Question

Find the area of the shaded region. All arcs are of radii 2.

Fairly easy to do with trig, which is why it's a hard GEOMETRY problem

Answer 1

@Vocaloid

Answer 2

hm. I think I might have an idea of how to do this.

Answer 3

I know of two non trig ways to solve it

Answer 4

oh btw there's a hint for this one if you want one

Answer 5

i'll give you the hint, it has to do with an equilateral triangle ;)

Answer 6

oh ok yeah then the middle one is also 30

Answer 7

hm. not a 30-60-90 triangle.

Answer 8

that's when you either give up and cheat and use trig (dirty), orrr, you rethink the triangles you drew :P

Answer 9

and mayyybe which part of the figure you attempt to get the area from

Answer 10

well the inside of the shape isn't practical w/o trig so maybe I'll try the outside

Answer 11

alright I think i got it

Answer 12

honestly i spent hours thinking about this problem back in middle school when i was first introduced to it haha

Answer 13

alright i borked up the area of the smaller section

(area of the quarter circle - area of the equilateral triangle)/2 =

(1/4pi*2^2 - 4sqrt(3)/4)/2 = (1/2)(pi - sqrt(3)) praying to god I typed that right b/c I just plugged this into wolframalpha

Answer 14

ugh wait that won't do it either

Answer 15

alright it's the area of the sector - the area of the triangle, not the area of the quarter circle that gives the small segment

(60/360)pi*2^2 - 4*3/sqrt(4) = 2pi/3 - 6

Answer 16

area of that circled region = area of quarter circle - area of the sector - the small crescent section

= (1/4)pi*2^2 - (60/360)pi*2^2 - (2pi/3 - 6) = 6 - pi/3

Answer 17

that area in the middle = area of the whole square - 4(area of that circled region)

I get a negative number which means I screwed up somewhere ugh

Answer 18

i'd check over your work but tbh i'm too sleepy to verify anything... are you curious about my approach by any chance :P

Answer 19

sure

Answer 20

solve the system for y

Answer 21

hm. alright.

Answer 22

the other way i knew how to solve it is the approach you were taking with the two sectors, but frankly i lied to in that this is not the original problem; the original problem merely asked for a different shaded part of the figure. either way though i think you'd be able to get it with the two sectors