Find the area of the given vertices.?

Question

Okay. Find the area of the given vertices. (Whats a cooridinate grid?)
A: (1,1) (1,4) (-3,4)


B: (1,-3) (3,-1) (5,-3)

Answer 1

using scalene triangle rules the length of the 3 sides for the 2 triangles come to:
A:
3 [ (1,1) - (1,4)]
4 [ (1,4) - (-3,4)]
5[ (-3,4) - (1,1)]
but since this is a right angled triangle the area is simply...(1/2)*3*4=6 sq. units.

B:
sides are:
a=[ (1,-3) - (3,-1)] = 2*sq rt(2) {2 times the square root of 2}
b=[ (3,-1) - (5,-3)] = 2*sq rt(2)
c=[ (1,-3) - (5,-3)] = 4
this is an isosceles triangle and thus the midpoint of the base will be the mid point of c i.e. (3,-3)
thus base =4, height = [(3,-3) - (3,-1)] = 2
=> area = (1/2)*4*2=4

a coordinate grid is the grid formed by major perpendiculars of major units of coordinate axes(2 for 2D coordinate geometry, 3 for 3D coordinate geometry) which are used to plot a point.
e.g.
when you look at a graph sheet then you have the minimal unit as one small square and you'll see that 25 of these small sqares form one main unit's square.
Now according to the scale used on each axes the area represented by each square on that grid varies.
i hope this explained it rather than confused it...

Answer 2

A coordinate grid is a grid with ordered numbers assigned to both the horizontal and vertical lines. The intersection of the two lines labeled 0 is called the origin, and has co-ordinates (0, 0).

A.
(1,1) and (1,4) give you a vertical line 3 units long.
(1,4) and (-3,4) give you a horizontal line 4 units long, so you have a right triangle (a 3, 4, 5 right triangle, at that.)
The area is bh/2 = 3*4/2 = 6 square units

B.
(1,-3) and (5,-3) gives you a horizontal line 4 units long.
The (3,-1) point is 2 units above your horizontal line.
The area of this triangle is 2*4/2 = 4 square units.

Answer 3

the respond is: you won't be in a position to! think of a cube on a table and positioned over it a tetragonal pyramid with a base the better face of the cube (like a house, lined with a roof). This convex polyhedron has 9 vertexes and 9 faces. Now get rid of the decrease face of the cube (the backside of the residing house) and connect its 4 vertexes with the right of the roof, for this reason making 4 extra triangular faces interior the former good (turn the 'residing house' the opposite direction up and dig a pyramidal hollow). you have now yet another closed concave polyhedron with the comparable 9 VERTEXES (and 12 faces) whose floor section isn't comparable to the component of the former! you are able to rather grasp different comparable examples. the element is that the records supplied are no longer adequate. a set of things does no longer define unambiguously a polyhedron (as different solutions wisely point out), till they are 4, defining a distinctive tetrahedron, and the main suitable that is executed is to discover the exterior component of the so called 'convex envelope' - the smallest convex set, containing the standards given (it rather is a convex polyhedron in our case). looking the convex envelope is in many circumstances a complicated situation, there are algorithms for that yet no formula (different than tetrahedron of direction). Sorry if now you experience fairly disenchanted!

Answer 4

area= (1/2)*[1(4-4)+1(4-1)-3(1-4)]=6 sq units

area=(1/2)*
[1(3-5)+3(-3+3)-5(1-3)]
=13/2 sq units