Given a triangle $ABC$, with $AC = 3.5$ cm, $BE = 4.2$ cm, $DE = 2.1$ cm, and $\angle BAC = \angle BDE$. a) Name a triangle that is similar to triangle $ABC$. b) Calculate i) $AB$ and ii) the area of triangle $ABC$, given that the area of triangle $BDE$ is $22.5 \text{ cm}^2$.

1. Problem Description

Given a triangle $ABC$, with $AC = 3.5$ cm, $BE = 4.2$ cm, $DE = 2.1$ cm, and $\angle BAC = \angle BDE$.

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

b) Calculate i) $AB$ and ii) the area of triangle $ABC$, given that the area of triangle $BDE$ is $22.5 \text{ cm}^2$.

2. Solution Steps

a) Since $\angle BAC = \angle BDE$ and $\angle ABC$ is common to both triangles $ABC$ and $DBE$, we have that triangle $ABC$ is similar to triangle $DBE$ by the AA similarity criterion.

b) i) Since triangles $ABC$ and $DBE$ are similar, the ratio of corresponding sides must be equal. Therefore:

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

We are given $AC = 3.5$ cm, $BE = 4.2$ cm, and $DE = 2.1$ cm. Substituting these values into the equation:

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

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

$AB = \frac{14.7}{2.1}$

$AB = 7$ cm

b) ii) Since triangles $ABC$ and $DBE$ are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides. Therefore:

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

We are given that the area of triangle $BDE$ is $22.5 \text{ cm}^2$, $AC = 3.5$ cm, and $DE = 2.1$ cm. Substituting these values into the equation:

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

$\frac{\text{Area of } ABC}{22.5} = (\frac{5}{3})^2$

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

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

$\text{Area of } ABC = 2.5 \times 25$

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

3. Final Answer

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

b) i) $AB = 7$ cm

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