
Ananya is designing the window shown in the figure. She wants to make $\Delta PRQ$ congruent to $\Delta PRS$. She designs the window so that $PR \bot QS$. Which of the following conditions will make the two triangles congruent?
A) RQ = RS
B) PQ = PS
C) Both A and B
D) None of these

Answer
511.5k+ views
Hint: First of all , we will have to find the common things between those two triangles which Ananya wants to make congruent i.e. $\Delta PRQ$ congruent to $\Delta PRS$ , when two triangles are congruent they will have similar three sides and exactly the same three angles. The equal sides & angles may not be in the same position (if there is a turn or a flip), but they are there.
Complete step-by-step answer:
In $\Delta PRQ$ and $\Delta PRS$,
\[PR\] is common to both.
\[QR{\text{ }} = {\text{ }}RS\] [Since the point R divides the line \[QS\] in two equal parts]
Now,
$\because $\[QR{\text{ }} = {\text{ }}RS\] and \[PR\] being common,
\[PQ{\text{ }} = {\text{ }}PS\][ height & base being same Hypotenuse should also be same by Pythagoras theorem]
$\therefore $ Hence, the three sides of $\Delta PRQ$ and $\Delta PRS$ are similar.
Now as we know that,
$\because PR \bot QS$
$\therefore \angle PRQ = \angle PRS = {90^\circ }$
$\because $PR divides the angle $\angle SPQ$ as bisector,
$\therefore $$\angle RPQ = \angle RPS$
Hence, $\angle PQR = \angle PSR$ (As PQ=PS, so angles opposite to equal sides are equal)
So, if in $\Delta PRQ\;and{\text{ }}\Delta PRS$
$
\Rightarrow PQ = PS \\
\Rightarrow \angle PQR = \angle PSR \\
\Rightarrow RQ = RS \\
$
Then $\Delta PRQ$ and $\Delta PRS$can be congruent.
So, $\Delta PRQ \cong \Delta PRS(by{\text{ }}SAS{\text{ }}congruence)$
Therefore, correct answer option C.
Note: Two triangles are said to be congruent if all their three sides and three angles are equal or when two figures are similar (not same) to each other based on their shape and size.
Complete step-by-step answer:
In $\Delta PRQ$ and $\Delta PRS$,
\[PR\] is common to both.
\[QR{\text{ }} = {\text{ }}RS\] [Since the point R divides the line \[QS\] in two equal parts]
Now,
$\because $\[QR{\text{ }} = {\text{ }}RS\] and \[PR\] being common,
\[PQ{\text{ }} = {\text{ }}PS\][ height & base being same Hypotenuse should also be same by Pythagoras theorem]
$\therefore $ Hence, the three sides of $\Delta PRQ$ and $\Delta PRS$ are similar.
Now as we know that,
$\because PR \bot QS$
$\therefore \angle PRQ = \angle PRS = {90^\circ }$
$\because $PR divides the angle $\angle SPQ$ as bisector,
$\therefore $$\angle RPQ = \angle RPS$
Hence, $\angle PQR = \angle PSR$ (As PQ=PS, so angles opposite to equal sides are equal)
So, if in $\Delta PRQ\;and{\text{ }}\Delta PRS$
$
\Rightarrow PQ = PS \\
\Rightarrow \angle PQR = \angle PSR \\
\Rightarrow RQ = RS \\
$
Then $\Delta PRQ$ and $\Delta PRS$can be congruent.
So, $\Delta PRQ \cong \Delta PRS(by{\text{ }}SAS{\text{ }}congruence)$
Therefore, correct answer option C.
Note: Two triangles are said to be congruent if all their three sides and three angles are equal or when two figures are similar (not same) to each other based on their shape and size.
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