Join the center of the hexagon with the 2 base angles.
An equilateral triangle, with side length x, is formed.
(remark: a regular hexagon is made up of 6 equilateral triangles with equal length)
The height
![2 \sqrt{3}](https://tex.z-dn.net/?f=2%20%5Csqrt%7B3%7D%20)
forms 2 congruent right triangles with :
hypotenuse= x, side_1=x/2, and side_2=
![2 \sqrt{3}](https://tex.z-dn.net/?f=2%20%5Csqrt%7B3%7D%20)
.
From the pythagorean theorem we have:
![x^{2} = ( \frac{x}{2} )^{2}+(2 \sqrt{3})^{2}](https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%3D%20%28%20%5Cfrac%7Bx%7D%7B2%7D%20%29%5E%7B2%7D%2B%282%20%5Csqrt%7B3%7D%29%5E%7B2%7D%20%20)
![x^{2} = \frac{ x^{2} }{4} +12](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%3D%20%20%5Cfrac%7B%20x%5E%7B2%7D%20%7D%7B4%7D%20%2B12)
![\frac{3}{4} x^{2} =12](https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B4%7D%20x%5E%7B2%7D%20%3D12)
![x^{2} = \frac{12*4}{3}=4*4](https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%3D%20%5Cfrac%7B12%2A4%7D%7B3%7D%3D4%2A4%20)
thus, x=4.
The area of the triangle is 1/2 * 4 *
![2 \sqrt{3}](https://tex.z-dn.net/?f=2%20%5Csqrt%7B3%7D)
=6.93 (mm squared)
The area of the hexagon is 6* the area of the triangle = 42 (mm squared)
Answer: a. 42 (mm squared)
Answer:
21
hope this helped you
please mark as the brainliest (ㆁωㆁ)
Answer:
3xyz
Step-by-step explanation:
The required sum = 7xyz+(−5xyz)+9xyz+(−8xyz)
=7xyz–5xyz+9xyz–8xyz
=(7–5+9–8)xyz
=(16–13)xyz
=3xyz
Answer: xy1 = (6,-5) and xy2 = (-5,6), pls give brainliest if this solution helped you
Step-by-step explanation:
Y = kx - <span>y varies directly as x in this function
</span>
![y= \frac{k}{x}](https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7Bk%7D%7Bx%7D%20)
- y varies inversely as x in this function
For
![y= \frac{1}{4} x](https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7B1%7D%7B4%7D%20x)
y varies directly as x
For
![y= \frac{1}{4x} =\frac{\frac{1}{4}}{x}](https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7B1%7D%7B4x%7D%20%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B4%7D%7D%7Bx%7D)
y varies inversely as x