Question

1. $\triangle bcd \sim \triangle fge$

1. $klmn \sim pqrs$

1. $\triangle rst \sim \triangle tsz$

1. $\triangle cde \sim \triangle fge$

1. if $\triangle klm \sim \triangle pqr$ with a scale factor of $3:5$, find the perimeter of $\triangle pqr$.

Explanation:

Answer:

To solve the problem, we first need to find the perimeter of \(\triangle KLM\) and then use the scale factor to find the perimeter of \(\triangle PQR\).

Step 1: Calculate the perimeter of \(\triangle KLM\)

The sides of \(\triangle KLM\) are \(KL = 6\), \(LM = 15\), and \(KM = 12\).
The perimeter of \(\triangle KLM\) is the sum of its sides:

$$ \text{Perimeter of } \triangle KLM = 6 + 15 + 12 = 33 $$

Step 2: Use the scale factor to find the perimeter of \(\triangle PQR\)

The scale factor of \(\triangle KLM\) to \(\triangle PQR\) is \(3:5\). Let the perimeter of \(\triangle PQR\) be \(P\).
We know that the ratio of the perimeters of similar triangles is equal to the scale factor. So:

$$ \frac{\text{Perimeter of } \triangle KLM}{\text{Perimeter of } \triangle PQR} = \frac{3}{5} $$

Substituting the known perimeter of \(\triangle KLM\):

$$ \frac{33}{P} = \frac{3}{5} $$

Cross - multiply to solve for \(P\):

$$ 3P = 33\times5 $$

$$ 3P = 165 $$

Divide both sides by 3:

$$ P=\frac{165}{3}=55 $$

The perimeter of \(\triangle PQR\) is \(\boxed{55}\).