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,

6. The 4 triangles formed by the diagonal are congruent, so the area of the rhombus ABCD = 4 Area (triangle DOC)=4*

=

(

)
Answer:
DF = 14
Step-by-step explanation:
Andrew was wrong in the assumption that DE = EF
Segments of tangents to a circle from the same external point are congruent
That is
EF = FG = 8
Then
DF = DE + EF = 6 + 8 = 14
Refrection across the y-axis changes the function to y = |-x|.
Vertical stretch by a factor of 1.5 changes the graph to y = 1.5|-x|
Shifting the graph by 4 units downward changes the graph to y = 1.5|-x| - 4
Required equation is y = 1.5|-x| - 4