To solve this problem we will use the rule
The 2 tangents drawn from a point outside the circle are equal in length
From the given picture we can see
A circle D is inscribed in the triangle ABC and touches its sides at points H, F, and G
By using the rule above
AH = AG
BH = BF
CF = CG
Since BF = 6 cm, then BH = 6cm
Since CG = 9 cm, then CF = 9 cm
Since AB = 17 cm
Since AB = BH + HA
Since BH = 6 cm, then
[tex]17=6+HA[/tex]
Subtract 6 from both sides
[tex]\begin{gathered} 17-6=6-6+HA \\ 11=HA \end{gathered}[/tex]
Since HA = AG, then AG = 11 cm
Now, we can find the length of BC and AC
[tex]\begin{gathered} BC=BF+FC \\ BC=6+9 \\ BC=15\text{ cm} \end{gathered}[/tex][tex]\begin{gathered} AC=AG+GC \\ AC=11+9 \\ AC=20\text{ cm} \end{gathered}[/tex]
Since the perimeter of the triangle = the sum of the lengths of its sides, then
[tex]\begin{gathered} P=17+15+20 \\ P=52\text{ cm} \end{gathered}[/tex]
The answer is the 1st choice
