We are given a triangle $ABC$ with $AC = 3.5$ cm and $BE = 4.2$ cm. Also, $DE = 2.1$ cm and $\angle BAC = \angle BED$. We need to find: a) A triangle similar to $\triangle ABC$. b) The length of $AB$. c) 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 $AC = 3.5$ cm and $BE = 4.2$ cm. Also, $DE = 2.1$ cm and $\angle BAC = \angle BED$. We need to find:

a) A triangle similar to $\triangle ABC$.

b) The length of $AB$.

c) 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 $\triangle ABC$ and $\triangle BDE$, we can say that $\triangle ABC \sim \triangle EBD$ by the Angle-Angle (AA) similarity criterion.

b) Since $\triangle ABC \sim \triangle EBD$, the corresponding sides are proportional. Therefore, we have:

$\frac{AB}{EB} = \frac{AC}{ED} = \frac{BC}{BD}$

We are given $AC = 3.5$, $BE = 4.2$, and $DE = 2.1$.

We can use the proportion $\frac{AB}{EB} = \frac{AC}{ED}$ to find $AB$.

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

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

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

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

$AB = 3.5 \times 2$

$AB = 7$ cm

c) The ratio of areas of similar triangles is the square of the ratio of their corresponding sides.

$\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle BDE} = \left(\frac{AC}{DE}\right)^2$

$\frac{\text{Area of } \triangle ABC}{22.5} = \left(\frac{3.5}{2.1}\right)^2$

$\frac{\text{Area of } \triangle ABC}{22.5} = \left(\frac{35}{21}\right)^2 = \left(\frac{5}{3}\right)^2 = \frac{25}{9}$

$\text{Area of } \triangle ABC = 22.5 \times \frac{25}{9}$

$\text{Area of } \triangle ABC = \frac{22.5 \times 25}{9} = \frac{562.5}{9}$

$\text{Area of } \triangle ABC = 62.5$ cm$^2$

3. Final Answer

a) $\triangle EBD$ is similar to $\triangle ABC$.

b) $AB = 7$ cm

c) Area of $\triangle ABC = 62.5$ cm$^2$