ok i have 2 problems and id appreciate it if someone can help :)
1. given hexagon ABCDEF with diagonal AC being 2 calculate the area and perimeter of the hexagon.
2. the perimeter of an equilateral triangle is equal to its area. find the side length
Draw line BF, perpendicular to AC
/_GAC = 30º and /_ABG = 30º
AG = GC = 1
AB=1/cos30º=2/sqrt3 BC = CD = DE = EF = FA
BG = ABsin30º = 1/sqrt3
Perimeter P = 6AB
Area = (2)/sqrt3 + 2(2)/sqrt3
meen
Sure! I can explain how to solve both of these problems.
1. To find the area and perimeter of a hexagon, we will need the length of the sides. However, you have given the length of one of the diagonals, AC, which is not a side length. To solve this problem, we'll need some additional information, such as the length of a side or the value of any angles.
2. The perimeter of an equilateral triangle is equal to its area. Let's say the side length of the equilateral triangle is "s". The perimeter of the triangle is simply the sum of the lengths of its three sides, which is 3s.
Now, to find the area of the triangle, we can use the formula: Area = (sqrt(3) /4) * s^2, where "sqrt" represents the square root.
Since the problem states that the perimeter is equal to the area, we can set up an equation:
3s = (sqrt(3) / 4) * s^2
To find the side length, we can solve this equation. Rearrange the equation:
3s = (sqrt(3) / 4) * s^2
Multiply both sides by 4 to get rid of the fraction:
12s = sqrt(3) * s^2
Divide both sides by s to get rid of the "s" term:
12 = sqrt(3) * s
Divide both sides by sqrt(3) to solve for "s":
s = 12 / sqrt(3)
To simplify the expression, we rationalize the denominator by multiplying both the numerator and denominator by sqrt(3):
s = (12 / sqrt(3)) * (sqrt(3) / sqrt(3))
s = (12 * sqrt(3)) / 3
s = 4 * sqrt(3)
Therefore, the side length, "s," of the equilateral triangle is equal to "4 * sqrt(3)".