Monday, August 29, 2011

Two different ways to get the area of a triangle.


If we have three lengths, let's call them u, v and w, we have ways to find out if they could be the three sides of a triangle and if so, we can classify the triangle as isosceles, equilateral or scalene, as well as telling if the triangle is obtuse, right or acute.  With Heron's Formula (sometimes known as Hero's Formula), we can find the area.


Let p be the perimeter, u+v+w.  We will call the semi-perimeter s = ½(u+v+w). The formula for the area is sqrt(s(s-u)(s-v)(s-w)).  The simplest way to do this is to find the semi-perimeter and plug numbers in.

Example #1:  Let the sides be 5, 6 and 7.  The perimeter is 18, so s is 9.  The area is as follows.

sqrt(9(9-5)(9-6)(9-7)) =
sqrt(9 × 4 × 3 × 2) =
6 × sqrt(6) (square units)

Example #2:  Let the sides be 5, 6 and 5.  The perimeter is 16, so s is 8.  The area is as follows.

sqrt(8(8-5)(8-6)(8-5)) =
sqrt(8 × 3 × 2 × 3) =
12 (square units)

This takes more work that A = ½bh, but we aren't always given the height of a triangle.

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Here is a completely different way to define a triangle, three points on the xy-plane where one of the points is the origin (0, 0) and the other two are (x1, y1) and (x2, y2).  Now the formula for area is ½|x1y2 - x2y1|.  Let's do some examples of this.

Example #1: (0, 0) (3, -1) and (4, 2)
The area is ½|3×2 - -1×4| = ½|6 - -4| = ½|10| = 5 (square units) 


Example #2: (0, 0) (6, 3) and (4, 1)
The area is ½|6×1 4×3| = ½|6 - 12| = ½|-6| = 3 (square units) 

Practice for slope, distance and area of a triangle defined by three points.

We will deal with the following three points (3, 4), (7, 5) and (-2, 8) as well as the origin (0, 0)

Slope
a) slope from (3, 4) to (7, 5)
b) slope from (3, 4) to (-2, 8)
c) slope from (3, 4) to (0, 0)


d) slope from (7, 5) to (-2, 8)
e) slope from (7, 5) to (0, 0)
f) slope from (-2, 8) to (0, 0)

Distance




g) distance from (3, 4) to (7, 5)

h) distance from (3, 4) to (-2, 8)
i) distance from (3, 4) to (0, 0)


j) distance from (7, 5) to (-2, 8)
k) distance from (7, 5) to (0, 0)
l) distance from (-2, 8) to (0, 0)

Area of triangles
m) area of triangle with points (3, 4), (7, 5) and (0, 0)

n) area of triangle with points (3, 4), (-2, 8) and (0, 0)
o) area of triangle with points (-2, 8), (7, 5) and (0, 0)
p) area of triangle with points (3, 4), (7, 5) and (-2, 8)

 Answers in the comments.
 
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1 comment:

  1. Slope

    a) slope from (3, 4) to (7, 5)
    slope = 1/4

    b) slope from (3, 4) to (-2, 8)
    slope = -4/5

    c) slope from (3, 4) to (0, 0)
    slope = 4/3

    d) slope from (7, 5) to (-2, 8)
    slope = -3/9 = -1/3

    e) slope from (7, 5) to (0, 0)
    slope = 5/7

    f) slope from (-2, 8) to (0, 0)
    slope = 8/-2 = -4

    Distance

    g) distance from (3, 4) to (7, 5)
    distance = sqrt(17)

    h) distance from (3, 4) to (-2, 8)
    distance = sqrt(41)

    i) distance from (3, 4) to (0, 0)
    distance = sqrt(25) = 5

    j) distance from (7, 5) to (-2, 8)
    distance = sqrt(90) = 3sqrt(10)

    k) distance from (7, 5) to (0, 0)
    distance = sqrt(74)

    l) distance from (-2, 8) to (0, 0)
    distance = sqrt(68) = 2sqrt(17)



    Area of triangles

    m) area of triangle with points (3, 4), (7, 5) and (0, 0)
    ½|3*5 - 4*7| = 13/2 = 6.5

    n) area of triangle with points (3, 4), (-2, 8) and (0, 0)
    ½|3*8 - 4*(-2)| = 32/2 = 16

    o) area of triangle with points (-2, 8), (7, 5) and (0, 0)
    ½|-2*5 - 8*7| = 66/2 = 33

    p) area of triangle with points (3, 4), (7, 5) and (-2, 8)

    subtract (3, 4) from every point and get

    (0,0), (4, 1) and (-5, 4)
    ½|4*4 - 1*-5| = 21/2 = 10.5

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