It is given that ΔABC ~ ΔEDF and Area ΔABC : Area ΔEDF = 64 : 25. If AB = 16 cm, BC = 18 cm, CA = 20 cm. What is the value of EF (in cm)?
12.5
The correct answer is 12.5.
Mathematically, if $\Delta$ABC ~ $\Delta$EDF, then: $$ \frac{\text{Area}(\Delta \text{ABC})}{\text{Area}(\Delta \text{EDF})} = \left(\frac{\text{AB}}{\text{ED}}\right)^2 = \left(\frac{\text{BC}}{\text{DF}}\right)^2 = \left(\frac{\text{CA}}{\text{EF}}\right)^2 $$ Applying the Property to the Problem First, let's list the information given: Similarity: $\Delta$ABC ~ $\Delta$EDF Ratio of Areas: $\frac{\text{Area}(\Delta \text{ABC})}{\text{Area}(\Delta \text{EDF})} = \frac{64}{25}$ Sides of $\Delta$ABC: AB = 16 cm, BC = 18 cm, CA = 20 cm Target: Find the length of EF. By applying the theorem that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides, we found the length of the side EF. The calculation confirms that EF measures 12.5 cm.
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