Question

For the diagram given above,

• A. $△PRQ\cong ABC$
• B. $△PRQ\cong ACB$
• C. None of these

Answer 1

The correct option is A $△PRQ\cong ABC$


For the two right triangles $\Delta PRQ$ and $\Delta ABC$:
Here, we have $\angle R=\angle B$ (Right angle)
Also, $RQ=BC=4$ cm.
Now, applying angle sum theorem in $\Delta ABC$.
$\angle A+\angle B+\angle C={180}^{\circ }$
$\Rightarrow {60}^{\circ }+{90}^{\circ }+\angle C={180}^{\circ }$
$\Rightarrow \angle C={180}^{\circ }-({60}^{\circ }+{90}^{\circ })={30}^{\circ }$
Now, we can observe $\angle Q=\angle C$
$\therefore$ By A - S - A congruency, we can state
$\Delta PRQ\cong \Delta ABC$