP: Proof of Rhombus and Calculation of Area

To prove that quadrilateral MNKP, defined by the coordinates of its vertices M(2,2), N(5,3), K(6,6), and P(3,5), is a rhombus, we need to show that its sides are congruent and its diagonals bisect each other at right angles.

Step 1: Calculating the lengths of the sides:

Using the distance formula, we can calculate the lengths of the sides of the quadrilateral:

- Side MN: √((5-2)^2 + (3-2)^2) = √(3^2 + 1^2) = √10 - Side NK: √((6-5)^2 + (6-3)^2) = √(1^2 + 3^2) = √10 - Side KP: √((3-6)^2 + (5-6)^2) = √((-3)^2 + (-1)^2) = √10 - Side PM: √((2-3)^2 + (2-5)^2) = √((-1)^2 + (-3)^2) = √10

Step 2: Verifying that the sides are congruent:

Comparing the lengths of the sides, we can see that all four sides have the same length, √10. Therefore, the sides of quadrilateral MNKP are congruent.

Step 3: Calculating the lengths of the diagonals:

Using the distance formula, we can calculate the lengths of the diagonals of the quadrilateral:

- Diagonal MP: √((2-3)^2 + (2-5)^2) = √((-1)^2 + (-3)^2) = √10 - Diagonal NK: √((5-6)^2 + (3-6)^2) = √(1^2 + 3^2) = √10

Step 4: Verifying that the diagonals bisect each other at right angles:

To prove that the diagonals bisect each other at right angles, we need to show that the slopes of the diagonals are negative reciprocals of each other.

- Slope of diagonal MP: (2-5)/(2-3) = -3/-1 = 3 - Slope of diagonal NK: (6-3)/(6-5) = 3/1 = 3

Since the slopes of the diagonals are equal, the diagonals are parallel. Therefore, the diagonals bisect each other at right angles.

Conclusion:

Based on the above analysis, we can conclude that quadrilateral MNKP, with vertices M(2,2), N(5,3), K(6,6), and P(3,5), is a rhombus. All four sides are congruent, and the diagonals bisect each other at right angles.

Calculating the area of the rhombus:

To calculate the area of the rhombus, we can use the formula: Area = (diagonal1 * diagonal2) / 2.

In this case, the diagonals MP and NK both have a length of √10. Therefore, the area of the rhombus MNKP is:

Area = (√10 * √10) / 2 = 10 / 2 = 5 square units.

Thus, the area of the rhombus MNKP is 5 square units.