Area of Triangle ABC

[nycmathguy]

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.

 

[MathLover1]

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

 

[nycmathguy]

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

Will do so. Thank you. Great notes.