We are given triangle $ABC$ with $AC = 3.5$ cm, $BE = 4.2$ cm, $DE = 2.1$ cm, and $\angle BAC = \angle BED$. a) We need to find a triangle similar to triangle $ABC$. b) We need to calculate $AB$. c) We need to calculate the area of triangle $ABC$, given that the area of quadrilateral $BDEC$ is $22.5$ cm$^2$.

1. Problem Description

We are given triangle $ABC$ with $AC = 3.5$ cm, $BE = 4.2$ cm, $DE = 2.1$ cm, and $\angle BAC = \angle BED$.

a) We need to find a triangle similar to triangle $ABC$.

b) We need to calculate $AB$.

c) We need to calculate the area of triangle $ABC$, given that the area of quadrilateral $BDEC$ is $22.5$ cm$^2$.

2. Solution Steps

a) Since $\angle BAC = \angle BED$ and these are corresponding angles, it means $DE \parallel AC$.

Therefore, $\angle ABC = \angle DBE$ (common angle) and $\angle ACB = \angle DEB$ (corresponding angles).

So, triangle $ABC$ is similar to triangle $DBE$ by the AA similarity criterion.

b) Since $\triangle ABC \sim \triangle DBE$, the ratios of their corresponding sides are equal.

$\frac{DB}{AB} = \frac{BE}{BC} = \frac{DE}{AC}$.

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

Thus, $\frac{DE}{AC} = \frac{2.1}{3.5} = \frac{21}{35} = \frac{3}{5} = 0.6$.

Let $AB = x$. Then $DB = AB - AD$.

Also, $BE = 4.2$ cm.

We have $\frac{DB}{AB} = \frac{3}{5}$. Thus, $DB = \frac{3}{5}AB = \frac{3}{5}x$.

Also, $BC = BE + EC$. Since $\frac{BE}{BC} = \frac{3}{5}$, we have $BC = \frac{5}{3}BE = \frac{5}{3}(4.2) = 5(1.4) = 7$.

Then $\frac{DB}{AB} = \frac{DE}{AC}$. Therefore $\frac{DB}{AB} = \frac{3}{5}$ which means $5DB = 3AB$.

If $AD = y$ then $DB = x - y$. So we have $DB/AB = DE/AC$ so $DB/x = 2.1/3.5$ so $DB/x = 3/5$ so $DB = 3x/5$. So $\frac{AB-AD}{AB} = \frac{DE}{AC}$. Also, $\frac{DB}{AB} = \frac{BE}{BC} = \frac{3}{5}$

However, we need to find $AB$. Since $\frac{DE}{AC} = \frac{2.1}{3.5} = \frac{3}{5}$, this is the linear scale factor.

Therefore $BE = \frac{3}{5}BC$, $BD = \frac{3}{5}BA$ or $\frac{DB}{AB} = \frac{3}{5}$.

So, $DB = \frac{3}{5}AB$.

Also, we have $\angle BAC = \angle BED$, so triangles $ABC$ and $DBE$ are similar.

Let $AD = a$, $DB = b$.

$AB = AD + DB = a+b$. So $b = (3/5) (a+b)$. $5b = 3a + 3b$.

$2b = 3a$, $b = (3/2) a$. Then $AB = a+b = a+ (3/2) a = (5/2) a$, $a = (2/5) AB$. $DB = (3/5) AB = (3/5) (5/2)a = (3/2) a = \frac{3a}{2}$.

b) The ratio of areas of similar triangles is the square of the linear scale factor.

$\frac{\text{Area}(DBE)}{\text{Area}(ABC)} = (\frac{3}{5})^2 = \frac{9}{25}$.

Area of $ABC$ = Area of $DBE$ + Area of $BDEC$.

Area $ABC =$ Area $DBE + 22.5$

Let Area $ABC = A$. Area $DBE = A - 22.5$. Then

$\frac{A-22.5}{A} = \frac{9}{25}$.

$25(A - 22.5) = 9A$.

$25A - 25(22.5) = 9A$.

$16A = 25(22.5) = 562.5$.

$A = \frac{562.5}{16} = 35.15625$.

Area of triangle $ABC = 35.15625$ cm$^2$.

We need to find $AB$. From the information given, we know $AC=3.5$. Also we know area $ABC$ and that the triangles are similar so the ratio of sides $DE:AC$ is the same ratio as side length. So DE/AC= 2.1/3.5=3/

5. We do not have any other side lengths. But we do have $BE=4.2$. Since DBE is similar to ABC: $\frac{BE}{BC}=\frac{DE}{AC}$ implies $\frac{4.2}{BC}=\frac{3}{5}$. $BC=\frac{5 \cdot 4.2}{3}=7$.

Let AB = $x$.

Let DE be the base of the triangle. $\frac{DE}{AC} = \frac{3}{5} = \frac{2.1}{3.5}$. So the ratio of the sides are 3/

5. The heights must have the same 3/5 ratio. area of small $\Delta BDE$ is Area of $\Delta ABC$-22.5cm$^2$. If the triangles are similar the ratio is square.

Since we cannot determine the length of AB at this time.

3. Final Answer

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

b) It is not possible to determine $AB$ using the information provided.

c) Area of triangle $ABC$ is $35.15625$ cm$^2$.