## anonymous one year ago Find the area of the shaded portion https://flyingeaglesnv.sooschools.com/media/g_geo_2014/8/groupi60.gif Answer box would only allow this as the answer: A = x x (square root)x - xxPi

1. jim_thompson5910

|dw:1437528564446:dw|

2. jim_thompson5910

if we cut the quadrilateral into 2 pieces, we get 2 congruent triangles |dw:1437528642207:dw| the 120 degree angle is also cut in half

3. jim_thompson5910

this triangle is a 30-60-90 triangle what is the value of x? |dw:1437528698301:dw|

4. anonymous

the x is the hypotenuse and that is 12. cuz the short leg (6) is 1/2 the size of the hypotenuse. right?

5. jim_thompson5910

x is the long leg actually. The hypotenuse is 2*6 = 12

6. anonymous

so then the long leg is 6(square root 3)

7. jim_thompson5910

it turns out that x = 6*sqrt(3) template for a 30-60-90 triangle |dw:1437529004578:dw|

8. jim_thompson5910

correct

9. jim_thompson5910

|dw:1437529045712:dw|

10. jim_thompson5910

What is the area of the triangle?

11. anonymous

area of the triangle is 1/2 (base x height) so it is 1/2 (6 square root (3)) (6) = 18 square root 3.

12. jim_thompson5910

correct

13. jim_thompson5910

there are 2 of these triangles (one just mirrored over the hypotenuse to get the other) so the total area of this quadrilateral is 2*(18*sqrt(3)) = 36*sqrt(3)

14. jim_thompson5910

|dw:1437529313280:dw|

15. jim_thompson5910

to find the area of this shaded portion |dw:1437529356277:dw| use the formula A = (x/360)*pi*r^2 x = central angle r = radius leave the area in terms of pi

16. anonymous

so the area of the sector would be 12pi. So the area of the segment = area of sector - area of triangle. so A(segment) = 12pi - 36(sqrt3) am I on the right track??

17. jim_thompson5910

you subtracted in the wrong order the "12pi" area is smaller, so it needs to come second $\large \text{area of shaded region} = \text{large area} - \text{small area}$ $\large \text{area of shaded region} = \text{area of quadrilateral} - \text{area of sector}$ $\large \text{area of shaded region} = 36\sqrt{3} - 12\pi$

18. jim_thompson5910

and you should change "area of triangle" to "area of quadrilateral"

19. anonymous

Oh, so that is why i am coming up with a negative number. Thank you so much for your help. :)

20. jim_thompson5910

you're welcome