Asked by Anonymous
A(-11,8)B(2,-6) and C(-19,-8) are the vertices of Triangle ABC. N(X,Y) is a point on AC such the BN is Perpendicular to AC. Find Area of Triangle ABC. PLEASE HELP!!!
Answers
Answered by
bobpursley
The second sentence is not needed.
Draw the triangle. Compute the side lengths
s1=AB= sqrt ((bx-ax)^2+(by-ay)^2 )
= sqrt ( 13^2 + (-14)^2 )= sqrt (169+256)=20.6
and do that for all sides. Then knowing the sides, calculate the perimeter, divide by 2. Let that be p.
Area=sqrt (p(p-a)(p-b)(p-c))
Or you can do it here
http://www.analyzemath.com/Geometry_calculators/perimeter_area_tri_verti.html
Draw the triangle. Compute the side lengths
s1=AB= sqrt ((bx-ax)^2+(by-ay)^2 )
= sqrt ( 13^2 + (-14)^2 )= sqrt (169+256)=20.6
and do that for all sides. Then knowing the sides, calculate the perimeter, divide by 2. Let that be p.
Area=sqrt (p(p-a)(p-b)(p-c))
Or you can do it here
http://www.analyzemath.com/Geometry_calculators/perimeter_area_tri_verti.html
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