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what is reuleaux triangle???

Archit Murkunde
- 7 years, 3 months ago

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The shape of constant radius made by three overlapping circles. It's the figure that you find the area of in this problem.

Daniel Bortolussi
- 7 years, 3 months ago

great.. i nvr studied this kind of triangle before. thanks for the explanation :D

Heng Joe Kit
- 7 years, 3 months ago

its nice..

Vema Venki
- 7 years, 3 months ago

What is triangle?????

Rafaqat Ali
- 7 years, 3 months ago

reuleaux triangle??? plz explain..in detail...

Gv Nikhil
- 7 years, 3 months ago

it so simble if we think in right method

Melvin Mosq
- 7 years, 3 months ago

good

Melvin Mosq
- 7 years, 3 months ago

ans. 6.3429

Dean Clidoro
- 7 years, 3 months ago

(3Api*r^2)/360 ===3 sectors

then subtract 2 triangles inscribe in the shaded area

2(d1d1sinA/2) === area of 2 inscribed triangles in the shaded area

ans.6.3429

Dean Clidoro
- 7 years, 3 months ago

what is meant by reuleaux triangle??

Dhanya Nedungadi
- 7 years, 2 months ago

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A Reuleaux triangle is the simplest and best known Reuleaux polygon. It is a curve of constant width, meaning that the separation of two parallel lines tangent to the curve is independent of their orientation. Because all diameters are the same, the Reuleaux triangle is one answer to the question "Other than a circle, what shape can a manhole cover be made so that it cannot fall down through the hole?" The term derives from Franz Reuleaux, a 19th-century German engineer who did pioneering work on ways that machines translate one type of motion into another, although the concept was known before his time. With a compass, sweep an arc sufficient to enclose the desired figure. With radius unchanged, sweep a sufficient arc centred at a point on the first arc to intersect that arc. With the same radius and the centre at that intersection sweep a third arc to intersect the other arcs. The result is a curve of constant width. Equivalently, given an equilateral triangle T of side length s, take the boundary of the intersection of the disks with radius s centered at the vertices of T. By the Blaschke–Lebesgue theorem, the Reuleaux triangle has the least area of any curve of given constant width. This area is {1\over2}(\pi - \sqrt3)s^2, where s is the constant width. The existence of Reuleaux polygons shows that diameter measurements alone cannot verify that an object has a circular cross-section. The area of Reuleaux triangle is smaller than that of the disk of the same width (i.e. diameter); the area of such a disk is \pi s^2 \over 4.

Melvin Mosq
- 7 years, 2 months ago

ughh i suck at this..... i guess its because im 11 its just my dream is hardvard

bob jerry
- 7 years, 2 months ago

nice solution ivan martinez

Muhammed Attaul Gani Chowdhury
- 7 years, 1 month ago

what is reuleaux triangle???

Christian Cernechez
- 7 years, 3 months ago

You can connect the center of each circle to make a triangle inside the shaded area. Since the sides of the triangle are the radius of the circles (measuring 3 each), you have an equilateral triangle, with all angles measuring 60 degrees.

You can also get an arc from each circle inside the same area. The angle of the arc is the same as the angle of the triangle, which is 60 degrees. By putting the arcs together, you get to the shaded area you want, with the following equation for the area:

A=3
*
(pi
*
r^2*angle)/360

But there's a problem; the arc will have overlapping areas. Luckily, these areas are equivalent to the equilateral triangle from earlier, so all you have to do is subtract the area of the triangle twice to get rid of the overlapping, and you have the final equation to solve the problem:

A=[3
*
(pi
*
r^2
*
angle)/360]-[2
*
(r^2*sqrt3)/4]

Plug in the numbers and you'll have 9/2 (pi-sqrt3)

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very nice explanation bro......

Sanjeev Prasad
- 7 years, 3 months ago

very good.

Jess Toni Bautista
- 7 years, 2 months ago

Very nice & clear explanation thank u

Matheswari selvaraj
- 7 years, 2 months ago

I bumped into the answer this way:-

We know that, Segment = Sector - Corresponding Triangle.

Area of the shaded region is
= Area of ONE Sector + Area of TWO Segments

= Area of THREE Sectors - Area of TWO Triangles.

Hope you understand. Am not able to type-in the calculations here. See Ana Paula Mello's answer below.

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ur method is good

Mian Fahad
- 7 years, 3 months ago

Really cool method dude :D

Malhar Savale
- 7 years, 3 months ago

I did same too!

Kou$htav Chakrabarty
- 7 years, 3 months ago

same here

Ankit Tiwari
- 7 years, 3 months ago

i got the same solution

David Lancaster
- 7 years, 3 months ago

Very nice

Matheswari selvaraj
- 7 years, 2 months ago

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I not the only one!

Julian Poon
- 7 years, 2 months ago

The area of a reuleaux triangle is: 1/2(π-3^1/2)r^2 therefore: 1/2(π-3^1/2)3^2 = 9/2(π-3^1/2)

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WRONG

Jaan Khan
- 7 years, 3 months ago

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see the angle coversd in one circle it is 60.

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The needed figure shows 3 overlapping sectors.

To get the area of it simply get the area of 3 sectors minus the area of 2 equilateral triangle. Theta or the angle to be used is 60 ° = π/3 in radians

Use the formula below

Area of a sector = (1/2 r^2∅) or (1/2 r^2 π/3)

Multiplying this to 3 we get

1/2 r^2 π

The area of an equilateral triangle is

(r^2 √3)/4

Getting 2 of this will result to

(r^2 √3)/2

Area of the reuleaux triangle is

(r^2 π)/2 - (r^2 √3)/2

Simplify

r^2/2 (π-√3)

Substituting the value of the radius we get the answer

9^2/2 (π-√3)

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that is simply using the equation 9/2 (pi-sq. root of 3)

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3(9pi/6-9sqrt(3)/4)+9sqrt(3)/4

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*
theta & a=1/2
*
r^2*theta formula along wid area of equilateral triangle formula & compare.. ans:9/2(pi-root(3))

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1/2(π-3^1/2)3^2 = 9/2(π-3^1/2)

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(π.r^2)/2-2.( ( r^2.3^(1/2) )/4) = (π.9)/2 - ((9.3^(1/2))/2 = 9/2(π - 3^(1/2))

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1/2(π-3^1/2)r^2 so, 1/2(π-3^1/2)3^2 = 9/2(π-3^1/2)

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The area of a reuleaux triangle is: 1/2(π-3^1/2)r^2 therefore: 1/2(π-3^1/2)3^2 = 9/2(π-3^1/2)

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*
(3
*
3) = 9
*
root(3)/4
now if u see the triangle carefully, you will see an arc of 60 degree in each triangle, so lets first find out the area subtended by this arc to the centre of the circle=pi
*
3
*
3
*
(60/360). note that u are considering one circle.
subtract the area of the triangle from this area and multiply it by 3 to do it for all the three circles.so we get =
3
*
(9
*
pi/6-9*root(3)/4) . now add the area of the triangle to get the shaded area=9/2(pi-root(3)).

Thanks

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As all circle intersect at the center of other, by joining three point an equilateral triangle will be formed whose area is (3^1/2)
*
(3^2)/4. If we consider one circle at time considering point intersection, it will form a sector of 60 degree angle whose area is (60/360)
*
pi
*
(3^2). Now, area of shaded region = 3
*
area of sector - 2*area of equilateral triangle

Balram Suhane
- 7 years, 3 months ago

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The area of a reuleaux triangle is: 1/2(π-3^1/2)r^2 therefore: 1/2(π-3^1/2)3^2 = 9/2(π-3^1/2)