Answer:
![Orange\ area = A^2(\sqrt{3}/4 - \pi/8) = 0.0403A^2](https://tex.z-dn.net/?f=Orange%5C%20area%20%20%3D%20A%5E2%28%5Csqrt%7B3%7D%2F4%20%20-%20%5Cpi%2F8%29%20%3D%200.0403A%5E2)
Step-by-step explanation:
First let's find the area of the triangle, using the formula:
![Area\_triangle = side^2\sqrt{3}/4](https://tex.z-dn.net/?f=Area%5C_triangle%20%3D%20side%5E2%5Csqrt%7B3%7D%2F4)
![Area\_triangle = A^2\sqrt{3}/4](https://tex.z-dn.net/?f=Area%5C_triangle%20%3D%20A%5E2%5Csqrt%7B3%7D%2F4)
Now, let's find the area of each circular sector of 60° (internal angle of a equilateral triangle):
![Area\_sector = \pi*radius^2*60/360](https://tex.z-dn.net/?f=Area%5C_sector%20%3D%20%5Cpi%2Aradius%5E2%2A60%2F360)
![Area\_sector = \pi*(A/2)^2/6](https://tex.z-dn.net/?f=Area%5C_sector%20%3D%20%5Cpi%2A%28A%2F2%29%5E2%2F6)
![Area\_sector = \pi*A^2/24](https://tex.z-dn.net/?f=Area%5C_sector%20%3D%20%5Cpi%2AA%5E2%2F24)
Now, To calculate the orange area in the center, we have:
![Orange\ area = Area\_triangle - 3*Area\_sector](https://tex.z-dn.net/?f=Orange%5C%20area%20%20%3D%20Area%5C_triangle%20-%203%2AArea%5C_sector)
![Orange\ area = A^2\sqrt{3}/4 - \pi*A^2/8](https://tex.z-dn.net/?f=Orange%5C%20area%20%20%3D%20A%5E2%5Csqrt%7B3%7D%2F4%20%20-%20%5Cpi%2AA%5E2%2F8)
![Orange\ area = A^2(\sqrt{3}/4 - \pi/8)](https://tex.z-dn.net/?f=Orange%5C%20area%20%20%3D%20A%5E2%28%5Csqrt%7B3%7D%2F4%20%20-%20%5Cpi%2F8%29)
![Orange\ area = 0.0403A^2](https://tex.z-dn.net/?f=Orange%5C%20area%20%20%3D%200.0403A%5E2)
Answer: WY = 27
Step-by-step explanation:
Due to WX being a Perpendicular Bisector WY = WZ
4x - 5 = 2x + 11
Add 5 to both sides
4x = 2x + 16
Subtract 2x from both sides
2x = 16
Divide by 2
x = 8
So 4x - 5 of x = 8
4*8 = 32 - 5 =27
Answer:
b = 15.7 -2a
Step-by-step explanation:
You are asking us to solve for b
2a +b = 15.7
Subtract 2a from each side
2a-2a+b = 15.7 -2a
b = 15.7 -2a