Question

ABCD is a square and P,Q,R are points on AB,BC and CD respectively such that AP=BQ=CR and Angle PQR=90°,then Angle QPR=?

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Answer 1

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Answer 2

Answer:

The measure of angle QPR is 45°.

Step-by-step explanation:

Given a square ABCD.

P, Q, R are the points on AB, BC and CD respectively.

The measures of lengths of AP, BQ and CR are same i.e., AP = BQ = CR

And Angle PQR =

Since ABCD is a square, AB = BC

Given AP = BQ

Thus AB − AP = BC − BQ

$\implies PB=CQ$                          

In the triangles PBQ and CQR, $\angle PBQ~\&~\angle QCR$ are the angles of squares, hence equal to $\angle PBQ=\angle QCR=90^{o}$

Since PB = CQ and given BQ = CR

From Side-Angle-Side(SAS) rule, the two triangles are said to be congruent if any two sides and the angle included between the sides of two triangle are equal.

Therefore, $\triangle PBQ\cong\triangle CQR$

From corresponding parts of congruent triangles (CPCT) rule, if two triangles are congruent, then all of their corresponding angles and sides are congruent.

we get $PQ=QR$

And hence the triangle PQR is an isosceles triangle.

If PQ = QR then, $\angle QPR=\angle PRQ$

Given $\angle PQR=90^o$

In a triangle, sum of angles is equal to 180°.

$\therefore \angle PQR+\angle QPR+\angle QRP=180^o$

$\implies 90+2\angle QPR+=180$

$\implies 2\angle QPR+=180-90=90$

$\implies \angle QPR=\dfrac{90}{2}=45^o$

Therefore, the measure of angle QPR is 45°.