We are given a triangle $ABC$ with a line segment $DE$ inside. We are given the lengths $AC = 3.5$ cm, $BE = 4.2$ cm, and $DE = 2.1$ cm. We are also given that angle $BAC$ is equal to angle $BED$. We need to: a) Name a triangle that is similar to triangle $ABC$. b) Calculate the length of $AB$. c) Calculate the area of triangle $ABC$, given that the area of triangle $BDE$ is $22.5$ cm$^2$.

1. Problem Description

We are given a triangle $ABC$ with a line segment $DE$ inside. We are given the lengths $AC = 3.5$ cm, $BE = 4.2$ cm, and $DE = 2.1$ cm. We are also given that angle $BAC$ is equal to angle $BED$. We need to:

a) Name a triangle that is similar to triangle $ABC$.

b) Calculate the length of $AB$.

c) Calculate the area of triangle $ABC$, given that the area of triangle $BDE$ is $22.5$ cm$^2$.

2. Solution Steps

a) Since $\angle BAC = \angle BED$ and $\angle B$ is common to both triangles $BDE$ and $BAC$, then by the Angle-Angle (AA) similarity criterion, triangle $BDE$ is similar to triangle $BAC$.

b) Since triangle $BDE$ is similar to triangle $BAC$, the ratios of their corresponding sides are equal. Thus:

$\frac{BE}{BA} = \frac{DE}{AC}$

We have $BE = 4.2$, $DE = 2.1$, and $AC = 3.5$. We want to find $AB$.

$\frac{4.2}{AB} = \frac{2.1}{3.5}$

$AB = \frac{4.2 \times 3.5}{2.1} = \frac{4.2}{2.1} \times 3.5 = 2 \times 3.5 = 7$

Therefore, $AB = 7$ cm.

c) Since triangle $BDE$ is similar to triangle $BAC$, the ratio of their areas is the square of the ratio of their corresponding sides.

$\frac{Area(BDE)}{Area(ABC)} = (\frac{DE}{AC})^2$

We are given $Area(BDE) = 22.5$ cm$^2$, and we found $DE = 2.1$ and $AC = 3.5$.

$\frac{22.5}{Area(ABC)} = (\frac{2.1}{3.5})^2 = (\frac{3}{5})^2 = \frac{9}{25}$

$Area(ABC) = \frac{22.5 \times 25}{9} = \frac{22.5}{9} \times 25 = 2.5 \times 25 = 62.5$

Therefore, the area of triangle $ABC$ is $62.5$ cm$^2$.

3. Final Answer

a) Triangle $BDE$ is similar to triangle $ABC$.

b) $AB = 7$ cm.

c) $Area(ABC) = 62.5$ cm$^2$.