SOLUTION: What is the area of a triangle whose vertices are R(−4,2) , S(1,2) , and T(−5,−4)?

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Question 1064651: What is the area of a triangle whose vertices are R(−4,2) , S(1,2) , and T(−5,−4)?


Found 2 solutions by KMST, Alan3354:
Answer by KMST(5328)   (Show Source): You can put this solution on YOUR website!
That triangle has a horizontal base with y=2.
The length of that base is RS=1-(-4)=1+4=5.
The height of the triangle is the vertical distance
from line y=2 to T(-5,-4).
That height is
2-(-4)=2+4=6.
The area of the triangle is
.
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
What is the area of a triangle whose vertices are R(-4,2), S(1,2), and T(-5,-4)?
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You can find the 3 side lengths, then use Heron's Law.
-----
Or,
 R   S   T   R
-4   1  -5  -4
 2   2  -4   2

Add the diagonal products starting at the upper left:
= -8 - 4 - 10 = -22
---
Add the diagonal products starting at the lower left:
= 2 - 10 + 16 = 8
|-22 - 8| = 30
The area is 1/2 of that, = 15 sq units.
=====================
This method works for any polygon, any # of sides.
The points have to be in order around the perimeter.

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