Problem 156

Solution 1

Three vertices of the square are labelled $P$, $Q$, and $R$ such that $PR$ is the diagonal and $\angle PRQ$ measures $x\degree$. [[IMAGE0]] Since the given figure is a square, then $PQ=QR$ and $\angle PQR=90\degree$. Since $PQ=QR$, $\triangle PQR$ is isosceles and so $\angle QPR=\angle QRP= x\degree$. The three angles in any triangle add to $180\degree$ and since $\angle PQR=90\degree$, then $\angle QPR+\angle QRP=180\degree-90\degree=90\degree$. Since $\angle QPR=\angle QRP$, then $\angle QRP=90\degree\div2=45\degree$, and so $x=45$. Solution 2 The vertices of the square are labelled $P$, $Q$, $R$, and $S$ such that $PR$ is the diagonal and $\angle PRQ$ measures $x\degree$. [[IMAGE1]] Diagonal $PR$ divides square $PQRS$ into two identical triangles: $\triangle PQR$ and $\triangle PSR$. Since these triangles are identical, $\angle PRS=\angle PRQ=x\degree$. Since $PQRS$ is a square, then $\angle QRS=90\degree$. That is, $\angle PRS+\angle PRQ=90\degree$ or $x\degree+x\degree=90\degree$ or $2x=90$ and so $x=45$.