The lengths of the sides of a triangle are 6 cm, 7 cm and 9 cm.  In a similar triangle whose perimeter is 110 cm, what is the length of the longest side, in centimeters?
Let the ratio of side lengths between the similar triangle and the given triangle be $x$, so the lengths of the similar triangle are $6x$, $7x$, and $9x$.  We are given that $6x+7x+9x=110$; solving yields $x=\frac{110}{(6+7+9)} = \frac{110}{22}=5$.  The length of the longest side is thus $9x = 9 \cdot 5 = \boxed{45}$.
45
Level 1
Geometry
Problem: The lengths of the sides of a triangle are 6 cm, 7 cm and 9 cm.  In a similar triangle whose perimeter is 110 cm, what is the length of the longest side, in centimeters?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the longest side of the similar triangle, we first need to determine the perimeter of the original triangle. The perimeter is calculated by adding the lengths of all sides: 6 cm + 7 cm + 9 cm = 22 cm. 

Next, we establish the ratio of the perimeters of the similar triangles. The perimeter of the similar triangle is given as 110 cm. The ratio of the perimeters is therefore 110 cm / 22 cm = 5. 

Since the triangles are similar, the lengths of the corresponding sides are also in the same ratio. We can find the lengths of the sides of the similar triangle by multiplying each side of the original triangle by the ratio of 5. 

Calculating the lengths:
- The side corresponding to 6 cm: 6 cm * 5 = 30 cm
- The side corresponding to 7 cm: 7 cm * 5 = 35 cm
- The side corresponding to 9 cm: 9 cm * 5 = 45 cm

Among these calculated lengths, the longest side is 45 cm. Thus, the length of the longest side in the similar triangle is 45.<|END_OF_LATENT|> The final answer is $45$.