Area = 0.5*10*4 = 20 square units
Area = 1/2*5.5*4 = 11 square units
6 square feet
it is 24in The area of the first triangle is 24 [(6*8)/2]. The area of the smaller triangle is 12 [24/2]. The two legs will have a relationship of [4/3x to x], the same as any right triangle.
Information about the lengths of two sides of a triangle is insufficient to determine its area.
Find the perimeter of a right triangle with legs measuring 3 and 4
The area of a right triangle that has legs 7 cm and 4 cm long can be calculated using the fact that a right triangle is half of a rectangle. The area of a rectangle is l*h, so the area of a right triangle is l*h/2. In this case, the area is 14 cm^2.
Area = 1/2*5.5*4 = 11 square units
6 square feet
it is 24in The area of the first triangle is 24 [(6*8)/2]. The area of the smaller triangle is 12 [24/2]. The two legs will have a relationship of [4/3x to x], the same as any right triangle.
Information about the lengths of two sides of a triangle is insufficient to determine its area.
A triangle is half a square The area of a square is height × width(base) So, the area of a triangle is height × base ÷ 2 5×4÷2 = 10
Given the legs a and b of a triangle are 3 and 4, the hypotenuse is: 5
Find the perimeter of a right triangle with legs measuring 3 and 4
the square root of 116
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.
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.
Ah, what a lovely question! To find the area of a triangle, you simply multiply the base by the height and divide by 2. So, for a triangle with a base of 4 and a height of 10, the area would be 20 square units. Just remember, there are no mistakes, just happy little accidents in math!