ΔABC ~ ΔEDF and area (ΔABC) : area (ΔEDF) = 9 : 4. If AB = 5 cm, BC = 8 cm and CA = 10 cm, then DF is equal to:

The correct answer is $5\frac{1}{3}$ cm.

Step-by-step explanation:

1. We are given two similar triangles: ΔABC ~ ΔEDF.

2. We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Therefore:

$\frac{\text{area}(\Delta \text{ABC})}{\text{area}(\Delta \text{EDF})}={\left(\frac{\text{AC}}{\text{EF}}\right)}^2={\left(\frac{\text{BC}}{\text{DF}}\right)}^2={\left(\frac{\text{AB}}{\text{ED}}\right)}^2$

3. Identifying corresponding sides from the similarity order ΔABC ~ ΔEDF:

Side BC of ΔABC corresponds to side DF of ΔEDF.

4. Setting up the proportion using the given ratio of areas (9 : 4) and the length of side BC (8 cm):

$\frac{\text{area}(\Delta \text{ABC})}{\text{area}(\Delta \text{EDF})}={\left(\frac{\text{BC}}{\text{DF}}\right)}^2$

$\frac{9}{4}={\left(\frac{8}{\text{DF}}\right)}^2$

5. Taking the square root on both sides:

$\frac{3}{2}=\frac{8}{\text{DF}}$

6. Solving for DF:

$3\times \text{DF}=8\times 2$

$3\times \text{DF}=16$

$\text{DF}=\frac{16}{3}\text{cm}$

7. Converting the improper fraction $\frac{16}{3}$ to a mixed fraction:

$16÷3=5$ with a remainder of 1.

Hence, $\text{DF}=5\frac{1}{3}\text{cm}$.