<h3>Answer:</h3>
There are more red marbles in the bag.
<h3>Explanation:</h3>
The fraction of red marbles is 1/3 = 3/9. That is a larger fraction than the fraction of green marbles, which is 2/9.
1. Check the drawing of the rhombus ABCD in the picture attached.
2. m(CDA)=60°, and AC and BD be the diagonals and let their intersection point be O.
3. The diagonals:
i) bisect the angles so m(ODC)=60°/2=30°
ii) are perpendicular to each other, so m(DOC)=90°
4. In a right triangle, the length of the side opposite to the 30° angle is half of the hypothenuse, so OC=3 in.
5. By the pythagorean theorem,
![=\sqrt{9}* \sqrt{3} =3\sqrt{3} (in)](https://tex.z-dn.net/?f=%3D%5Csqrt%7B9%7D%2A%20%5Csqrt%7B3%7D%20%3D3%5Csqrt%7B3%7D%20%20%28in%29)
6. The 4 triangles formed by the diagonal are congruent, so the area of the rhombus ABCD = 4 Area (triangle DOC)=4*
![\frac{DO*OC}{2}=4 \frac{3 \sqrt{3} *3}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7BDO%2AOC%7D%7B2%7D%3D4%20%5Cfrac%7B3%20%5Csqrt%7B3%7D%20%2A3%7D%7B2%7D%20)
=
![=2*9 \sqrt{3}=18 \sqrt{3}](https://tex.z-dn.net/?f=%3D2%2A9%20%5Csqrt%7B3%7D%3D18%20%5Csqrt%7B3%7D%20%20)
(
![in^{2}](https://tex.z-dn.net/?f=%20in%5E%7B2%7D%20)
)