What is the area of a quadrilateral with vertices at (1,6),(7,6),(1,12),(7,12)?

Question

What is the area of a quadrilateral with vertices at (1,6),(7,6),(1,12),(7,12)?

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Trúc Chi 1 year 2021-09-04T13:47:34+00:00 1 Answers 9 views 0

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    2021-09-04T13:48:40+00:00

    Answer:

    Let us consider the quadrilateral ABCD shown below. 

    Let us consider the quadrilateral ABCD shown below. 

    In the above quadrilateral, A(x1, y1), B(x2, y2), C(x3, y3) and D(x4, y4) are the vertices.

    To find area of the quadrilateral ABCD, now we have take the vertices A(x1, y1), B(x2, y2), C(x3, y3) and D(x4, y4) of the quadrilateral ABCD in order (counter clockwise direction) and write them column-wise as shown below. 

    Add the diagonal products x1y2, x2y3, x3y4 and x4y1 are shown in the dark arrows.

    (x1y2 + x2y3 + x3y4 + x4y1) —–(1)

    Add the diagonal products x2y1, x3y2, x4y3 and x1y4 are shown in the dotted arrows.

    (x2y1 + x3y2 + x4y3 + x1y4) —–(2)

    Subtract (2) from (1) and multiply the difference by 1/2 to get area of the quadrilateral ABCD.

    So, area of the quadrilateral ABCD is

    =  (1/2) ⋅ {(x1y2 + x2y3 + x3y4 + x4y1)

    – (x2y1 + x3y2 + x4y3 + x1y4)}

    Problem :

    Find the area of the quadrilateral whose vertices are

    (-4, -2), (-3, -5), (3, -2) and (2, 3)

    Solution : 

    Let A(-4, -2), B(-3, -5), C(3, -2) and (2, 3).

    Plot A, B, C and D in a rough diagram and take them in counter-clockwise order.

    Then,

    (x1, y1)  =  (-4, -2)

    (x2, y2)  =  (-3, -5)

    (x3, y3)  =  (3, -2)

    (x4, y4)  =  (2, 3)

    Area of triangle ABC is 

    =  (1/2) ⋅ {(x1y2 + x2y3 + x3y4 + x4y1)

    – (x2y1 + x3y2 + x4y3 + x1y4)}

    =  (1/2) x  {[20 + 6 + 9 – 4] – [6 – 15 – 4 – 12]}

    =  (1/2) x  {[31] – [-25]}

    =  (1/2) x  {31 + 25}

    =  (1/2) x  56

    =  28

    So, area of the given quadrilateral is 28 square units.

    In the above quadrilateral, A(x1, y1), B(x2, y2), C(x3, y3) and D(x4, y4) are the vertices.

    To find area of the quadrilateral ABCD, now we have take the vertices A(x1, y1), B(x2, y2), C(x3, y3) and D(x4, y4) of the quadrilateral ABCD in order (counter clockwise direction) and write them column-wise as shown below. 

    Add the diagonal products x1y2, x2y3, x3y4 and x4y1 are shown in the dark arrows.

    (x1y2 + x2y3 + x3y4 + x4y1) —–(1)

    Add the diagonal products x2y1, x3y2, x4y3 and x1y4 are shown in the dotted arrows.

    (x2y1 + x3y2 + x4y3 + x1y4) —–(2)

    Subtract (2) from (1) and multiply the difference by 1/2 to get area of the quadrilateral ABCD.

    So, area of the quadrilateral ABCD is

    =  (1/2) ⋅ {(x1y2 + x2y3 + x3y4 + x4y1)

    – (x2y1 + x3y2 + x4y3 + x1y4)}

    Problem :

    Find the area of the quadrilateral whose vertices are

    (-4, -2), (-3, -5), (3, -2) and (2, 3)

    Solution : 

    Let A(-4, -2), B(-3, -5), C(3, -2) and (2, 3).

    Plot A, B, C and D in a rough diagram and take them in counter-clockwise order.

    Then,

    (x1, y1)  =  (-4, -2)

    (x2, y2)  =  (-3, -5)

    (x3, y3)  =  (3, -2)

    (x4, y4)  =  (2, 3)

    Area of triangle ABC is 

    =  (1/2) ⋅ {(x1y2 + x2y3 + x3y4 + x4y1)

    – (x2y1 + x3y2 + x4y3 + x1y4)}

    =  (1/2) x  {[20 + 6 + 9 – 4] – [6 – 15 – 4 – 12]}

    =  (1/2) x  {[31] – [-25]}

    =  (1/2) x  {31 + 25}

    =  (1/2) x  56

    =  28

    So, area of the given quadrilateral is 28 square units.

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