Let's begin by identifying key information given to us:
Triangle HGF is equilateral
a.
[tex]\begin{gathered} \angle F=\angle H=\angle G \\ \angle F=(x+32)^{\circ} \\ \angle H=(2x+4)^{\circ} \\ \Rightarrow(x+32)^{\circ}=(2x+4)^{\circ} \\ x+32=2x+4 \\ \text{Put like terms together, we have:} \\ 32-4=2x-x \\ 28=x\Rightarrow x=28 \\ x=28 \end{gathered}[/tex]
Since we know that all the angles are equal, we will substitute the value of x into angle F or H. We have:
[tex]\begin{gathered} \angle H=(2x+4)^{\circ} \\ \angle H=2(28)+4=56+4 \\ \angle H=60^{\circ} \\ But,\angle H=\angle G=60^{\circ} \\ \\ \therefore m\angle G=60^{\circ} \end{gathered}[/tex]
b.
[tex]\begin{gathered} Perimeter=HG+GF+FH \\ Perimeter=6y+24 \\ HG=3y-7 \\ \text{The three sides are equivalent, thus we have:} \\ 6y+24=3(3y-7) \\ 6y+24=9y-21 \\ \text{Put like terms together, we have:} \\ 9y-6y=24+21 \\ 3y=45 \\ y=\frac{45}{3}=15 \\ y=15 \\ \\ Perimeter=6(15)+24=90+24=114 \\ \therefore Perimeter=114 \end{gathered}[/tex]
