a. (2,6). you draw it on a graph (x,y) b. c. d. make the triangle, a. lies between b. and c.
b
The area of a triangle is (1/2) x (length of the base) x (height of the triangle). You ought to be able to handle it from this point.
2,0
The area of triangle is : 4.0
In short, the orthocenter really has no purpose. There are 4 points of Concurrency in Triangles: 1) The Centroid - the point of concurrency where the 3 medians of a triangle meet. This point is also the triangle's center of gravity. 2) The Circumcenter - the point of concurrency where the perpendicular bisectors of all three sides of the triangle meet. This point is the center of the triangle's circumscribed circle. 3) The Incenter - the point of concurrency where the angle bisectors of all three angles of the triangle meet. Like the circumcenter, the incenter is the center of the inscribed circle of a triangle. 4) The Orthocenter - the point of concurrency where the 3 altitudes of a triangle meet. Unlike the other three points of concurrency, the orthocenter is only there to show that altitudes are concurrent. Thus, bringing me back to the initial statement.
A triangle with sides measuring ; 4 feet , 6 feet and 9 feet is a right triangle. A triangle is a right triangle as long as it has one 90 degree point.
area of triangle =(1/2)(base)(height) 16=(1/2)(4)h h=8
Area of triangle = 1/2 base times height1/2 * 12 * 4 = 24
if hypotenuse is 4 then each side is 4 divided by sqrt 2 = 2.828 with two angles 45 degrees The triangle height is 2 so area is 1/2 base x height =1/2 times 4 times 2 = 4
1. Isoceles triangle 2. saclane triangle 3.equlateral triangle
3 units2
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.
If it is a right triangle its 6*4/2=12squared if not the its simply 6*4=24sqaured