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What is the area of the triangle ABC with vertices A at -1 4 B at 1 -2 and C at 5 1?
Wiki User
∙ 7y ago
The area is calculated easily using the determinant of the matrix of coordinates, or Heron's formula and is 15 square units.
Wiki User
∙ 7y ago
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How many vertices and edges does a triangle have?
3 vertices, 3 edges and 1 face.
How can you solve an area of a triangle?
Area of a triangle = (1/2)(base)(height)
How do you rotate triangle 180 degrees about the origin?
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.
Triangle ABC below will be dilated with the origin as the center of dilation and scale factor of 1/2?
0.5
what is the perimeter of the triangle ABC with point A(-3,2), B(4,4), C(1,-2)?
I need help
What is the area of the triangle represented by the vertices A(1 1) B(7 1) and C(7 9?
It is a right angle triangle with an area of: 0.5*7*9 = 31.5 square units
How do you find the area of a triangle whose vertices have the coordinates ( -1 -1) (-13) and (5 -1)?
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
What is the area of a triangle with vertices at (0 -2) (8 -2) (9 1)?
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.
Triangle abc has vertices a 1 2 3 b 4 0 5 and c 3 6 4 Show that abc is a right triangle?
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.
How can it be proved that the circumradius of a triangle is the product of three sides divided by four times the area of the triangle?
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)
Find the ratio of area of the triangles ABC and PQR such that AB = BC=CA = 6m each and PQ, QR, RP are half of the sides AB BC and CA respectively?
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
How many vertices and edges does a triangle have?
3 vertices, 3 edges and 1 face.
How many edges and faces and vertices does a triangle have?
Faces: 1 Vertices: 3 Edges: 3
S 6 5 T 8 3 U 12 -1 coordinates are the vertices of a triangle?
The coordinates are the vertices of a triangle since they form three points.
What is the area of an equilateral triangle inscribed in a circle of radius 4cm?
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.
Triangle ABC is reflected over the x-axis. Triangle ABC has points A (-6, -1), B (-2, -1), and C (-5, -6). What is the C' coordinate?
(-5, 6)
How many faces does a triangle have?
One - the section of the plane in which it lies.
