# Weird radians question

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There's a square ABCD with side length r. Arc with centre A drawn from B to D, and arc with centre B drawn from A to C. The area required is those on the 2 sides. The answers says the length from the top to the intersection is root3/2*r. Why's that? Sorry idk how to upload pics so I have to awkwardly describe it.

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#2

(Original post by

There's a square ABCD with side length r. Arc with centre A drawn from B to D, and arc with centre B drawn from A to C. The area required is those on the 2 sides. The answers says the length from the top to the intersection is root3/2*r. Why's that? Sorry idk how to upload pics so I have to awkwardly describe it.

**dumb2020**)There's a square ABCD with side length r. Arc with centre A drawn from B to D, and arc with centre B drawn from A to C. The area required is those on the 2 sides. The answers says the length from the top to the intersection is root3/2*r. Why's that? Sorry idk how to upload pics so I have to awkwardly describe it.

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Ahh, I see why it's root 3 /2 now, but how could I have known it's an equilateral triangle? It sure looks like one, but I suppose I shouldn't say that in an exam.

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I finally got to my desktop so I can post the picture. The (root 3)/2*r side doesn't go all the way down, so I can't see anything that tells me what length it has

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#6

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I finally got to my desktop so I can post the picture. The (root 3)/2*r side doesn't go all the way down, so I can't see anything that tells me what length it has

**dumb2020**)I finally got to my desktop so I can post the picture. The (root 3)/2*r side doesn't go all the way down, so I can't see anything that tells me what length it has

Not sure what you mean? The original square is of length r.

Note it looks like a typo in the final line, the shaded area 2U should be double the answer. The given answer evaluates to a bit less than a quarter of the square. It should be double that.

There is also a bit more direct way of getting the (correct) ans without using R. Just use S is a 30 degree sector so

U = S - 60 degree segment

60 degree segment = 60 degree sector - equilateral triangle area

So

2U = 2 equilateral triangles - 60 degree sector

The areas of an equilateral triangle and a 60 degree sector can be just written down.

Last edited by mqb2766; 4 weeks ago

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Sorry I was a bit busy. I'm also very sorry I don't really have any idea what you meant. How do I know the vertical height from the top to where the arcs intersect is (root 3)/2*r?

B is the centre of a circle but how do we know S is a 30 degree sector?

U is just a tiny bit smaller compared to S. I don't see how it's a 60 degree segment smaller than S.

And then since it's hard without diagrams, I lost you about the Us and Ss and which equilateral triangles and sectors you meant.

Sorry! This is taking too much of your time.

B is the centre of a circle but how do we know S is a 30 degree sector?

U is just a tiny bit smaller compared to S. I don't see how it's a 60 degree segment smaller than S.

And then since it's hard without diagrams, I lost you about the Us and Ss and which equilateral triangles and sectors you meant.

Sorry! This is taking too much of your time.

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#8

(Original post by

Sorry I was a bit busy. I'm also very sorry I don't really have any idea what you meant. How do I know the vertical height from the top to where the arcs intersect is (root 3)/2*r?

B is the centre of a circle but how do we know S is a 30 degree sector?

U is just a tiny bit smaller compared to S. I don't see how it's a 60 degree segment smaller than S.

And then since it's hard without diagrams, I lost you about the Us and Ss and which equilateral triangles and sectors you meant.

Sorry! This is taking too much of your time.

**dumb2020**)Sorry I was a bit busy. I'm also very sorry I don't really have any idea what you meant. How do I know the vertical height from the top to where the arcs intersect is (root 3)/2*r?

B is the centre of a circle but how do we know S is a 30 degree sector?

U is just a tiny bit smaller compared to S. I don't see how it's a 60 degree segment smaller than S.

And then since it's hard without diagrams, I lost you about the Us and Ss and which equilateral triangles and sectors you meant.

Sorry! This is taking too much of your time.

having said that, please read up about equilateral triangles, circles segments, sectors. If you're happy with those, the question isn't too bad.

Last edited by mqb2766; 4 weeks ago

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(Original post by

Which answer do you want explained, the more complex model answer or the simpler geometry one?

having said that, please read up about equilateral triangles, circles segments, sectors. If you're happy with those, the question isn't too bad.

**mqb2766**)Which answer do you want explained, the more complex model answer or the simpler geometry one?

having said that, please read up about equilateral triangles, circles segments, sectors. If you're happy with those, the question isn't too bad.

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#12

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Hmm just walk me through the simpler geometry one. I've always been horrendous at geometry, but I'd like to think I know enough already. I just don't know when to do what.

**dumb2020**)Hmm just walk me through the simpler geometry one. I've always been horrendous at geometry, but I'd like to think I know enough already. I just don't know when to do what.

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#13

**dumb2020**)

Hmm just walk me through the simpler geometry one. I've always been horrendous at geometry, but I'd like to think I know enough already. I just don't know when to do what.

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**mqb2766**)

Which answer do you want explained, the more complex model answer or the simpler geometry one?

having said that, please read up about equilateral triangles, circles segments, sectors. If you're happy with those, the question isn't too bad.

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#15

(Original post by

Oh right. It took me so long to see how the triangle is equilateral. How does this get me the answer?

**dumb2020**)Oh right. It took me so long to see how the triangle is equilateral. How does this get me the answer?

Last edited by mqb2766; 3 weeks ago

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Ok...? If the triangle you meant was B and the 2 Ps, which is the only triangle you outlined, I thought the 60 degree segment is something you should add, and you had to subtract the two small segment looking things one the sides.

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#17

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Ok...? If the triangle you meant was B and the 2 Ps, which is the only triangle you outlined, I thought the 60 degree segment is something you should add, and you had to subtract the two small segment looking things one the sides.

**dumb2020**)Ok...? If the triangle you meant was B and the 2 Ps, which is the only triangle you outlined, I thought the 60 degree segment is something you should add, and you had to subtract the two small segment looking things one the sides.

So you want to find:

the triangle area - 60 degree segment area.

Last edited by mqb2766; 3 weeks ago

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