To construct triangle PQR and measure the lengths of $PQ$ and $PR$: Given: • $QR = 6 \text{ cm}$ • $\angle PQR = 75^\circ$ • $\angle PRQ = 30^\circ$ Construction Steps: Step 1: Draw the base $QR$. Draw a straight line and mark a point Q on it. Using a ruler, measure $6 \text{ cm}$ from Q and mark point R. So, $QR = 6 \text{ cm}$. Step 2: Construct $\angle PRQ = 30^\circ$ at point R. • With R as the center, draw an arc of any convenient radius to intersect $QR$ at a point, say A. • With A as the center and the same radius, draw another arc to intersect the first arc at a point, say B. $\angle BRQ = 60^\circ$. • Bisect $\angle BRQ$. With A and B as centers, draw arcs of the same radius (larger than half the distance AB) to intersect at a point, say C. • Draw a line from R through C. This line forms an angle of $30^\circ$ with $QR$. Step 3: Construct $\angle PQR = 75^\circ$ at point Q. • With Q as the center, draw an arc of any convenient radius to intersect $QR$ at a point, say D. • With D as the center and the same radius, draw an arc to intersect the first arc at E ($60^\circ$). • With E as the center and the same radius, draw an arc to intersect the first arc at F ($120^\circ$). • Bisect the angle between E and F to construct $90^\circ$. With E and F as centers, draw arcs of the same radius to intersect at G. • Draw a dashed line from Q through G. This line represents $90^\circ$. Let the intersection of this line with the initial arc be H. • Now, bisect the angle between E ($60^\circ$) and H ($90^\circ$). With E and H as centers, draw arcs of the same radius to intersect at I. • Draw a line from Q through I. This line forms an angle of $60^\circ + 15^\circ = 75^\circ$ with $QR$. Step 4: Locate point P. The intersection of the line from R (from Step 2) and the line from Q (from Step 3) is point P. Connect P to Q and P to R to complete triangle PQR. Measurement of $|PQ|$ and $|PR|$: Since physical measurement cannot be performed by an AI, we will calculate the lengths using trigonometry. First, find the third angle $\angle QPR$: The sum of angles in a triangle is $180^\circ$. $$ \angle QPR = 180^\circ - \angle PQR - \angle PRQ $$ $$ \angle QPR = 180^\circ - 75^\circ - 30^\circ $$ $$ \angle QPR = 180^\circ - 105^\circ $$ $$ \angle QPR = 75^\circ $$ Since $\angle PQR = \angle QPR = 75^\circ$, triangle PQR is an isosceles triangle with sides opposite these angles being equal. Therefore, $PR = QR$. Given $QR = 6 \text{ cm}$, $$ PR = 6 \text{ cm} $$ Now, use the Sine Rule to find the length of $PQ$: $$ \frac{PQ}{\sin(\angle PRQ)} = \frac{QR}{\sin(\angle QPR)} $$ $$ \frac{PQ}{\sin(30^\circ)} = \frac{6 \text{ cm}}{\sin(75^\circ)} $$ $$ PQ = \frac{6 \text{ cm} \times \sin(30^\circ)}{\sin(75^\circ)} $$ We know $\sin(30^\circ) = 0.5$ and $\sin(75^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4} \approx 0.9659$. $$ PQ = \frac{6 \text{ cm} \times 0.5}{0.9659} $$ $$ PQ = \frac{3 \text{ cm}}{0.9659} $$ $$ PQ \approx 3.1058 \text{ cm} $$ Rounding to two decimal places: $$ PQ \approx 3.11 \text{ cm} $$ The measured lengths are: $|PQ| = \boxed{\text{3.11 cm}}$ $|PR| = \boxed{\text{6.00 cm}}$