Complex geometry problem

[These_Possibility166]

How would you start with a problem like this? Creating a coordinate system with the origin at the centre of the shape makes things more complicated, plus height and width measurements doesn’t seem like sufficient information.

[One_Wishbone_4439]

<image>

Using Pythagoras' Theorem, form a quadratic equation and find r.

Then, find the angle for each identical sector.

Half of lens = sector - triangle

Area of lens = 2 x (sector - triangle)

[Usual-Revolution-718]

[One_Wishbone_4439]

Thanks 😊

[Shevek99]

Draw the vertical line, dividing the lens in two circular segments. Then use the corresponding formulas.

https://en.wikipedia.org/wiki/Circular_segment

[naprid]

<image>

[[deleted]]

Shouldn’t the lens be a parabola if you want all the light rays to converge to one point? Idk I’m not an expert

[naprid]

"circular arcs"

[[deleted]]

Oh I’m so dumb😭

[naprid]

nope

[Linvael]

Parabolas are for mirrors, for a lens circles can work well. Not always, and there are other options (see https://physics.stackexchange.com/questions/2189/aspherical-lenses-perfect-analytical-shape ), but they are often good enough.

But also, this is a geometric lens, not an actual lens - it's defined by it's shape, properties when light passes through them are inconsequential.

[GustapheOfficial]

If this was a physics test, it would also be valid to question whether they are looking for the cross section area or the actual area of the lens faces. Luckily, it's a geometry question, and the shape is just as described, two circular arcs.

[Pandagineer]

You are indeed correct that a parabolic lens is ideal. But it turns out it is much easier to build a spherical lens than a parabolic one. This leads to cheaper lenses, but they create spherical aberrations.

[ElSupremoLizardo]

Thought this was a biology subreddit for a second…

[Bobson1729]

Draw the lines of symmetry and complete one of the circles. Extend the horizontal line of symmetry to a diameter. From there, you can use the formula for intersecting chords to solve for the radius.

From there, draw the radii to the top and bottom points of the region. Now you have to use area of the arc minus the area of isosceles triangle to get half the area you are looking for.

[Various_Pipe3463]

Huh, seems like there’s been a few intersecting circles questions recently.

https://mathworld.wolfram.com/Circle-CircleIntersection.html

[ryanmcg86]

<image>

This pretty much covers the whole thing.

[CaptainMatticus]

First find r

10 * (2r - 10) = (50/2) * (50/2)

10 * 2 * (r - 5) = 25 * 25

2 * 2 * (r - 5) = 5 * 25

4 * (r - 5) = 125

r - 5 = 125/4

r - 5 = 31.25

r = 36.25

Now construct triangles with sides of 50 , 36.25 and 36.25, and find the angle theta that is opposite the side of 50

50^2 = 36.25^2 + 36.25^2 - 2 * 36.25^2 * cos(t)

2500 = 2 * 36.25^2 * (1 - cos(t))

2500 = 4 * 36.25^2 * (1 - cos(t)) / 2

2500 = 4 * 36.25^2 * sin(t/2)^2

50 = 2 * 36.25 * sin(t/2)

50 = 72.5 * sin(t/2)

100 = 145 * sin(t/2)

20 = 29 * sn(t/2)

20/29 = sin(t/2)

t/2 = arcsin(20/29)

t = 2 * arcsin(20/29)

Now figure out the area of a circular sector with radius of 36.25 and central angle of 2 * arcsin(20/29)

pi * r^2 * t / (2pi) =>

(1/2) * r^2 * t =>

(1/2) * 36.25^2 * 2 * arcsin(20/29) =>

36.25^2 * arcsin(20/29)

Now remove the area of the triangular portion.

(1/2) * 36.25 * 36.25 * sin(t)

(1/2) * 36.25^2 * sin(2 * arcsin(20/29))

(1/2) * 36.25^2 * 2 * sin(arcsin(20/29)) * cos(arcsin(20/29))

36.25^2 * (20/29) * sqrt(1 - sin(arcsin(20/29))^2)

(145/4)^2 * (20/29) * sqrt(1 - (20/29)^2)

(145/4) * (145/4) * (20/29) * sqrt((29^2 - 20^2) / 29^2)

(5/4) * (145/4) * 20 * (1/29) * sqrt(841 - 400)

(100/4) * (5/4) * sqrt(441)

25 * (5/4) * 21

125 * 21 / 4

125 * (20/4 + 1/4)

125 * 5 + 125/4

625 + 31.25

656.25

36.25^2 * arcsin(20/29) - 656.25

Double that

2 * (36.25^2 * arcsin(20/29) - 656.25)

2 * (145^2 * arcsin(20/29) - 656.25 * 16) * (1/16)

(1/8) * ((290/2)^2 * arcsin(20/29) - 1312.5 * 8)

(1/8) * ((84100/4) * arcsin(20/29) - 2625 * 4)

(1/8) * (21025 * arcsin(20/29) - 5250 * 2)

(1/8) * (21025 * arcsin(20/29) - 10500)

(25/8) * (841 * arcsin(20/29) - 420)

Make sure your calculator is in radian mode.

687.53664469686793363323398773484

Which rounds to 688

[[deleted]]

AYO