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Luba_88 [7]
1 year ago
13

Find the area P(4,6), Q(8,5), and R(5,9)

Mathematics
1 answer:
RSB [31]1 year ago
7 0

The area of the triangle PQR is 6.5 unit square.

<h3>What is defined as the triangles?</h3>

Triangles are three-sided polygons with three vertices.

  • The angles of the triangle are established by connecting the three sides end to end at a point.
  • The sum of the triangle's three angles corresponds 180 degrees.

The three coordinates of the points are P(4,6), Q(8,5), and R(5,9).

The formula for the area of the triangle when the coordinates are given is;

area of ΔABC = (1/2)[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

(x₁, y₁) = P(4,6)

(x₂,y₂) = Q(8,5)

(x₃, y₃) = R(5,9)

Put the given values in the formula;

area of ΔABC = (1/2)[4(5 - 9) + 8(9 - 6) + 5(6 - 5)]

On solving;

area of ΔABC = (1/2)[-16 + 24 + 5]

area of ΔABC = (1/2)[13]

area of ΔABC = 6.5

Therefore, the area for the triangle formed by the coordinates is 6.5 unit square.

To know more about the triangles, here

brainly.com/question/1058720

#SPJ1

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Answer:

B. 21.2

Step-by-step explanation:

Perimeter of ∆ABC = AB + BC + AC

A(-4, 1)

B(-2, 3)

C(3, -4)

✔️Distance between A(-4, 1) and B(-2, 3):

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

AB = \sqrt{(-2 - (-4))^2 + (3 - 1)^2} = \sqrt{(2)^2 + (2)^2)}

AB = \sqrt{4 + 4}

AB = \sqrt{16}

AB = 4 units

✔️Distance between B(-2, 3) and C(3, -4):

BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

BC = \sqrt{(3 - (-2))^2 + (-4 - 3)^2} = \sqrt{(5)^2 + (-7)^2)}

BC = \sqrt{25 + 49}

BC = \sqrt{74}

BC = 8.6 units (nearest tenth)

✔️Distance between A(-4, 1) and C(3, -4):

AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

AC = \sqrt{(3 - (-4))^2 + (-4 - 1)^2} = \sqrt{(7)^2 + (-5)^2)}

AC = \sqrt{47 + 25}

AC = \sqrt{74}

AC = 8.6 units (nearest tenth)

Perimeter of ∆ABC = 4 + 8.6 + 8.6 = 21.2 units

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