Triangle Pqr Cb8394

1. **Problem Statement:** Draw triangle $PQR$ where side $PQ = 4.5$ cm, side $PR = 11.7$ cm, and angle $\angle PQR = 90^\circ$. 2. **Understanding the problem:** We have two sides and one angle given. The angle $\angle PQR = 90^\circ$ means the triangle has a right angle at vertex $Q$. 3. **Steps to draw:** - Draw segment $PQ$ of length $4.5$ cm. - At point $Q$, construct a $90^\circ$ angle. - From $Q$, draw a line perpendicular to $PQ$. - Locate point $R$ on this perpendicular line such that $PR = 11.7$ cm. 4. **Finding coordinates (optional for accuracy):** - Place $P$ at origin $(0,0)$. - Place $Q$ at $(4.5,0)$ since $PQ=4.5$ cm along x-axis. - Since $\angle PQR=90^\circ$, $R$ lies on vertical line through $Q$ at $(4.5,y)$. - Use distance formula for $PR=11.7$ cm: $$ PR = \sqrt{(4.5-0)^2 + (y-0)^2} = 11.7 $$ $$ \Rightarrow \sqrt{4.5^2 + y^2} = 11.7 $$ $$ 4.5^2 + y^2 = 11.7^2 $$ $$ 20.25 + y^2 = 136.89 $$ $$ y^2 = 136.89 - 20.25 = 116.64 $$ $$ y = \pm \sqrt{116.64} = \pm 10.8 $$ 5. **Conclusion:** Point $R$ is at $(4.5, 10.8)$ or $(4.5, -10.8)$. 6. **Draw the triangle with points:** - $P(0,0)$ - $Q(4.5,0)$ - $R(4.5,10.8)$ (choosing positive y for convenience) This completes the triangle $PQR$ with the given conditions.