Question

triangle area

Answer 1

u can use this formula
\[1/2 \left| y1(x2-x3)+y2(x3-x1)+y3(x2-x1) \right|\]

Answer 2

I'm not what to do with that formula...

Answer 3

For you to find the area of the triangle, you can use the following formulas:
1.)A = 1/2b*h
where b is the length of the base and
h is the length of the altitude or the height perpendicular to the horizontal
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By inspection,
b = 6 units
h = 4 units
Substituting the values to the equation, you'll have:
A = 1/2(6)(4)
A = 12 sq. units
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2.) Heron's Formula:
\[A = \sqrt{s(s-a)(s-b)(s-c)}\]
where s is the semi-perimeter (perimeter divided by 2)
a,b and c are the length of the sides of the triangle
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Using distance formula to find the length of the sides you'll have:
\[EC = \sqrt{(5-1)^2 +(1-(-2))^2}\]
\[EC = 5 units\]
\[CD = \sqrt {(5-1)^2 + (1-4)^2}\]
\[CD = 5 units\]
\[ED = 4-(-2)\]
\[ED = 6 units\]
\[s = \frac{CD+ED+EC}{2}\]
\[s = \frac{5+6+5}{2}\]
\[s=8\]
Substituting the values to Heron's Formula you'll have:
\[A = \sqrt{8(8-6)(8-5)(8-5)}\]
\[A = \sqrt{8(2)(3)(3)}\]
\[A = \sqrt{144}\]
\[A = 12 sq. units\]
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3.) Using angle:
\[A = \frac{1}{2} a b \sin \theta\]
where a and b are the length of the sides adjacent to a known angle
theta is the angle in between a and b
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Let's have angle CED
\[\tan \theta = \frac{4}{3}\]
\[\theta = \tan^{-1} \frac {4}{3}\]
\[\theta = 53.15^o\]
Substituting to the formula:
where a and b are the lengths of CE and ED, 5 and 6 units respectively
\[A = \frac{1}{2} (5)(6)\sin 53.13^o\]
\[A = 12 sq. units\]

Answer 4

wow that's a lot. I really appreciate you typing all that out. I'm going to read through it so I understand.