To find the perimeter of triangle \( \Delta BCA \), we will use the given measurements and properties of tangents to the inscribed circle.
Let:
- \( BF \) be the tangent from point \( B \) to the incircle.
- \( CF \) be the tangent from point \( C \).
- \( AG \) be the tangent from point \( A \).
Since the lengths of tangents drawn from an external point to a circle are equal:
- Let \( BF = BH = x \)
- Let \( CF = CG = 6 \) cm (as given)
- Let \( AG = AH = 9 \) cm (as given)
Now, using the tangents, we can express the sides of triangle \( \Delta BCA \):
- Side \( BC = BF + CF = x + 6 \)
- Side \( AC = AG + CG = 9 + 6 = 15 \)
- Side \( AB = AH + BF = 9 + x \)
Now we can find the perimeter \( P \) of triangle \( \Delta BCA \):
\[ P = AB + BC + AC = (9 + x) + (x + 6) + 15 \]
Simplifying,
\[ P = 9 + x + x + 6 + 15 = 2x + 30 \]
To find \( x \) (which represents the length of \( BF \)), we can make use of the fact that in this triangle, \( CB = 11 \) cm.
We already know \( CF \) is \( 6 \) cm, so we can write:
\[ CB + CF = 11 + 6 = 17 \text{ cm} \] where \( CB = CF + BF = 11 \).
Thus,
\[ x + 6 = 11 \implies x = 5 \text{ cm} \]
Now substituting \( x \) back into the perimeter equation:
\[ P = 2(5) + 30 = 10 + 30 = 40 \text{ cm} \]
Thus, the perimeter of triangle \( \Delta BCA \) is 40 cm.
The answer is 40 cm.
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