Area of Triangle ABC

Discussion in 'Algebra' started by nycmathguy, Jan 7, 2022.

  1. nycmathguy

    nycmathguy

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    Set 1.4
    David Cohen

    1. Do Part (a) using the hint provided. I will then use your work in (a) to do Part (b) using the same hint.

    2. What is the set up for the work requested by Part (c)?

    3. Explain Cohen's "Remark" in Part (c). What exactly is he saying there?

    Note: This weekend we go back to our Ron Larson Precalculus textbook as we come to the end of Section 5.5. As you know, Section 6.1 (we will start on Sunday) is all about the Law of Sines.



    Screenshot_20220107-041840_Samsung Notes.jpg
     
    nycmathguy, Jan 7, 2022
    #1
  2. nycmathguy

    MathLover1

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    a.

    as you can see, the area of the triangle is equal to the difference between the areas of the rectangle and the sum of areas of three triangles with blue dashed legs (note: all three are right triangles)

    the length of rectangle is distance between x coordinates of the points A and B, which is 5
    the width of rectangle is distance between y coordinates of the points A and C, which is 4

    so, the area of rectangle is
    A=5*4
    A=20 square units

    one of these three triangles has hypothenuse AC, the length of one leg is distance between x coordinates of the points A and C which is 4 units, the other leg is distance between y coordinates which is 4 units
    the area will be: (1/2)*4*4=8 sq. units

    second triangle , hypothenuse BC,
    one leg=1unit
    other leg=3 units
    the area will be: (1/2)*1*3=(3/2) sq. units

    third triangle, hypothenuse AB
    one leg=5 unit
    other leg=1 units
    the area will be: (1/2)*5*1=(5/2) sq. units

    then, the area of the triangle ABC is

    area=20-(8+3/2+5/2)
    area=20-12
    area=8 sq. units

    apply same method to do b
     
    MathLover1, Jan 7, 2022
    #2
    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    Will do so. Thank you. Great notes.
     
    nycmathguy, Jan 8, 2022
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