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Q: Which point is outside of the triangle A. (2 , 6) B. (2 , 5) C. (2 , 7) D. (4 , 6)?

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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

area of triangle =(1/2)(base)(height) 16=(1/2)(4)h h=8

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 times height1/2 * 12 * 4 = 24

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.

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

If it is a right triangle its 6*4/2=12squared if not the its simply 6*4=24sqaured

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.

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