question_answer

It is given that, \[\Delta ABC\tilde{\ }\Delta EDF\]such that \[\text{AB}=\text{5 cm},\] \[\text{AC}=\text{7 cm},\] \[\text{DF}=\text{15 cm}\] and \[\text{DE}=\text{12 cm},\]then the sum of the remaining sides of the triangles is:

A) \[\text{23}.0\text{5 cm}\]

B) \[\text{16}.\text{8 cm}\]

C) \[\text{6}.\text{25 cm}\]

D) \[\text{24 cm}\]

Correct Answer: A

Solution :

[a] Given, \[\Delta ABC\tilde{\ }\Delta EDF\]

Since,    \[\Delta ABC\tilde{\ }\Delta EDF\]

\[\frac{5}{12}=\frac{7}{EF}\]

On taking first and second ratios, we get

\[\frac{5}{12}=\frac{7}{EF}\,\,\,\Rightarrow \,\,\,EF=\frac{7\times 12}{5}=16.8\,cm\]

On taking first and third ratios, we get

\[\frac{5}{12}=\frac{BC}{15}\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,BC=\frac{5\times 15}{12}=6.25\,cm\]

Now, sum of the remaining sides of triangle,

\[=EF+BC=16.8+6.25=23.05\,\,cm\]