MCQ
If ABC and DEF are two triangles such that $\triangle\text{ABC}\cong\triangle\text{FDE}$ and $\text{AB}=5\text{m},\angle\text{B}=40^\circ$ and $\angle\text{A}=80^\circ.$ Then, which of the following is true?
  • A
    $\text{DF}=5\text{cm},\angle\text{F}=60^\circ$
  • B
    $\text{DE}=5\text{cm},\angle\text{E}=60^\circ$
  • C
    $\text{DF}=5\text{cm},\angle\text{E}=60^\circ$
  • D
    $\text{DE}=5\text{cm},\angle\text{D}=40^\circ$

Answer

  1. $\text{DF}=5\text{cm},\angle\text{E}=60^\circ$
    Solution:
    In $\triangle\text{ABC},$
    $\angle\text{C}=180^\circ-\angle\text{A}+\angle\text{B}=180^\circ-80^\circ-40^\circ=60^\circ$
    $\triangle\text{ABC}\cong\triangle\text{FDE}$
    $\Rightarrow\text{AB}=\text{FD}=5\text{cm}$
    $\Rightarrow\angle\text{B}=\angle\text{D}=40^\circ$
    $\Rightarrow\angle\text{A}=\angle\text{F}=80^\circ$
    $\Rightarrow\angle\text{C}=\angle\text{E}=60^\circ$
    $\Rightarrow\text{DF}=\text{FD}=5\text{cm}$ and $\angle\text{E}=60^\circ$
    Hence, correct option is (c).

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