Two triangles ABC and DEF are similar. The smallest side of ABC is equal to 15 units. If the sides of ABC are in the ratio 3 ∶ 4 ∶ 5, and the area of DEF is half of the area of ABC, then what is the largest side of DEF (in units)?

Correct option is D

Given:
∆ABC ~ ∆ DEF
​$$\text{Ratio of } \frac{\text{ar}(\triangle DEF)}{\text{ar}(\triangle ABC)} = \frac{1}{2}$$​​
Smallest side of ABC=15 units
Sides of ABC are in the ratio 3∶4 ∶ 5
Formula Used:
​$$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \frac{AB^2}{DE^2} = \frac{BC^2}{EF^2} = \frac{AC^2}{DF^2}$$​​
Solution:
Let AB be smallest side of ∆ABC and AC be largest.
Then,
​$$\frac{\text{ar}(\triangle DEF)}{\text{ar}(\triangle ABC)} = \frac{DE^2}{AB^2}\\\frac{1}{2} = \frac{DE^2}{15^2}\\\sqrt{\frac{1}{2}} = \frac{DE}{15}\\DE = \frac{15}{\sqrt{2}}\\$$​​
If sides of ABC are in the ratio 3∶4 ∶ 5then sides of ∆ DEF are also in the same ratio.
Hence DE: EF: DF = 3:4:5
​$$\frac{DE}{DF} = \frac{3}{5}\\DE \times \frac{5}{3} = DF\\\frac{15}{\sqrt{2}} \times \frac{5}{3} = \frac{25}{\sqrt{2}}\\$$​​