Someone help me w/ this math question ?The area of a triangle with vertices a(0,2) b(8,2) and c(4,-3)?

20 sq units





Base of the triangle is the side from (0.2) to (8,2) which has lenght of 8.





"Height" of triangle is 5 - from the point (4,-3) up to the base which is 2 units above the x axis.





so area = 1/2 bh


area = 1/2 (8)(5)


area = 20 sqare units

Someone help me w/ this math question ?The area of a triangle with vertices a(0,2) b(8,2) and c(4,-3)?
find the lenghts of each side by Pythagorean theorem.





after knowing the lengths of each side (a, b %26amp; c) you can find the area using Heron's theorem.
Reply:First the area of a triangle is a half times base times height. 1/2xbxh.


Therefore substituting the figures in you get this.


=1/2 x 6 x 5


=1/2 x 30


=15
Reply:1/2 |(b-a) cross (c-a)|





b-a = (8,0)


c-a = (4,-5)





(b-a) cross (c-a)


=


| i j k |


| 8 0 0 |


| 4 -5 0 |


=


i(0) - j(0) + k(-40)





| -40k | = 40





1/2 * 40 = 20
Reply:The base of your triangle is the segment AB, with length 8. (If this is not clear from the coordinates, plot them and you'll see this easily.)





The height of your triangle is the segment from point C perpendicular to AB; hence, from (4, -3) to (4,2); the distance of this is 5.





Thus, A = (1/2)*b*h = 20 sq units.





Alternatively, you can use the distance formula and Heron's formula, which would work for any triangle that isn't as easily decomposed into a "base" and "height" by inspection.
Reply:use the equation


area= 1/2 * I x1(y2-y3)+x2(y3-y1)+x3(y1-y2) I





* =multiply


the I.....I= means modulus so if the ans.comes negative ignore it since area can't be -ve right?


(x1,y1) =(0,2) so x1=0 and y1=2


(x2.y2)= (8,2)


(x3,y3)=(4,-3)





area= 1/2 * I 0(2+3) + 8(-3-2)+4(2-2) I


=0.5 * I -40 I


= 0.5 * 40


=20
Reply:Take side a-b, which is horizontal, as the base of the triangle. This distance is 8 - 0 = 8.





Drop a perpendicular line from point c to a point d in side a-b. The height of the triangle is the distance c-d. Since c-d is vertical, the distance is 2 - (-3) = 5.





The area = 1/2*b*h = (1/2)*8*5 = 20
Reply:The lengths of the sides are 8, sqrt(41), sqrt(41).


Seimi-perimeter is 4 + sqrt(41).


Area is sqrt ( (sqrt(41)+4) * (sqrt(41)-4) * 4 * 4 )


= sqrt (25 * 4 * 4)


= 20.
Reply:We know that A and B are on a straight line 8 units apart.





c is 4 units right of A and 5 below, c is 4 units left of B and 5 below.





Therefore we have the perpendicular height, which is 5.


We know the base (from A to B is 8)





Area of Triangle = 1/2 x Base x Perpendicular Height


Area = 1/2 x 8 x 5 = 20 units^2