Answer:
![23\sqrt{3}\ un^2](https://tex.z-dn.net/?f=23%5Csqrt%7B3%7D%5C%20un%5E2)
Step-by-step explanation:
Connect points I and K, K and M, M and I.
1. Find the area of triangles IJK, KLM and MNI:
![A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\](https://tex.z-dn.net/?f=A_%7B%5Ctriangle%20IJK%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%20IJ%5Ccdot%20JK%5Ccdot%20%5Csin%20120%5E%7B%5Ccirc%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%202%5Ccdot%203%5Ccdot%20%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%3D%5Cdfrac%7B3%5Csqrt%7B3%7D%7D%7B2%7D%5C%20un%5E2%5C%5C%20%5C%5C%20%5C%5CA_%7B%5Ctriangle%20KLM%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%20KL%5Ccdot%20LM%5Ccdot%20%5Csin%20120%5E%7B%5Ccirc%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%208%5Ccdot%202%5Ccdot%20%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%3D4%5Csqrt%7B3%7D%5C%20un%5E2%5C%5C%20%5C%5C%20%5C%5CA_%7B%5Ctriangle%20MNI%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%20MN%5Ccdot%20NI%5Ccdot%20%5Csin%20120%5E%7B%5Ccirc%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%203%5Ccdot%208%5Ccdot%20%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%3D6%5Csqrt%7B3%7D%5C%20un%5E2%5C%5C%20%5C%5C%20%5C%5C)
2. Note that
![A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2](https://tex.z-dn.net/?f=A_%7B%5Ctriangle%20IJK%7D%3DA_%7B%5Ctriangle%20IAK%7D%3D%5Cdfrac%7B3%5Csqrt%7B3%7D%7D%7B2%7D%5C%20un%5E2%20%5C%5C%20%5C%5C%20%5C%5CA_%7B%5Ctriangle%20KLM%7D%3DA_%7B%5Ctriangle%20KAM%7D%3D4%5Csqrt%7B3%7D%5C%20un%5E2%20%5C%5C%20%5C%5C%20%5C%5CA_%7B%5Ctriangle%20MNI%7D%3DA_%7B%5Ctriangle%20MAI%7D%3D6%5Csqrt%7B3%7D%5C%20un%5E2)
3. The area of hexagon IJKLMN is the sum of the area of all triangles:
![A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2](https://tex.z-dn.net/?f=A_%7BIJKLMN%7D%3D2%5Ccdot%20%5Cleft%28%5Cdfrac%7B3%5Csqrt%7B3%7D%7D%7B2%7D%2B4%5Csqrt%7B3%7D%2B6%5Csqrt%7B3%7D%5Cright%29%3D23%5Csqrt%7B3%7D%5C%20un%5E2)
Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.
Answer:
Use a compass to draw two equal arcs ...
Step-by-step explanation:
The attached diagram shows the construction of an angle bisector. First you draw arc DE, then arcs from D and E with the same radius that cross at F.
So, the step on your list is, "use a compass to draw two equal arcs from the intersection points of a previous arc and the legs."
The square of 60 is closest to -8 and 8 since 7x7=49 which is 11 away while 8x8=64 only 4 away.
Answer:
50-3 x 12 - 9
Step-by-step explanation: