Question

I need help with this ( but I need to send another picture that has the answers )

Answer 1

Step 1:

Draw the figure

The diagonals of a rhombus bisect each other,

therefore

FJ = JH

this implies that

JH = 4.

Opposite angles of a rhombus are equal, so

The diagonals of a rhombus bisect the interior angles.

Therefore,

`\text{ angle JHG = }(120^0)/(2)=60^0`

The diagonals of a rhombus are perpendicular

`\begin{gathered} Hence, \\ \cos 60^0=(4)/(GH) \\ \Rightarrow GH=(4)/(\cos60^0)=(4)/((1)/(2))=4*2=8 \end{gathered}`

GH = 8 (b)

Also,

Using the Pythagorean rule,

`\begin{gathered} JG^2+4^2=GH^2 \\ \Rightarrow JG^2+16=8^2 \\ \Rightarrow JG=\sqrt[]{64-16}=\sqrt[]{48}=4\sqrt[]{3}=6.928\text{ } \\ \text{right choice is C} \end{gathered}`

FH = FJ + JH

this implies that

FH = 4 + 4 = 8

FH = 8

right choice B

The diagonals of a rhombus bisect the interior angles.

therefore

`\begin{gathered} angle\text{ JFG = }\frac{angle\text{ IFG}}{2}=(120^0)/(2)=60^0 \\ \text{ right choice is h} \end{gathered}`

Sum of the interior angles of a rhombus is 360degrees.

And opposite interior angles of a polygon are congruent.

Therefore

`2(\text{angle FGH) +2(angle IFG) = 360degr}ees`
`\begin{gathered} \text{this implies that} \\ 2(\text{angle FGH) = 360 - 2(angle IFG)} \\ \Rightarrow\text{angle FGH = 180 -angle IFG = 180 - 120= }60^0 \\ \text{ right choice is h} \end{gathered}`
`\begin{gathered} \text{ but }FGJ\text{ = }(FGH)/(2)\text{ (diagonals bisect interior angles)} \\ \Rightarrow\text{FGJ =}(60^0)/(2)=30^0 \\ \text{right choice is j} \end{gathered}`

Also,