## anonymous one year ago ***WILL MEDAL AND FAN!!!*** A square with sides of https://suwannee.owschools.com/media/g_geo_2013/8/3rad2.gif is inscribed in a circle. What is the area of one of the sectors formed by the radii to the vertices of the square? a) 1.5pi b) 2.25pi c) 4.5pi

1. anonymous

@phi

2. anonymous

|dw:1437403588550:dw|

3. anonymous

so i think thats what the question is saying, and you need to find the area of the coloured in section. s is the side of the square which is 3sqrt(2). r is the radius of the circle, which is also equal to half of the diagonal line of the square. so first, we have to figure out the diagonal line of the square. that can be done by using pythagorian equation: (side1)^2+(side2)^2 = diagonal length^2

4. phi

|dw:1437403532240:dw|

5. anonymous

Hence, (3sqrt(2))^2 + (3sqrt(2))^2 = d^2 Hence, d=6 (you can do the math following the equation above) so divide this by 2 in order to get the radius of the circle. therefore, r = 3 Next, you will notice that the coloured area we are looking for is only a quarter of the entire circle minus the area of a quarter of the entire square. So knowing that, first, you must the area of the circle using the equation: pi*r^2 = pi*(3)^2 = 9pi you need a quarter of that so divide it by 4 to get 9pi/4. Next, find the area of the square. length x width. So, 2*(3sqrt(2)) = 6sqrt(2) but similarly, you only need a quarter of that, so divide that by 4 to get 3sqrt(2)/2 Therefore, your final answer of the area of the coloured region will be the subtraction of the two areas: [(9pi/4)-(3sqrt(2)/2)] = (9pi-6sqrt(2))/4

6. anonymous

Is this considered to be a Trigonometry question?

7. phi

A sector is the shape of a piece of pie. The first step is to find the radius. The triangle is a 45-45-90 triangle, and you are given the hypotenuse

8. anonymous

which is 3 and the sqare root of 2

9. phi

the hypotenuse is $$3 \sqrt{2}$$ Do you remember (or have in your notes) how long is a side ? in a 45-45-90 triangle?

10. anonymous

Honestly no I don't. That's been almost a year ago since I was learning about that

11. phi

oh. math builds on itself and you end up using the old stuff... you can also use Kim's idea |dw:1437404053629:dw| and use the pythagorean theorem to find the diameter, and then find the radius

12. phi

pythagoras is very famous a^2 + b^2 = c^2 where a and b are the legs, and c is the hypotenuse

13. anonymous

I know I hear about the theorem every which way I turn! It's like the plague

14. phi

It is amazing how often it shows up. (but I would not call it a plague) Look on the bright side, if you know it, you know a *lot*

15. anonymous

Yeah

16. phi

can you find the diameter?

17. anonymous

Wouldn't that be the same as the hypotenuse?

18. phi

yes, of the last figure I posted.

19. anonymous

Sorry I left you hanging here. It was time to go

20. phi

did you finish this or is it too late?

21. anonymous

I didn't finish, but I really do need to though

22. anonymous

Can anybody help me?

23. Loser66

|dw:1437482709311:dw|

24. anonymous

Finally somebody to help. Thanks!

25. Loser66

26. Loser66

That gives us the area of the WHOLE circle is 9pi

27. Loser66

what you need? Because this guy is 1/4 of the circle |dw:1437482954457:dw|

28. Loser66

|dw:1437483048616:dw|

29. phi

fyi a sector is the two radius and the arc http://www.mathopenref.com/arcsector.html

30. anonymous

Let me look. I'll be right back. Thanks!

31. anonymous

@Loser66

32. anonymous

@AaronAndyson

33. anonymous

@dinosour1480

34. anonymous

@jamesr

35. anonymous

a² R² = ----- where a is the length of the side of the rectangle 2 In this case (3√2)² 9*2 R² = ------ = ----- = 9 2 2 so R = √9 => R = 3

36. anonymous

Now radii to the vertices form an angle of 90° so we have to find the area of a sector of angle 90° angle 90° area of sector = -------- * πR² = ------ * 3² * π 360° 360° area of sector =2.25π

37. anonymous

You made it sound so easy! Wow! Thanks!`