A flower in a hexagon3D Geometry Contest Math ProblemFind the area of the region $ABCD$.How do you calculate...
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A flower in a hexagon
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A flower in a hexagon
3D Geometry Contest Math ProblemFind the area of the region $ABCD$.How do you calculate the smallest cycle possible for a given tile shape?A circle is divided into equal arcs by $n$ diamtrs. Prove the feet of the $perp$s from a point $M$ in circle to them are vertices of a regular n-gonDesign a tangramDividing a polygon in desired wayTricky Geometry Problem involving rectangle rotation.What's the word for same-length-sided-ness? equilarity? equilaterity? equilateralness?What 'uniform' shapes can be used to build an approximated spherical object?Area of a regular hexagon via area of triangles
$begingroup$
The area of the flower
So I had a geometry problem in my maths exam test and I couldn't solve it, though it didn't seem that hard. I tried to connect the points from center to end and create an equilateral triangle but still was unable to find the solution.
Would be thankful if you could help me out.
The question is the following:
We have an equal hexagon with sides equal to $1$ and six circular arcs with radius equal to $1$ from each vertices of the regular hexagon in it create a flower-shaped-like object. Find the area of the flower.
Here are the possible answers ↓
1. $pi$
3. $3pi/2$
4. $4sqrt2-pi$
5. $pi/2+sqrt3$
6. $2pi-3sqrt3$
geometry logic triangle area
$endgroup$
add a comment |
$begingroup$
The area of the flower
So I had a geometry problem in my maths exam test and I couldn't solve it, though it didn't seem that hard. I tried to connect the points from center to end and create an equilateral triangle but still was unable to find the solution.
Would be thankful if you could help me out.
The question is the following:
We have an equal hexagon with sides equal to $1$ and six circular arcs with radius equal to $1$ from each vertices of the regular hexagon in it create a flower-shaped-like object. Find the area of the flower.
Here are the possible answers ↓
1. $pi$
3. $3pi/2$
4. $4sqrt2-pi$
5. $pi/2+sqrt3$
6. $2pi-3sqrt3$
geometry logic triangle area
$endgroup$
2
$begingroup$
I count six ellipses
$endgroup$
– Ross Millikan
4 hours ago
$begingroup$
Are the ellipses tangent to the sides at the vertices and meeting without overlap at the center? If so, they are not ellipses. Do you mean pairs of circular arcs?
$endgroup$
– Oscar Lanzi
3 hours ago
add a comment |
$begingroup$
The area of the flower
So I had a geometry problem in my maths exam test and I couldn't solve it, though it didn't seem that hard. I tried to connect the points from center to end and create an equilateral triangle but still was unable to find the solution.
Would be thankful if you could help me out.
The question is the following:
We have an equal hexagon with sides equal to $1$ and six circular arcs with radius equal to $1$ from each vertices of the regular hexagon in it create a flower-shaped-like object. Find the area of the flower.
Here are the possible answers ↓
1. $pi$
3. $3pi/2$
4. $4sqrt2-pi$
5. $pi/2+sqrt3$
6. $2pi-3sqrt3$
geometry logic triangle area
$endgroup$
The area of the flower
So I had a geometry problem in my maths exam test and I couldn't solve it, though it didn't seem that hard. I tried to connect the points from center to end and create an equilateral triangle but still was unable to find the solution.
Would be thankful if you could help me out.
The question is the following:
We have an equal hexagon with sides equal to $1$ and six circular arcs with radius equal to $1$ from each vertices of the regular hexagon in it create a flower-shaped-like object. Find the area of the flower.
Here are the possible answers ↓
1. $pi$
3. $3pi/2$
4. $4sqrt2-pi$
5. $pi/2+sqrt3$
6. $2pi-3sqrt3$
geometry logic triangle area
geometry logic triangle area
edited 46 mins ago
J. W. Tanner
2,6161217
2,6161217
asked 5 hours ago
DAVODAVO
85
85
2
$begingroup$
I count six ellipses
$endgroup$
– Ross Millikan
4 hours ago
$begingroup$
Are the ellipses tangent to the sides at the vertices and meeting without overlap at the center? If so, they are not ellipses. Do you mean pairs of circular arcs?
$endgroup$
– Oscar Lanzi
3 hours ago
add a comment |
2
$begingroup$
I count six ellipses
$endgroup$
– Ross Millikan
4 hours ago
$begingroup$
Are the ellipses tangent to the sides at the vertices and meeting without overlap at the center? If so, they are not ellipses. Do you mean pairs of circular arcs?
$endgroup$
– Oscar Lanzi
3 hours ago
2
2
$begingroup$
I count six ellipses
$endgroup$
– Ross Millikan
4 hours ago
$begingroup$
I count six ellipses
$endgroup$
– Ross Millikan
4 hours ago
$begingroup$
Are the ellipses tangent to the sides at the vertices and meeting without overlap at the center? If so, they are not ellipses. Do you mean pairs of circular arcs?
$endgroup$
– Oscar Lanzi
3 hours ago
$begingroup$
Are the ellipses tangent to the sides at the vertices and meeting without overlap at the center? If so, they are not ellipses. Do you mean pairs of circular arcs?
$endgroup$
– Oscar Lanzi
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We can't easily find the area exactly - we just don't have enough information for that. But then, what are the estimates we're looking at? (1) $approx 3.1$. (3) $approx 4.7$. (4) $approx 2.5$. (5) $approx 3.3$. (6) $approx 1.1$.
If the area's close to $3$, we might have difficulty deciding. If it's farther away, it'll be an easy choice.
Now, what's the area of the hexagon? That's something we can calculate exactly; an equilateral triangle of side $1$ has area $frac{sqrt{3}}{4}$, and the hexagon is six of those stuck together for a total area of $frac32sqrt{3}approx 2.6$. The flower is clearly much less than that. Only one of the answers is plausible. It has to be (6), $2pi-3sqrt{3}$.
And, reverse-engineering from the answer... the curves that define the "petals" are the six circles centered at the vertices of the hexagon with radius $1$. That bit about "five ellipses" is just wrong.
$endgroup$
add a comment |
$begingroup$
Assuming those are circular arcs and not "ellipses", you can indeed find the area exactly.
The "flower" has six "petals" Each of those petals has axial symmetry and can be divided into two halves. Each of those halves is the segment subtending a central angle of $frac{pi}{3}$ (radian measure) of a circle of radius $1$.
The area of one such segment is $frac 12 r^2(theta - sintheta) = frac 12(frac{pi}{3} - frac{sqrt{3}}{2})$.
There are $12$ such segments, yielding the total area of the "flower" as $2pi - 3sqrt 3$.
$endgroup$
$begingroup$
Since the shape of each petal is claimed to be ellipses, your answer is not valid in assuming that they consist of circular arcs. However, I do reckon that they are intented to be arcs.
$endgroup$
– Trebor
3 hours ago
$begingroup$
@Trebor I had edited my answer (I think before your comment) to clarify that assumption. However, I think it's a valid assumption, and just a very poor question.
$endgroup$
– Deepak
3 hours ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We can't easily find the area exactly - we just don't have enough information for that. But then, what are the estimates we're looking at? (1) $approx 3.1$. (3) $approx 4.7$. (4) $approx 2.5$. (5) $approx 3.3$. (6) $approx 1.1$.
If the area's close to $3$, we might have difficulty deciding. If it's farther away, it'll be an easy choice.
Now, what's the area of the hexagon? That's something we can calculate exactly; an equilateral triangle of side $1$ has area $frac{sqrt{3}}{4}$, and the hexagon is six of those stuck together for a total area of $frac32sqrt{3}approx 2.6$. The flower is clearly much less than that. Only one of the answers is plausible. It has to be (6), $2pi-3sqrt{3}$.
And, reverse-engineering from the answer... the curves that define the "petals" are the six circles centered at the vertices of the hexagon with radius $1$. That bit about "five ellipses" is just wrong.
$endgroup$
add a comment |
$begingroup$
We can't easily find the area exactly - we just don't have enough information for that. But then, what are the estimates we're looking at? (1) $approx 3.1$. (3) $approx 4.7$. (4) $approx 2.5$. (5) $approx 3.3$. (6) $approx 1.1$.
If the area's close to $3$, we might have difficulty deciding. If it's farther away, it'll be an easy choice.
Now, what's the area of the hexagon? That's something we can calculate exactly; an equilateral triangle of side $1$ has area $frac{sqrt{3}}{4}$, and the hexagon is six of those stuck together for a total area of $frac32sqrt{3}approx 2.6$. The flower is clearly much less than that. Only one of the answers is plausible. It has to be (6), $2pi-3sqrt{3}$.
And, reverse-engineering from the answer... the curves that define the "petals" are the six circles centered at the vertices of the hexagon with radius $1$. That bit about "five ellipses" is just wrong.
$endgroup$
add a comment |
$begingroup$
We can't easily find the area exactly - we just don't have enough information for that. But then, what are the estimates we're looking at? (1) $approx 3.1$. (3) $approx 4.7$. (4) $approx 2.5$. (5) $approx 3.3$. (6) $approx 1.1$.
If the area's close to $3$, we might have difficulty deciding. If it's farther away, it'll be an easy choice.
Now, what's the area of the hexagon? That's something we can calculate exactly; an equilateral triangle of side $1$ has area $frac{sqrt{3}}{4}$, and the hexagon is six of those stuck together for a total area of $frac32sqrt{3}approx 2.6$. The flower is clearly much less than that. Only one of the answers is plausible. It has to be (6), $2pi-3sqrt{3}$.
And, reverse-engineering from the answer... the curves that define the "petals" are the six circles centered at the vertices of the hexagon with radius $1$. That bit about "five ellipses" is just wrong.
$endgroup$
We can't easily find the area exactly - we just don't have enough information for that. But then, what are the estimates we're looking at? (1) $approx 3.1$. (3) $approx 4.7$. (4) $approx 2.5$. (5) $approx 3.3$. (6) $approx 1.1$.
If the area's close to $3$, we might have difficulty deciding. If it's farther away, it'll be an easy choice.
Now, what's the area of the hexagon? That's something we can calculate exactly; an equilateral triangle of side $1$ has area $frac{sqrt{3}}{4}$, and the hexagon is six of those stuck together for a total area of $frac32sqrt{3}approx 2.6$. The flower is clearly much less than that. Only one of the answers is plausible. It has to be (6), $2pi-3sqrt{3}$.
And, reverse-engineering from the answer... the curves that define the "petals" are the six circles centered at the vertices of the hexagon with radius $1$. That bit about "five ellipses" is just wrong.
answered 4 hours ago
jmerryjmerry
11.6k1527
11.6k1527
add a comment |
add a comment |
$begingroup$
Assuming those are circular arcs and not "ellipses", you can indeed find the area exactly.
The "flower" has six "petals" Each of those petals has axial symmetry and can be divided into two halves. Each of those halves is the segment subtending a central angle of $frac{pi}{3}$ (radian measure) of a circle of radius $1$.
The area of one such segment is $frac 12 r^2(theta - sintheta) = frac 12(frac{pi}{3} - frac{sqrt{3}}{2})$.
There are $12$ such segments, yielding the total area of the "flower" as $2pi - 3sqrt 3$.
$endgroup$
$begingroup$
Since the shape of each petal is claimed to be ellipses, your answer is not valid in assuming that they consist of circular arcs. However, I do reckon that they are intented to be arcs.
$endgroup$
– Trebor
3 hours ago
$begingroup$
@Trebor I had edited my answer (I think before your comment) to clarify that assumption. However, I think it's a valid assumption, and just a very poor question.
$endgroup$
– Deepak
3 hours ago
add a comment |
$begingroup$
Assuming those are circular arcs and not "ellipses", you can indeed find the area exactly.
The "flower" has six "petals" Each of those petals has axial symmetry and can be divided into two halves. Each of those halves is the segment subtending a central angle of $frac{pi}{3}$ (radian measure) of a circle of radius $1$.
The area of one such segment is $frac 12 r^2(theta - sintheta) = frac 12(frac{pi}{3} - frac{sqrt{3}}{2})$.
There are $12$ such segments, yielding the total area of the "flower" as $2pi - 3sqrt 3$.
$endgroup$
$begingroup$
Since the shape of each petal is claimed to be ellipses, your answer is not valid in assuming that they consist of circular arcs. However, I do reckon that they are intented to be arcs.
$endgroup$
– Trebor
3 hours ago
$begingroup$
@Trebor I had edited my answer (I think before your comment) to clarify that assumption. However, I think it's a valid assumption, and just a very poor question.
$endgroup$
– Deepak
3 hours ago
add a comment |
$begingroup$
Assuming those are circular arcs and not "ellipses", you can indeed find the area exactly.
The "flower" has six "petals" Each of those petals has axial symmetry and can be divided into two halves. Each of those halves is the segment subtending a central angle of $frac{pi}{3}$ (radian measure) of a circle of radius $1$.
The area of one such segment is $frac 12 r^2(theta - sintheta) = frac 12(frac{pi}{3} - frac{sqrt{3}}{2})$.
There are $12$ such segments, yielding the total area of the "flower" as $2pi - 3sqrt 3$.
$endgroup$
Assuming those are circular arcs and not "ellipses", you can indeed find the area exactly.
The "flower" has six "petals" Each of those petals has axial symmetry and can be divided into two halves. Each of those halves is the segment subtending a central angle of $frac{pi}{3}$ (radian measure) of a circle of radius $1$.
The area of one such segment is $frac 12 r^2(theta - sintheta) = frac 12(frac{pi}{3} - frac{sqrt{3}}{2})$.
There are $12$ such segments, yielding the total area of the "flower" as $2pi - 3sqrt 3$.
edited 1 hour ago
Anirban Niloy
609218
609218
answered 3 hours ago
DeepakDeepak
17.2k11536
17.2k11536
$begingroup$
Since the shape of each petal is claimed to be ellipses, your answer is not valid in assuming that they consist of circular arcs. However, I do reckon that they are intented to be arcs.
$endgroup$
– Trebor
3 hours ago
$begingroup$
@Trebor I had edited my answer (I think before your comment) to clarify that assumption. However, I think it's a valid assumption, and just a very poor question.
$endgroup$
– Deepak
3 hours ago
add a comment |
$begingroup$
Since the shape of each petal is claimed to be ellipses, your answer is not valid in assuming that they consist of circular arcs. However, I do reckon that they are intented to be arcs.
$endgroup$
– Trebor
3 hours ago
$begingroup$
@Trebor I had edited my answer (I think before your comment) to clarify that assumption. However, I think it's a valid assumption, and just a very poor question.
$endgroup$
– Deepak
3 hours ago
$begingroup$
Since the shape of each petal is claimed to be ellipses, your answer is not valid in assuming that they consist of circular arcs. However, I do reckon that they are intented to be arcs.
$endgroup$
– Trebor
3 hours ago
$begingroup$
Since the shape of each petal is claimed to be ellipses, your answer is not valid in assuming that they consist of circular arcs. However, I do reckon that they are intented to be arcs.
$endgroup$
– Trebor
3 hours ago
$begingroup$
@Trebor I had edited my answer (I think before your comment) to clarify that assumption. However, I think it's a valid assumption, and just a very poor question.
$endgroup$
– Deepak
3 hours ago
$begingroup$
@Trebor I had edited my answer (I think before your comment) to clarify that assumption. However, I think it's a valid assumption, and just a very poor question.
$endgroup$
– Deepak
3 hours ago
add a comment |
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$begingroup$
I count six ellipses
$endgroup$
– Ross Millikan
4 hours ago
$begingroup$
Are the ellipses tangent to the sides at the vertices and meeting without overlap at the center? If so, they are not ellipses. Do you mean pairs of circular arcs?
$endgroup$
– Oscar Lanzi
3 hours ago