A hexagon with sides of 60 m will have an area greater than 0 m2 and less than or equal to 5400√3 m2 (≈ 9353.1 m2).
The maximum area is a regular hexagon with sides 60 m which has an area of:
area_regular_hexagon = 1/2 x 3 x √3 x side2
= 1/2 x 3 x √3 x (60 m)2vv
= 5400√3m2
≈ 9353.1 m2
The side lengths of a square with an area of 60 square meters is: 7.746 meters.
The formula for a hexagon is 6x.5xsin(60)edge2 = 3xsin(60)edge2=2.598edge2.
Assuming that the "5 and 12" refer to side lengths, the answer is 5*12 = 60 square units.
if r = perpendicular distance center of hexagon to a side, and r^2 = r squared, then AREA = 6x r^2 x tan 30 degrees = 3.464 r^2 or if R = distance center of hexagon to a corner, AREA = 3 R^2 sin 60 = 2.598 R^2
The hexagon will consist of 6 equilateral triangles having sides of 20 cm and so:- Area of the regular hexagon: 0.5*20*20*sin(60 degrees)*6 = 1039 square cm rounded
The side lengths of a square with an area of 60 square meters is: 7.746 meters.
A triangle with sides of 50, 50 and 60 units has an area of 1200 sq units.
A hexagon has 6 sides. The area of a regular hexagon that has a perimeter of 60 cm is 259.81 cm squared.
1, 3, 5 and 15.
A square with an area of 3,600 square units has side lengths of 60 units each.
The formula for a hexagon is 6x.5xsin(60)edge2 = 3xsin(60)edge2=2.598edge2.
Assuming that the "5 and 12" refer to side lengths, the answer is 5*12 = 60 square units.
There is only one equilateral triangle with a perimeter of 60 units. Its side lengths are integers.
The square root of 225 is 15. Since the side length of a square is just the square root of the area, the side length must be 15. As a side note, the perimeter is four times the square root of the area, or in this case, 60.
if r = perpendicular distance center of hexagon to a side, and r^2 = r squared, then AREA = 6x r^2 x tan 30 degrees = 3.464 r^2 or if R = distance center of hexagon to a corner, AREA = 3 R^2 sin 60 = 2.598 R^2
12*5 (sides) =60 inches.
Let the vertices of the hexagon be A, B, C, D, E and F. Join AD, BE and CF. These lines intersect at point M. The external angle of a hexagon =360/6 =60° . The corresponding internal angle is 120°. The lines drawn bisect the internal angle at each vertex. The angles of ∆ABM are 60°, 60° and thus the angle at M is also 60° - the triangle is therefore equilateral with a side length of 6 cm. There are 6 congruent equilateral triangles forming the hexagon. Draw a line MN where N is the mid point of AB then ∆MNA is a right angled triangle with a hypotenuse (MA) length of 6 cm and a base (AN) length of 3 cm. By Pythagoras, the length of the altitude (MN) is √(62 - 32) = √27 cm The area of ∆ABM = 2 x area of ∆MNA = 3√27 The area of the hexagon = 6 x area of ∆ABM = 6 x 3√27 = 18√27 = 93.53 sq cm (2dp)