In a rectangular coordinate system, what is the area of a triangle whose vertices have the coordinates(4,0),(6,3)and(6,-3)?
a)6
b)7
c)7.5
d)6.5

• Area of a triangle if their co-ordinates are given is,
  
  area = |(A(x)*[B(y)-C(y)]+B(x)*[C(y)-A(y)]+C(x)*[A(y)-B(y)])/2|
  So let the co-ordinates be A(4,0), B(6,3), C(6,-3), now the area will be,
  Area = |[4 * (3+3) + 6 * (-3-0) + 6 * (0-3)] / 2 |
  Area = | -12/2 | = 6
  
• 12 years agoHelpfull: Yes(48) No(4)
• 6
  
  In this case , base = 6 units
  and
  height = 2 units
  Area = 6*2*1/2 = 6 sq units
• 12 years agoHelpfull: Yes(27) No(4)
• Formula for rectangular coordiantes is =A(4,0) B(6,3) C(6,-3)
  =1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
  on soving
  =1/2[4(3-(-3))+6(-3-0)+6(0-3)]
  =1/2[24-18-18}
  =1/2[24-36]
  =1/2*12
  ans =6
  Note- negative is neglected in finding the area of the triangle.
• 10 years agoHelpfull: Yes(10) No(0)
• area of the triangle is 1/2* {x1*(y2-y3)-y1*(x2-x3)+1(x2y3-x3y2)}
  after substituting values we will get ans :6
• 12 years agoHelpfull: Yes(8) No(7)
• area=1/2(l*b)
  l=6-4=2
  b=3(+y)+3(-y)=6
  area=1/2*2*6=6
• 12 years agoHelpfull: Yes(8) No(5)
• NEED MORE EXPLAINANTION
• 12 years agoHelpfull: Yes(6) No(5)
• Answer is Option d) 6.5
  Base=sqrt13 and height=sqrt13
  so area of triangle=1/2*sqrt13*sqrt13=6.5
• 10 years agoHelpfull: Yes(0) No(5)