Area of a regular hexagon via area of triangles
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Problem: Find the area of a regular hexagon whose sides measures 5 cm
Sol'n 1: I can cut the hexagon into 6 small triangles, so the area of triangle times 6 will be equal to the area of the polygon. Since the triangles are equilateral I can use the formula for it.
Area of triangle = $fracsqrt34(5)^2= 10.83$
Area of hexagon = (6)(10.83) = 64.95
Sol'n 2: Using the formula $frac12(base)(height)$ for the area of triangle. The angle of triangle (angle at the radius) is equal to 60 degrees (From 360 / 6). Cutting the triangle into half (to get the base and height) will result to an angle of 30 degrees and base of 2.5.
To get the height: $tan(30) = frac2.5height = 4.33$, so the area of the triangle: $frac12(2.5)(4.33) = 5.41$
Area of hexagon = (6)(5.41) = 32.475 which does not equal to the area in solution 1.
Question: I noticed that the area calculated on solution 2 is half the area calculated in solution 1. I don't know why. I don't think plugging in the original length of the base is right or logical? I have answered a related problem like this and I have gotten the polygon's area with 1/2(base)(height) and not plugging the original base length back to the formula of area after getting the height. Topic is easy but I don't get why I get so confused. :( Any help will be appreciated.
geometry area
edited Aug 17 at 23:31
HugoTeixeira
21919
asked May 19 at 14:11
Jayce
1668
add a comment |Â
up vote
2
down vote
favorite
Problem: Find the area of a regular hexagon whose sides measures 5 cm
Sol'n 1: I can cut the hexagon into 6 small triangles, so the area of triangle times 6 will be equal to the area of the polygon. Since the triangles are equilateral I can use the formula for it.
Area of triangle = $fracsqrt34(5)^2= 10.83$
Area of hexagon = (6)(10.83) = 64.95
Sol'n 2: Using the formula $frac12(base)(height)$ for the area of triangle. The angle of triangle (angle at the radius) is equal to 60 degrees (From 360 / 6). Cutting the triangle into half (to get the base and height) will result to an angle of 30 degrees and base of 2.5.
To get the height: $tan(30) = frac2.5height = 4.33$, so the area of the triangle: $frac12(2.5)(4.33) = 5.41$
Area of hexagon = (6)(5.41) = 32.475 which does not equal to the area in solution 1.
Question: I noticed that the area calculated on solution 2 is half the area calculated in solution 1. I don't know why. I don't think plugging in the original length of the base is right or logical? I have answered a related problem like this and I have gotten the polygon's area with 1/2(base)(height) and not plugging the original base length back to the formula of area after getting the height. Topic is easy but I don't get why I get so confused. :( Any help will be appreciated.
geometry area
edited Aug 17 at 23:31
HugoTeixeira
21919
asked May 19 at 14:11
Jayce
1668
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Problem: Find the area of a regular hexagon whose sides measures 5 cm
Sol'n 1: I can cut the hexagon into 6 small triangles, so the area of triangle times 6 will be equal to the area of the polygon. Since the triangles are equilateral I can use the formula for it.
Area of triangle = $fracsqrt34(5)^2= 10.83$
Area of hexagon = (6)(10.83) = 64.95
Sol'n 2: Using the formula $frac12(base)(height)$ for the area of triangle. The angle of triangle (angle at the radius) is equal to 60 degrees (From 360 / 6). Cutting the triangle into half (to get the base and height) will result to an angle of 30 degrees and base of 2.5.
To get the height: $tan(30) = frac2.5height = 4.33$, so the area of the triangle: $frac12(2.5)(4.33) = 5.41$
Area of hexagon = (6)(5.41) = 32.475 which does not equal to the area in solution 1.
Question: I noticed that the area calculated on solution 2 is half the area calculated in solution 1. I don't know why. I don't think plugging in the original length of the base is right or logical? I have answered a related problem like this and I have gotten the polygon's area with 1/2(base)(height) and not plugging the original base length back to the formula of area after getting the height. Topic is easy but I don't get why I get so confused. :( Any help will be appreciated.
geometry area
edited Aug 17 at 23:31
HugoTeixeira
21919
asked May 19 at 14:11
Jayce
1668
Problem: Find the area of a regular hexagon whose sides measures 5 cm
Sol'n 1: I can cut the hexagon into 6 small triangles, so the area of triangle times 6 will be equal to the area of the polygon. Since the triangles are equilateral I can use the formula for it.
Area of triangle = $fracsqrt34(5)^2= 10.83$
Area of hexagon = (6)(10.83) = 64.95
Sol'n 2: Using the formula $frac12(base)(height)$ for the area of triangle. The angle of triangle (angle at the radius) is equal to 60 degrees (From 360 / 6). Cutting the triangle into half (to get the base and height) will result to an angle of 30 degrees and base of 2.5.
To get the height: $tan(30) = frac2.5height = 4.33$, so the area of the triangle: $frac12(2.5)(4.33) = 5.41$
Area of hexagon = (6)(5.41) = 32.475 which does not equal to the area in solution 1.
Question: I noticed that the area calculated on solution 2 is half the area calculated in solution 1. I don't know why. I don't think plugging in the original length of the base is right or logical? I have answered a related problem like this and I have gotten the polygon's area with 1/2(base)(height) and not plugging the original base length back to the formula of area after getting the height. Topic is easy but I don't get why I get so confused. :( Any help will be appreciated.
geometry area
edited Aug 17 at 23:31
HugoTeixeira
21919
asked May 19 at 14:11
Jayce
1668
edited Aug 17 at 23:31
HugoTeixeira
21919
edited Aug 17 at 23:31
HugoTeixeira
21919
edited Aug 17 at 23:31
HugoTeixeira
21919
21919
asked May 19 at 14:11
Jayce
1668
asked May 19 at 14:11
Jayce
1668
asked May 19 at 14:11
Jayce
1668
1668
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
You've created your diagram like this, calculated the area of the triangle I've labelled, and then multiplied that area by $6$. Notice however, by your divisions, we've split the hexagon into $12$ triangles, so we would need to multiply the area by $12$, not $6$, which explains why your area for Sol $2$ is half the area for Sol $1$ (which is correct)
The area of one triangle is given by $frac 12cdot2.5cdot2.5tan(60)=frac25sqrt38$. Multiplying this area by $12$ (the amount of triangles) gives us $frac75sqrt32approx64.9519$
answered May 19 at 14:24
Rhys Hughes
4,0081227
Please excuse the awful quality of the diagram but I hope that it is clear enough
– Rhys Hughes
May 19 at 14:29
I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I also dissected the octagon into 8 triangles and like in the problem above, I formed a right triangle with hypotenuse = 12, angle = 22.5 and base = x/2. The calculated area of the octagon using this method equals the calculated area using the other triangle formula: $absintheta$ and I did not need to multiply the area by 16 but just by 8 which confuses me.
– Jayce
May 19 at 14:36
Typo on my comment above. Should be $frac12absintheta$.
– Jayce
May 19 at 14:52
Octagons have a regular angle of $135^0$. When you divide the octagon into eight triangles, you get an angle of $45^0$ at the centre. Thus for the whole octagon you should use $8cdotfrac12cdot12^2cdotsin45=288sqrt2 approx407.2935$
– Rhys Hughes
May 19 at 14:58
You are right. Try solving it via $frac12(base)(height)$. It will result to 407.29 even though the area is only multiplied by 8 which should be by 16 as per your explanation. I used 22.5 as the angle for the right triangle, "x/2" for the base and the hypotenuse, 12. Then Via "soh-cah-toa", base is 9.184 and height is 11.087. Plugging it in: $frac12(9.184)(11.087) = 50.911$ then multiplied by 8 which equals to 407.29, the same with $absintheta$ but I only multiplied it by 8 and not by 16. I need an answer for this, so I can move on. :(
– Jayce
May 19 at 15:14
 |Â
show 2 more comments
up vote
2
down vote
Solution 1 gives the correct answer. In solution 2, the triangle that you're calculating is actually a right-angled triangle with interior angles $30°$, $60°$ and $90°$, and it is in fact half of the triangle calculated in solution 1. This explains why the numerical result obtained in solution 2 is half of that in solution 1.
answered May 19 at 14:18
GNU Supporter
11.8k72143
I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I used 2 formulas of triangle: $frac12absintheta$ and $frac12(base)(height)$. For $frac12(base)(height)$, I did the same thing, I formed a right triangle to get the height then calculated the area then multiplied it by 8 to get the octagon area. This equals the area calculated using $frac12absintheta$. It equals even though I multiplied it just by 8 which is supposed to be by 16 just like you explained. :(
– Jayce
May 19 at 14:50
The solution for the above, @The right triangle: angle = 22.5 (From 360/8 = 45/2 = 22.5), side = 12 (From the given radius) and base = x/2. Via "soh-cah-toa", base = 9.184 and height = 11.087. Plugging it into $8*[frac12(base)(height)] = 407.89$ which is the area of octagon. Via $frac12absintheta$ where a = b = 12 and angle = 45 (from 360/8). Plugging it in the formula and multiplying it by 8 to get the octagon area, we get 407.29. The areas are equal even though in solution 1, I just multiplied it by 8.
– Jayce
May 19 at 15:05
@Jayce Sorry for late reply as I was outside. It seems that another answerer is following your problem. Please leave a comment if you need further help :-)
– GNU Supporter
May 19 at 19:52
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
You've created your diagram like this, calculated the area of the triangle I've labelled, and then multiplied that area by $6$. Notice however, by your divisions, we've split the hexagon into $12$ triangles, so we would need to multiply the area by $12$, not $6$, which explains why your area for Sol $2$ is half the area for Sol $1$ (which is correct)
The area of one triangle is given by $frac 12cdot2.5cdot2.5tan(60)=frac25sqrt38$. Multiplying this area by $12$ (the amount of triangles) gives us $frac75sqrt32approx64.9519$
answered May 19 at 14:24
Rhys Hughes
4,0081227
Please excuse the awful quality of the diagram but I hope that it is clear enough
– Rhys Hughes
May 19 at 14:29
I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I also dissected the octagon into 8 triangles and like in the problem above, I formed a right triangle with hypotenuse = 12, angle = 22.5 and base = x/2. The calculated area of the octagon using this method equals the calculated area using the other triangle formula: $absintheta$ and I did not need to multiply the area by 16 but just by 8 which confuses me.
– Jayce
May 19 at 14:36
Typo on my comment above. Should be $frac12absintheta$.
– Jayce
May 19 at 14:52
Octagons have a regular angle of $135^0$. When you divide the octagon into eight triangles, you get an angle of $45^0$ at the centre. Thus for the whole octagon you should use $8cdotfrac12cdot12^2cdotsin45=288sqrt2 approx407.2935$
– Rhys Hughes
May 19 at 14:58
You are right. Try solving it via $frac12(base)(height)$. It will result to 407.29 even though the area is only multiplied by 8 which should be by 16 as per your explanation. I used 22.5 as the angle for the right triangle, "x/2" for the base and the hypotenuse, 12. Then Via "soh-cah-toa", base is 9.184 and height is 11.087. Plugging it in: $frac12(9.184)(11.087) = 50.911$ then multiplied by 8 which equals to 407.29, the same with $absintheta$ but I only multiplied it by 8 and not by 16. I need an answer for this, so I can move on. :(
– Jayce
May 19 at 15:14
 |Â
show 2 more comments
up vote
5
down vote
accepted
You've created your diagram like this, calculated the area of the triangle I've labelled, and then multiplied that area by $6$. Notice however, by your divisions, we've split the hexagon into $12$ triangles, so we would need to multiply the area by $12$, not $6$, which explains why your area for Sol $2$ is half the area for Sol $1$ (which is correct)
The area of one triangle is given by $frac 12cdot2.5cdot2.5tan(60)=frac25sqrt38$. Multiplying this area by $12$ (the amount of triangles) gives us $frac75sqrt32approx64.9519$
answered May 19 at 14:24
Rhys Hughes
4,0081227
Please excuse the awful quality of the diagram but I hope that it is clear enough
– Rhys Hughes
May 19 at 14:29
I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I also dissected the octagon into 8 triangles and like in the problem above, I formed a right triangle with hypotenuse = 12, angle = 22.5 and base = x/2. The calculated area of the octagon using this method equals the calculated area using the other triangle formula: $absintheta$ and I did not need to multiply the area by 16 but just by 8 which confuses me.
– Jayce
May 19 at 14:36
Typo on my comment above. Should be $frac12absintheta$.
– Jayce
May 19 at 14:52
Octagons have a regular angle of $135^0$. When you divide the octagon into eight triangles, you get an angle of $45^0$ at the centre. Thus for the whole octagon you should use $8cdotfrac12cdot12^2cdotsin45=288sqrt2 approx407.2935$
– Rhys Hughes
May 19 at 14:58
You are right. Try solving it via $frac12(base)(height)$. It will result to 407.29 even though the area is only multiplied by 8 which should be by 16 as per your explanation. I used 22.5 as the angle for the right triangle, "x/2" for the base and the hypotenuse, 12. Then Via "soh-cah-toa", base is 9.184 and height is 11.087. Plugging it in: $frac12(9.184)(11.087) = 50.911$ then multiplied by 8 which equals to 407.29, the same with $absintheta$ but I only multiplied it by 8 and not by 16. I need an answer for this, so I can move on. :(
– Jayce
May 19 at 15:14
 |Â
show 2 more comments
up vote
5
down vote
accepted
up vote
5
down vote
accepted
You've created your diagram like this, calculated the area of the triangle I've labelled, and then multiplied that area by $6$. Notice however, by your divisions, we've split the hexagon into $12$ triangles, so we would need to multiply the area by $12$, not $6$, which explains why your area for Sol $2$ is half the area for Sol $1$ (which is correct)
The area of one triangle is given by $frac 12cdot2.5cdot2.5tan(60)=frac25sqrt38$. Multiplying this area by $12$ (the amount of triangles) gives us $frac75sqrt32approx64.9519$
answered May 19 at 14:24
Rhys Hughes
4,0081227
You've created your diagram like this, calculated the area of the triangle I've labelled, and then multiplied that area by $6$. Notice however, by your divisions, we've split the hexagon into $12$ triangles, so we would need to multiply the area by $12$, not $6$, which explains why your area for Sol $2$ is half the area for Sol $1$ (which is correct)
The area of one triangle is given by $frac 12cdot2.5cdot2.5tan(60)=frac25sqrt38$. Multiplying this area by $12$ (the amount of triangles) gives us $frac75sqrt32approx64.9519$
answered May 19 at 14:24
Rhys Hughes
4,0081227
answered May 19 at 14:24
Rhys Hughes
4,0081227
answered May 19 at 14:24
Rhys Hughes
4,0081227
answered May 19 at 14:24
Rhys Hughes
4,0081227
4,0081227
Please excuse the awful quality of the diagram but I hope that it is clear enough
– Rhys Hughes
May 19 at 14:29
I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I also dissected the octagon into 8 triangles and like in the problem above, I formed a right triangle with hypotenuse = 12, angle = 22.5 and base = x/2. The calculated area of the octagon using this method equals the calculated area using the other triangle formula: $absintheta$ and I did not need to multiply the area by 16 but just by 8 which confuses me.
– Jayce
May 19 at 14:36
Typo on my comment above. Should be $frac12absintheta$.
– Jayce
May 19 at 14:52
Octagons have a regular angle of $135^0$. When you divide the octagon into eight triangles, you get an angle of $45^0$ at the centre. Thus for the whole octagon you should use $8cdotfrac12cdot12^2cdotsin45=288sqrt2 approx407.2935$
– Rhys Hughes
May 19 at 14:58
You are right. Try solving it via $frac12(base)(height)$. It will result to 407.29 even though the area is only multiplied by 8 which should be by 16 as per your explanation. I used 22.5 as the angle for the right triangle, "x/2" for the base and the hypotenuse, 12. Then Via "soh-cah-toa", base is 9.184 and height is 11.087. Plugging it in: $frac12(9.184)(11.087) = 50.911$ then multiplied by 8 which equals to 407.29, the same with $absintheta$ but I only multiplied it by 8 and not by 16. I need an answer for this, so I can move on. :(
– Jayce
May 19 at 15:14
 |Â
show 2 more comments
Please excuse the awful quality of the diagram but I hope that it is clear enough
– Rhys Hughes
May 19 at 14:29
I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I also dissected the octagon into 8 triangles and like in the problem above, I formed a right triangle with hypotenuse = 12, angle = 22.5 and base = x/2. The calculated area of the octagon using this method equals the calculated area using the other triangle formula: $absintheta$ and I did not need to multiply the area by 16 but just by 8 which confuses me.
– Jayce
May 19 at 14:36
Typo on my comment above. Should be $frac12absintheta$.
– Jayce
May 19 at 14:52
Octagons have a regular angle of $135^0$. When you divide the octagon into eight triangles, you get an angle of $45^0$ at the centre. Thus for the whole octagon you should use $8cdotfrac12cdot12^2cdotsin45=288sqrt2 approx407.2935$
– Rhys Hughes
May 19 at 14:58
You are right. Try solving it via $frac12(base)(height)$. It will result to 407.29 even though the area is only multiplied by 8 which should be by 16 as per your explanation. I used 22.5 as the angle for the right triangle, "x/2" for the base and the hypotenuse, 12. Then Via "soh-cah-toa", base is 9.184 and height is 11.087. Plugging it in: $frac12(9.184)(11.087) = 50.911$ then multiplied by 8 which equals to 407.29, the same with $absintheta$ but I only multiplied it by 8 and not by 16. I need an answer for this, so I can move on. :(
– Jayce
May 19 at 15:14
Please excuse the awful quality of the diagram but I hope that it is clear enough
– Rhys Hughes
May 19 at 14:29
Please excuse the awful quality of the diagram but I hope that it is clear enough
– Rhys Hughes
May 19 at 14:29
I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I also dissected the octagon into 8 triangles and like in the problem above, I formed a right triangle with hypotenuse = 12, angle = 22.5 and base = x/2. The calculated area of the octagon using this method equals the calculated area using the other triangle formula: $absintheta$ and I did not need to multiply the area by 16 but just by 8 which confuses me.
– Jayce
May 19 at 14:36
I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I also dissected the octagon into 8 triangles and like in the problem above, I formed a right triangle with hypotenuse = 12, angle = 22.5 and base = x/2. The calculated area of the octagon using this method equals the calculated area using the other triangle formula: $absintheta$ and I did not need to multiply the area by 16 but just by 8 which confuses me.
– Jayce
May 19 at 14:36
Typo on my comment above. Should be $frac12absintheta$.
– Jayce
May 19 at 14:52
Typo on my comment above. Should be $frac12absintheta$.
– Jayce
May 19 at 14:52
Octagons have a regular angle of $135^0$. When you divide the octagon into eight triangles, you get an angle of $45^0$ at the centre. Thus for the whole octagon you should use $8cdotfrac12cdot12^2cdotsin45=288sqrt2 approx407.2935$
– Rhys Hughes
May 19 at 14:58
Octagons have a regular angle of $135^0$. When you divide the octagon into eight triangles, you get an angle of $45^0$ at the centre. Thus for the whole octagon you should use $8cdotfrac12cdot12^2cdotsin45=288sqrt2 approx407.2935$
– Rhys Hughes
May 19 at 14:58
You are right. Try solving it via $frac12(base)(height)$. It will result to 407.29 even though the area is only multiplied by 8 which should be by 16 as per your explanation. I used 22.5 as the angle for the right triangle, "x/2" for the base and the hypotenuse, 12. Then Via "soh-cah-toa", base is 9.184 and height is 11.087. Plugging it in: $frac12(9.184)(11.087) = 50.911$ then multiplied by 8 which equals to 407.29, the same with $absintheta$ but I only multiplied it by 8 and not by 16. I need an answer for this, so I can move on. :(
– Jayce
May 19 at 15:14
You are right. Try solving it via $frac12(base)(height)$. It will result to 407.29 even though the area is only multiplied by 8 which should be by 16 as per your explanation. I used 22.5 as the angle for the right triangle, "x/2" for the base and the hypotenuse, 12. Then Via "soh-cah-toa", base is 9.184 and height is 11.087. Plugging it in: $frac12(9.184)(11.087) = 50.911$ then multiplied by 8 which equals to 407.29, the same with $absintheta$ but I only multiplied it by 8 and not by 16. I need an answer for this, so I can move on. :(
– Jayce
May 19 at 15:14
 |Â
show 2 more comments
up vote
2
down vote
Solution 1 gives the correct answer. In solution 2, the triangle that you're calculating is actually a right-angled triangle with interior angles $30°$, $60°$ and $90°$, and it is in fact half of the triangle calculated in solution 1. This explains why the numerical result obtained in solution 2 is half of that in solution 1.
answered May 19 at 14:18
GNU Supporter
11.8k72143
I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I used 2 formulas of triangle: $frac12absintheta$ and $frac12(base)(height)$. For $frac12(base)(height)$, I did the same thing, I formed a right triangle to get the height then calculated the area then multiplied it by 8 to get the octagon area. This equals the area calculated using $frac12absintheta$. It equals even though I multiplied it just by 8 which is supposed to be by 16 just like you explained. :(
– Jayce
May 19 at 14:50
The solution for the above, @The right triangle: angle = 22.5 (From 360/8 = 45/2 = 22.5), side = 12 (From the given radius) and base = x/2. Via "soh-cah-toa", base = 9.184 and height = 11.087. Plugging it into $8*[frac12(base)(height)] = 407.89$ which is the area of octagon. Via $frac12absintheta$ where a = b = 12 and angle = 45 (from 360/8). Plugging it in the formula and multiplying it by 8 to get the octagon area, we get 407.29. The areas are equal even though in solution 1, I just multiplied it by 8.
– Jayce
May 19 at 15:05
@Jayce Sorry for late reply as I was outside. It seems that another answerer is following your problem. Please leave a comment if you need further help :-)
– GNU Supporter
May 19 at 19:52
add a comment |Â
up vote
2
down vote
Solution 1 gives the correct answer. In solution 2, the triangle that you're calculating is actually a right-angled triangle with interior angles $30°$, $60°$ and $90°$, and it is in fact half of the triangle calculated in solution 1. This explains why the numerical result obtained in solution 2 is half of that in solution 1.
answered May 19 at 14:18
GNU Supporter
11.8k72143
I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I used 2 formulas of triangle: $frac12absintheta$ and $frac12(base)(height)$. For $frac12(base)(height)$, I did the same thing, I formed a right triangle to get the height then calculated the area then multiplied it by 8 to get the octagon area. This equals the area calculated using $frac12absintheta$. It equals even though I multiplied it just by 8 which is supposed to be by 16 just like you explained. :(
– Jayce
May 19 at 14:50
The solution for the above, @The right triangle: angle = 22.5 (From 360/8 = 45/2 = 22.5), side = 12 (From the given radius) and base = x/2. Via "soh-cah-toa", base = 9.184 and height = 11.087. Plugging it into $8*[frac12(base)(height)] = 407.89$ which is the area of octagon. Via $frac12absintheta$ where a = b = 12 and angle = 45 (from 360/8). Plugging it in the formula and multiplying it by 8 to get the octagon area, we get 407.29. The areas are equal even though in solution 1, I just multiplied it by 8.
– Jayce
May 19 at 15:05
@Jayce Sorry for late reply as I was outside. It seems that another answerer is following your problem. Please leave a comment if you need further help :-)
– GNU Supporter
May 19 at 19:52
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Solution 1 gives the correct answer. In solution 2, the triangle that you're calculating is actually a right-angled triangle with interior angles $30°$, $60°$ and $90°$, and it is in fact half of the triangle calculated in solution 1. This explains why the numerical result obtained in solution 2 is half of that in solution 1.
answered May 19 at 14:18
GNU Supporter
11.8k72143
Solution 1 gives the correct answer. In solution 2, the triangle that you're calculating is actually a right-angled triangle with interior angles $30°$, $60°$ and $90°$, and it is in fact half of the triangle calculated in solution 1. This explains why the numerical result obtained in solution 2 is half of that in solution 1.
answered May 19 at 14:18
GNU Supporter
11.8k72143
answered May 19 at 14:18
GNU Supporter
11.8k72143
answered May 19 at 14:18
GNU Supporter
11.8k72143
answered May 19 at 14:18
GNU Supporter
11.8k72143
11.8k72143
I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I used 2 formulas of triangle: $frac12absintheta$ and $frac12(base)(height)$. For $frac12(base)(height)$, I did the same thing, I formed a right triangle to get the height then calculated the area then multiplied it by 8 to get the octagon area. This equals the area calculated using $frac12absintheta$. It equals even though I multiplied it just by 8 which is supposed to be by 16 just like you explained. :(
– Jayce
May 19 at 14:50
The solution for the above, @The right triangle: angle = 22.5 (From 360/8 = 45/2 = 22.5), side = 12 (From the given radius) and base = x/2. Via "soh-cah-toa", base = 9.184 and height = 11.087. Plugging it into $8*[frac12(base)(height)] = 407.89$ which is the area of octagon. Via $frac12absintheta$ where a = b = 12 and angle = 45 (from 360/8). Plugging it in the formula and multiplying it by 8 to get the octagon area, we get 407.29. The areas are equal even though in solution 1, I just multiplied it by 8.
– Jayce
May 19 at 15:05
@Jayce Sorry for late reply as I was outside. It seems that another answerer is following your problem. Please leave a comment if you need further help :-)
– GNU Supporter
May 19 at 19:52
add a comment |Â
I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I used 2 formulas of triangle: $frac12absintheta$ and $frac12(base)(height)$. For $frac12(base)(height)$, I did the same thing, I formed a right triangle to get the height then calculated the area then multiplied it by 8 to get the octagon area. This equals the area calculated using $frac12absintheta$. It equals even though I multiplied it just by 8 which is supposed to be by 16 just like you explained. :(
– Jayce
May 19 at 14:50
The solution for the above, @The right triangle: angle = 22.5 (From 360/8 = 45/2 = 22.5), side = 12 (From the given radius) and base = x/2. Via "soh-cah-toa", base = 9.184 and height = 11.087. Plugging it into $8*[frac12(base)(height)] = 407.89$ which is the area of octagon. Via $frac12absintheta$ where a = b = 12 and angle = 45 (from 360/8). Plugging it in the formula and multiplying it by 8 to get the octagon area, we get 407.29. The areas are equal even though in solution 1, I just multiplied it by 8.
– Jayce
May 19 at 15:05
@Jayce Sorry for late reply as I was outside. It seems that another answerer is following your problem. Please leave a comment if you need further help :-)
– GNU Supporter
May 19 at 19:52
I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I used 2 formulas of triangle: $frac12absintheta$ and $frac12(base)(height)$. For $frac12(base)(height)$, I did the same thing, I formed a right triangle to get the height then calculated the area then multiplied it by 8 to get the octagon area. This equals the area calculated using $frac12absintheta$. It equals even though I multiplied it just by 8 which is supposed to be by 16 just like you explained. :(
– Jayce
May 19 at 14:50
I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I used 2 formulas of triangle: $frac12absintheta$ and $frac12(base)(height)$. For $frac12(base)(height)$, I did the same thing, I formed a right triangle to get the height then calculated the area then multiplied it by 8 to get the octagon area. This equals the area calculated using $frac12absintheta$. It equals even though I multiplied it just by 8 which is supposed to be by 16 just like you explained. :(
– Jayce
May 19 at 14:50
The solution for the above, @The right triangle: angle = 22.5 (From 360/8 = 45/2 = 22.5), side = 12 (From the given radius) and base = x/2. Via "soh-cah-toa", base = 9.184 and height = 11.087. Plugging it into $8*[frac12(base)(height)] = 407.89$ which is the area of octagon. Via $frac12absintheta$ where a = b = 12 and angle = 45 (from 360/8). Plugging it in the formula and multiplying it by 8 to get the octagon area, we get 407.29. The areas are equal even though in solution 1, I just multiplied it by 8.
– Jayce
May 19 at 15:05
The solution for the above, @The right triangle: angle = 22.5 (From 360/8 = 45/2 = 22.5), side = 12 (From the given radius) and base = x/2. Via "soh-cah-toa", base = 9.184 and height = 11.087. Plugging it into $8*[frac12(base)(height)] = 407.89$ which is the area of octagon. Via $frac12absintheta$ where a = b = 12 and angle = 45 (from 360/8). Plugging it in the formula and multiplying it by 8 to get the octagon area, we get 407.29. The areas are equal even though in solution 1, I just multiplied it by 8.
– Jayce
May 19 at 15:05
@Jayce Sorry for late reply as I was outside. It seems that another answerer is following your problem. Please leave a comment if you need further help :-)
– GNU Supporter
May 19 at 19:52
@Jayce Sorry for late reply as I was outside. It seems that another answerer is following your problem. Please leave a comment if you need further help :-)
– GNU Supporter
May 19 at 19:52
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2787516%2farea-of-a-regular-hexagon-via-area-of-triangles%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
