A) 1:4
B) 1:2
C) 2:3
D) 4:5
Correct Answer: A
Solution :
[a] We know that the line segment joining the mid-points of two sides of the triangle is half the length of the third side. |
\[\therefore \,\,\,\,\,\,\,\,DE=\frac{1}{2}AB,\] |
\[EF=\frac{1}{2}BC\] and \[DF=\frac{1}{2}AC\] |
\[\Rightarrow \,\,\,\,\,\frac{DE}{AB}=\frac{EF}{BC}=\frac{DF}{AC}=\frac{1}{2}\] |
Thus, \[\Delta DEF\tilde{\ }\Delta ABC\] |
(BY SSS similarity criterion) |
\[\therefore \,\,\,\,\,\,\frac{ar(\Delta DEF)}{ar(\Delta ABC)}={{\left( \frac{DF}{AC} \right)}^{2}}={{\left( \frac{1}{2} \right)}^{2}}=\frac{1}{4}\] |
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