Question

if triangle CDE has sides that are tangent to the circle, and CF = 15 cm, BD = 12 cm, and EG = 4 cm, then what is the perimeter of triangle CDE

Answer 1

To find the perimeter of triangle CDE given that the sides are tangent to an incircle, we can use the property that tangents drawn from a point outside a circle to the circle are equal in length.

Let:
- \( CF \) be the tangent segment from point C to the incircle,
- \( BD \) be the tangent segment from point B to the incircle,
- \( EG \) be the tangent segment from point E to the incircle.

From the information provided:
- \( CF = 15 \, \text{cm} \)
- \( BD = 12 \, \text{cm} \)
- \( EG = 4 \, \text{cm} \)

Let:
- \( DE = x \) (the segment from D to E),
- \( CE = y \) (the segment from C to E),
- \( CD = z \) (the segment from C to D).

Since the lengths of tangents from a single external point are equal, we can denote:
- \( CE = CF = 15 \, \text{cm} \)
- \( BD = BE = 12 \, \text{cm} \)
- \( AG = AE = 4 \, \text{cm} \)

Now, we know that:
- \( CE + DE = CA + AB = CF + BD = 15 + 12 = 27 \, \text{cm} \)
- \( DE + CD = AG + EC = EG + CD = 4 + CD \) (crucial values need to be confirmed).

We can write the relationships using each segment:
\( CA = CF = 15 \)
\( AB = BD = 12 \)
\( BC = EG = 4 \)

We can use these equalities to determine the total side lengths of triangle CDE.

Now compute the perimeter:
The perimeter \( P = CA + AB + BC = CF + BD + EG \):
1. \( = 15 + 12 + 4 \)
2. \( = 31 \, \text{cm} \)

Thus, the perimeter of triangle CDE is \( \boxed{31 \, \text{cm}} \).