What is the area of the triangle whose vertices are D( 3, 3), E(3, -1), and F(-2, -5)? Show ALL your work! Your choice of method on ho

Question

What is the area of the triangle whose vertices are D( 3, 3), E(3, -1), and F(-2, -5)?
Show ALL your work! Your choice of method on how to find the area.

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    2021-09-14T18:50:45+00:00

    Answer-

    Area of the triangle is 10 sq.units

    Solution-

    We know that,

    \text{Area of the triangle}=\dfrac{1}{2}[{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]

    Taking,

    (x₁, y₁) = (3, 3)

    (x₂, y₂) = (3, -1)

    (x₃, y₃) = (-2, -5)

    Then putting these in the formula,

    \text{Area of the triangle}=\dfrac{1}{2}[3(-1+5)+3(-5-3)-2(3+1)]

    =\dfrac{1}{2}[3(4)+3(-8)-2(4)]

    =\dfrac{1}{2}[12-24-8]

    =\dfrac{1}{2}[-20]

    =-10

    As area can not be negative, ignoring negative sign,

    \text{Area of the triangle}=10

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