Answer:
rational
Step-by-step explanation:
Answer:
2 cups
Step-by-step explanation:
Answer:
3/4 is a bigger
Step-by-step explanation:
and 4/15 is a smaller
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.
This follows from the half-angle identity for cosine.