The area is calculated easily using the determinant of the matrix of coordinates, or Heron's formula and is 15 square units.
3 vertices, 3 edges and 1 face.
Area of a triangle = (1/2)(base)(height)
Negate each of the x and y components of all three vertices of the triangle. For example, a triangle with vertices (1,2), (8,3), and (5,6) would become (-1,-2), (-8,-3) and (-5,-6) when rotated 180 degrees about the origin.
0.5
I need help
It is a right angle triangle with an area of: 0.5*7*9 = 31.5 square units
If you mean vertices of: (-1, -1) (-1, 3) and (5, -1) then when plotted on the Cartesian plane it will form a right angle triangle with a base of 6 units and a height of 4 units. Area of triangle: 0.5*6*4 = 12 square units
If the vertices are at (0, -2) (8, -2) and (9, 1) on the Cartesian plane plane then by using the distance formulae and trigonometry the area of the triangle works out as 12 square units.
Your question doesn't seem to make any sense but if the sides of the triangle are 3, 4 and 5 then it is a right angle triangle because it complies with Pythagoras' theorem.
We know that R = a/2sinA area of triangle = 1/2 bc sinA sin A = 2(area of triangle)/bc R = (a/2)*2(area of triangle)/bc R = abc/4*(area of triangle)
3 vertices, 3 edges and 1 face.
Since the sides of triangle are equal, the triangles are equilateral. Just for your information, in this question, we do not require the length of sides. It is just additional information. :) The area of equilateral triangle is: (√3)/4 × a², where a is the side of the equilateral triangle. For triangle ABC, area will be = (√3)/4 × a² (Let 'a' is the side of triangle ABC) Since, side of triangle PQR is half that of ABC, it will be = a/2 Therefore, area of triangle PQR = (√3)/4 × (a/2)² = (√3)/16 × a² Take the ratio of areas of triangle ABC and PQR: [(√3)/4 × a²] / [(√3)/16 × a²] = 4:1
Faces: 1 Vertices: 3 Edges: 3
The coordinates are the vertices of a triangle since they form three points.
Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.And so the area of the inscribed triangle is 12 cm2.Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.And so the area of the inscribed triangle is 12 cm2.Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.And so the area of the inscribed triangle is 12 cm2.Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.And so the area of the inscribed triangle is 12 cm2.
(-5, 6)
Triangular prism has 6 vertices. Square pyramid has 5 vertices. Answer: 1 vertex.