In ΔBCA, CB = 11 cm, CG = 6 cm, AH = 9 cm. Find the perimeter of ΔBCA. (1 point)

Question

Triangle BCA with inscribed circle D. Segments BF and BH, CF and CG, and AG and GH are tangent to circle D.

32 cm

40 cm

35 cm

42 cm

Answer 1

To find the perimeter of triangle $\mathrm{\Delta }BCA$, 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:

1. Let $BF=BH=x$
2. Let $CF=CG=6$ cm (as given)
3. Let $AG=AH=9$ cm (as given)

Now, using the tangents, we can express the sides of triangle $\mathrm{\Delta }BCA$:

• 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 $\mathrm{\Delta }BCA$:

$$P=AB+BC+AC=(9+x)+(x+6)+15$$

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,

$$x+6=11⟹x=5\text{ cm}$$

Now substituting $x$ back into the perimeter equation:

$$P=2(5)+30=10+30=40\text{ cm}$$

Thus, the perimeter of triangle $\mathrm{\Delta }BCA$ is 40 cm.

The answer is 40 cm.