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    quadrilateral area and diagonals

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    7. Figure 4 shows a survey of a building which forms a quadrilateral ABCD
    Calculate a) the length of the diagonals - AC and BD
    b) the area of the plot ABCD

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    © BrainMass Inc. brainmass.com September 26, 2022, 8:47 am ad1c9bdddf
    https://brainmass.com/math/geometric-shapes/quadrilateral-area-diagonals-540023

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    SOLUTION This solution is FREE courtesy of BrainMass!

    7.a
    AC^2 = AB^2 + BC^2 - 2*AB*BC*cos(65)
    => AC = sqrt(21.4^2 + 29.6^2 - 2*21.4*29.6*cos(65)) = 28.26 --ANSWER

    To estimate BD:

    AC^2 = AD^2 + DC^2 - 2 * AD * DC * cos(ADC) = AB^2 + BC^2 - 2*AB*BC*cos(65)
    => 2*AD*DC*cos(ADC) = AD^2 + DC^2 - AB^2 - BC^2 + 2*AB*BC*cos(65)
    => cos(ADC) = (AD^2 + DC^2 - AB^2 - BC^2 + 2*AB*BC*cos(65))/(2*AD*DC)
    => cos(ADC) = (19.3^2 + 17.9^2 - 21.4^2 - 29.6^2 + 2*21.4*29.6*cos(65))/(2*19.3*17.9) = - 0.153
    => angle ADC = cos-1(-0.153) = 98.8 degree == 99 degree
    Hence,
    angle DAB + angle DCB = 360 - 65 - 99 = 196 degree

    From triangle ABC:
    AC/sin(ABC) = AB/sin(ACB)
    => sin(ACB) = AB*sin(ABC)/AC = 21.4 *sin(65)/28.26 = 0.686
    => angle ACB == 43.5 degree

    from triangle ADC:
    AC/sin(ADC) = AD/sin(ACD)
    => sin(ACD) = AD*sin(ADC)/AC = 19.3*sin(99)/28.26 = 0.675
    => angle ACD = 42.5 degree

    Therefore, angle BCD = angle ACB + angle ACD = 43.5+42.5 = 86 degree

    Triangle BCD:
    BD = sqrt(BC^2 + CD^2 - 2*BC*CD*cos(BCD))
    => BD = sqrt(29.6^2 + 17.9^2 - 2*29.6*17.9*cos(86)) = 33.5 --ANSWER.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com September 26, 2022, 8:47 am ad1c9bdddf>
    https://brainmass.com/math/geometric-shapes/quadrilateral-area-diagonals-540023

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