The points (-3, 7) and (10, 7) lie on the same horizontal line as the y-coordinates are the same), therefore it is best to take this as the base of the triangle; it is 10 - -3 = 10 + 3 = 13 units long.
The third point (2, 19) is 19 - 7 = 12 units above this line which is the height of the triangle.
Thus the area is ½ × base × height = ½ × 13 units × 12 units = 78 square units.
Another possible answer:-
Distance formula: square root of (x2 -x1) together with +(y2 -y1)
Using the distance formula for the given vertices it works out that it is an isosceles triangle with equal sides of 13, a base of 4 Times Square root of 13 and a height of 3 times square root of 13.
Therefore area is: 0.5*4 times sq rt 13*3 times sq rt 13 = 78 square units
Checking using Pythagoras' theorem will confirm that it is an isosceles triangle.
Vertices or points: (0, 0) (3, 4) and (6,0) Type of shape: an isosceles triangle Base: 6 units Height: 4 units Area: 0.5*6*4 = 12 square units
The points will form a right angle triangle in a circle whose hypotenuse is its diameterLength from (8, 5) to (5, 1) = 5Midpoint: (6.5, 3) which is the centre of the circleRadius: 2.5Equation of the circle: (x-6.5)^2+(y-3)^2 = 6.25
If: y = 5x and y = 3 -x Then: 5x = 3 -x => 5x +x = 3 => 6x = 3 => x = 1/2 By substitution point of contact is at: (1/2, 5/2)
Oh, dude, you're talking about triangles now? Alright, so, like, triangles can be classified based on their sides - you've got equilateral triangles (all sides are equal), isosceles triangles (two sides are equal), and scalene triangles (all sides are different lengths). It's like a little triangle party with different sides showing up!
Endpoints: (2, 2) and (10, -4) Midpoint: (6, -1) which is the centre of the circle Distance from (6, -1) to (2, 2) or (10, -4) = 5 which is the radius of the circle Therefore equation of the circle: (x-6)^2 + (y+1)^2 = 25
Vertices or points: (0, 0) (3, 4) and (6,0) Type of shape: an isosceles triangle Base: 6 units Height: 4 units Area: 0.5*6*4 = 12 square units
An adjacency matrix is a matrix showing which vertices of a graph are adjacent to which other vertices.
The given vertices works out as a right angle triangle when plotted on the Cartesian plane and using the distance formula its dimensions with respect to Q are:- Hypotenuse: 15 Adjacent: 6 times square root of 5 Opposite: 3 times square root of 5 Using any of the 3 trigonometry ratios angle Q works out as in the following:- sin-1(opp/hyp) = 26.6 degrees to 3 significant figures tan-1(opp/adj) = 26.6 degrees rounded to 1 decimal place cos-1(adj/hyp) = 26.6 degrees to the nearest tenth
it should be a triangle with no red showing
Plotting the coordinates it is an isosceles triangle with equal sides of 13 and a base of 4*sq rt of 13 with a height of 3*sq rt of 13 Area = 0.5*(4*sq rt of 13)*(3*sq rt of 13) = 78 square units
There cannot be a proof because the statement need not be true.
the illuminati shows a giant triangle on a pyramid showing only one eye
The points will form a right angle triangle in a circle whose hypotenuse is its diameterLength from (8, 5) to (5, 1) = 5Midpoint: (6.5, 3) which is the centre of the circleRadius: 2.5Equation of the circle: (x-6.5)^2+(y-3)^2 = 6.25
I won't explain it in depth but the simple answer is because if you take and identical right triangle then you can arrange them to make a rectangle. I've attached a link to an animation showing it rearranging a triangle to show the relationship.
Suppose the line meets the x and y axes at A and B (respectively). Then at A, x = 0 =-> -7y + 4 = 0 so y = 4/7. Therefore A = (0, 4/7) => |OA| = 4/7, and at B, y = 0 => 3x + 4 = 0 so x = -4/3. Therefore B = (-4/3, 0) => |OB| = 4/3. AOB is a right angled triangle with the right angle at O. Therefore area AOB = 1/2*|OA|*|OB| = 1/2*4/7*4/3 = 8/21 square units.
Love triangle, Annabeth's showing her jealous side, saving the world once again and the end.
Such a triangle would be impossible to construct with the given 3 dimensions because in order to construct a triangle the sum of its 2 shortest sides must be greater than the length of its longest side.