Is it possible to determine the area of a regular hexagon with only the apothem?

Question

The following problem has stumped me for the last 2 hours-

A regular hexagon has a line down the center perpendicular to the bottom side measuring 14. I have to find the area using only this. For reference, I am only in 10th grade so I have no experience with SIN/TAN so I cannot use those to find the answer, even if I use some formulae I found.

I checked the back of the book and found the answer is 98 x SQRT of 3 equals roughly 170.

As a recap, my question is how do I find out that the area of a regular hexagon is 170 sq. units knowing only that the apothem is 7? Thanks in advance. I'll be back to pick a best answer ASAP.

Answer 1

You can think of the hexagon as six equilateral triangles stuck together. So if you can find the area of one triangle, you just have to multiply it by 6.

The area of a triangle is 1/2 base times height. We already know that the height of the triangle is 7 (the apothem). The trick is to find the length of one of the sides of the equilateral triangles.

The apothem divides one of the sides into two equal parts. So lets say the length of one side is "2 a". The apothem, one half of one side of the triangle and a line from the center to a vertex form a right triangle. So we can say that (2 a)^2 = 7^2 + a^2 using Pythagorean's Theorem. From this we can solve for "a".

a = 7/sqrt(3)
1/2 base times height = (1/2) * (2 a) * 7 = (1/2) * (14/sqrt(3)) * 7
so the area of one square is 49/sqrt(3)

multiply by 6 triangles gives an area of 294/sqrt(3). Multiply this by sqrt(3)/sqrt(3) and you get 98 * sqrt(3)

Answer 2

Draw lines from the center to each of the vertices, dividing the hexagon into six identical equilateral triangles. Every one of these triangles is bisected by an apothem, which doubles as the altitude of the triangle. Let x be the length of one side of one of these triangles. Then the apothem divides it into two right triangles with legs of length x/2 and 7 and hypotenuse of length x. We can solve for x with the pythagorean theorem:
x²=(x/2)²+7² → 3x²/4 = 49 → x² = 4*49/3 → x = 2*7/√3 = 14√3/3

Thus, we know the base of each triangle is 14√3/3, and its height is 7. Thus, the area of each triangle is 1/2 * 14√3/3 * 7 = 49√3/3. There are six of these triangles in the hexagon, so multiply the area of each triangle by six to yield the area of the hexagon, 98√3.

Answer 3

The the perpendicular defines a 30-60-90 triangle and you know the length of one side and all of the angles. Let the length of the bottom side of the triangle be x. Then you have x/7=tan 30 from which x=7/sqrt(3). The area of the triangle is 7*7/2sqrt(3) and there are 12 such triangles. You need to factor the terms to get area = 7*7*4*3/(2*sqrt(3) = 98*sqrt(3)

Answer 4

OK, I'm not sure if you have this much experience at the moment, but there are two things you should be knowing:

1. A regular hexagon can be broken down into a pattern of six equilateral triangles, each of whose sides are of length 'a' which in turn is also equal to each side of the regular hexagon itself.

2. The area of an equilateral triangle equals (a^2)*sqrt(3)/4

Also, the area of one of those triangles = a*h/2

'h' is the height of the triangle, which in our case equals half the apothegm of the hexagon (if you look at the pattern). Thus, the height of our triangle equals h = 14/2 = 7

Therefore, we have:

a*h/2 = (a^2)*sqrt(3)/4
h/2 = a*sqrt(3)/4
h = a*sqrt(3)/2
a = h*2/sqrt(3)
a = 7*2/sqrt(3)
a = 14/sqrt(3)

Thus, the area of one of the triangles =
a*h/2 = {14/sqrt(3)}*(7/2) = 49/sqrt(3)

Thus, area of the hexagon = 6 times area of each triangle
A = 6*49/sqrt(3) = 2*3*49/sqrt(3) = 2*49*{3/sqrt(3)}
A = 2*49*sqrt(3)
A = 98*sqrt(3)

I hope you understand it now.

Answer 5

Radius? a known hexagon (6 sided ensure with equivalent length aspects) does not have a radius so a techniques as i comprehend, yet i ought to o.k. be mistaken. a known hexagon is only a variety of of 6 equilateral triangles of equivalent length which meet on the middle. If I draw lines to connect the corners with techniques from dealing with the middle, you'll see the 6 triangles. you artwork the component to the hexagon with techniques from first determining the component to a unmarried triangle and then only multiplying with techniques from 6. there is somewhat a formula which mixes like words and resolves to, section=(a million/2)(altitude)(perimeter) the position altitude is the area from the middle of a facet to the middle of the ensure and perimeter is the sum of the 6 section lengths. So, on the grounds that i do not comprehend what you're concerning with techniques out of your use of radius, i am going to't replace to discover the section. although, on the grounds that I have given you the formula as I discovered it, you ought to be able to ensure out some thing for your self.