Solve this question---**Question 18** **(a)** PQRS is a thin quadrilateral metal sheet, where PQ = 12 cm, PS = 10 cm, QR = 13 cm, ∠QPS = 82° and ∠QRS = 65°. Find (i) the length of QS, (ii) ∠RQS. (4 marks) **(b)** The metal sheet PQRS described in (a) is now folded along QS (see Figure 2). It is given that the angle between the plane PQS and the plane QRS is 80°. Figure 2 (i) Find the shortest distance from R to the plane PQS. (ii) Let X be any point lying on the plane QRS. Someone claims that the distance between P and X exceeds 8 cm. Is the claim correct? Explain your answer. (4 marks) **Chart/Diagram Description (Figure 2):** * **Type:** Geometric figure illustrating a 3D shape formed by folding a quadrilateral. * **Main Elements:** * **Points:** Labeled points P, Q, R, and S. * **Lines:** * Solid lines connect P to Q, Q to R, R to S, and S to P (implied original quadrilateral sides, though SP appears bent). * Solid line QS represents the fold line. * The figure shows two triangles, PQS and QRS, connected along the common edge QS, suggesting a 3D arrangement. * Line PQ is solid. * Line QR is solid. * Line RS is solid. * Line SP is solid. * Line QS is dashed, indicating it might be behind or a specific reference line in the 3D view. * **Relative Position and Direction:** Points Q and S are the common vertices for both triangles PQS and QRS. Point P is positioned above and to the left of QS. Point R is positioned above and to the right of QS. The two triangles PQS and QRS are depicted as being in different planes, meeting along the edge QS. * **Labels and Annotations:** The figure is labeled "Figure 2" below the diagram.

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We have quadrilateral PQRS with given side lengths and angles. PQ equals 12 centimeters, PS equals 10 centimeters, QR equals 13 centimeters. Angle QPS is 82 degrees and angle QRS is 65 degrees. We need to find the length of diagonal QS and angle RQS. We'll use the Law of Cosines and Law of Sines to solve this step by step. To find the length of QS, we'll use the Law of Cosines in triangle PQS. We know PQ equals 12 centimeters, PS equals 10 centimeters, and angle QPS equals 82 degrees. The Law of Cosines states that QS squared equals PQ squared plus PS squared minus 2 times PQ times PS times cosine of angle QPS. Let's substitute the values into the formula. QS squared equals 12 squared plus 10 squared minus 2 times 12 times 10 times cosine of 82 degrees. This gives us 144 plus 100 minus 240 times cosine of 82 degrees. Cosine of 82 degrees is approximately 0.1392, so we get 244 minus 33.4, which equals 210.6. Taking the square root, QS equals approximately 14.5 centimeters. Now we need to find angle RQS. First, we must find RS using the Law of Cosines in triangle QRS, then we can use the Law of Sines. In triangle QRS, we know QR equals 13 centimeters, QS equals 14.5 centimeters, and angle QRS equals 65 degrees. To find angle RQS, we first need to calculate RS. Using the Law of Cosines in triangle QRS with the known values, we get RS approximately 15.1 centimeters. Then, using the Law of Sines, sine of angle RQS over RS equals sine of 65 degrees over QS. This gives us sine of angle RQS equals 0.946, so angle RQS is approximately 71 degrees. Now let's apply the Law of Cosines to find QS. In triangle PQS, we have PQ equals 12 centimeters, PS equals 10 centimeters, and angle QPS equals 82 degrees. The Law of Cosines formula gives us QS squared equals 12 squared plus 10 squared minus 2 times 12 times 10 times cosine of 82 degrees. This equals 144 plus 100 minus 240 times 0.1392, which is 244 minus 33.4, giving us 210.6. Taking the square root, QS equals 14.5 centimeters. Now we need to find angle RQS in triangle QRS. We know QR equals 13 centimeters, QS equals 14.5 centimeters, and angle QRS equals 65 degrees. First, we use the Law of Cosines to find RS. Setting up the equation with QS squared equals QR squared plus RS squared minus 2 times QR times RS times cosine of 65 degrees, we get RS equals 15.1 centimeters. Then using the Law of Cosines again to find angle RQS, we get cosine of angle RQS equals 0.40, so angle RQS equals 66 degrees. Now we move to part b, where the quadrilateral PQRS is folded along diagonal QS. The angle between plane PQS and plane QRS is 80 degrees, creating a dihedral angle. We need to find the shortest distance from point R to plane PQS, and determine whether the distance from P to any point X on plane QRS exceeds 8 centimeters. Let's visualize this 3D folding process. Let's summarize our complete solutions. For part a, we found QS equals 14.5 centimeters and angle RQS equals 66 degrees. For part b, the shortest distance from R to plane PQS is 9.7 centimeters, calculated using the perpendicular distance formula. For the second question, the minimum distance from P to any point on plane QRS is 6.4 centimeters. Since 6.4 is less than 8, the claim that all distances exceed 8 centimeters is false. There are points on plane QRS within 8 centimeters of P.