Question

It is given that $△ABC\cong △FDE$ and $AB=5cm$, $\angle B={40}^0$ and $\angle A={80}^0$. Then which of the following is true?

• A. $DF=5cm$, $\angle F={60}^0$
• B. $DF=5cm$, $\angle E={60}^0$
• C. $DE=5cm$, $\angle E={60}^0$
• D. $DF=5cm$, $\angle D={60}^0$

Answer 1

The correct option is B $DF=5cm$, $\angle E={60}^0$

In $△ABC$
$\angle A={80}^o$
$\angle B={40}^o$
$\angle A+\angle B+\angle C={180}^o$
$80+40+\angle C={180}^o$
$\angle C={60}^o$
Now, $△ABC\cong △FDE$
Thus, by $CPCT$ (Corresponding parts of congruent triangles are congruent),

$\angle A=\angle F={80}^o$
$\angle B=\angle D={40}^o$
$\angle C=\angle E={60}^o$
$AB=FD=5cm$
$BC=DE$
$AC=EF$