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#1
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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1 month ago
#2
(Original post by 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.
The two radii for the arcs from an equilateral triangle with the intersection point and A and B. It's height is well known/easy to evaluate.
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#3
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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1 month ago
#4
All the sides are the same length.
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#5
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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1 month ago
#6
(Original post by 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
That is the vertical height from the top to the point where the arcs intersect?
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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#7
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.
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4 weeks ago
#8
(Original post by 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.
Which answer do you want explained, the more complex model answer or the simpler geometry one?
Last edited by mqb2766; 4 weeks ago
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4 weeks ago
#9
(Original post by mqb2766)
unilateral triangles
That's a new one.
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4 weeks ago
#10
(Original post by ghostwalker)
That's a new one.
Auto correct ffs.
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#11
(Original post by mqb2766)
Which answer do you want explained, the more complex model answer or the simpler geometry one?
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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4 weeks ago
#12
(Original post by 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.
You're happy with equilateral triangles (and the half 30-60-90 triangle), circles, sectors, segments?
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4 weeks ago
#13
(Original post by 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.
Are you happy with
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#14
(Original post by mqb2766)
Which answer do you want explained, the more complex model answer or the simpler geometry one?
Oh right. It took me so long to see how the triangle is equilateral. How does this get me the answer?
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3 weeks ago
#15
(Original post by dumb2020)
Oh right. It took me so long to see how the triangle is equilateral. How does this get me the answer?
The equilateral triangle is probably the most important step. Its how you construct them using compasses. To get the shaded area, you can (laboriously) calculate each region, or split the diagram (vertically) and reassemble and note the shaded area is just the same equilateral triangle - 60 degree segment.
Last edited by mqb2766; 3 weeks ago
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#16
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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3 weeks ago
#17
(Original post by 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.
That's correct you add 1 60 degree segment (on the triangle base) and also subtract 2 60 degree segments (on the triangle sides) so it's the same as just subtracting 1 60 degree segment. Just imagine using the bottom segment to fill in one of the missing segments on the side of the triangle.

So you want to find:
the triangle area - 60 degree segment area.
Last edited by mqb2766; 3 weeks ago
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