A regular tetrahedron is a three-dimensional shape with four faces, each of which is an equilateral triangle. All edges are of equal length, in this case, 4 cm.

To solve this problem, we need to visualize the tetrahedron's net and then find the shortest path on the unfolded surface. A regular tetrahedron is a three-dimensional shape with four faces, each of which is an equilateral triangle. All edges are of equal length, in this case, 4 cm. The path from P to S passes over the edge QR at its midpoint T. This means the path lies on two faces: PQR and SQR. To find the straight path, we need to unfold these two faces into a single plane. Step 1: Draw the net of the relevant faces. 1. Draw an equilateral triangle PQR with side length 4 cm. 2. Draw another equilateral triangle SQR, sharing the edge QR, such that vertex S is on the opposite side of QR from vertex P. This forms a larger triangle (or a rhombus if P, Q, S, R were coplanar, but here P and S are vertices of two triangles sharing a base). The net will look like this: ` P / \ / \ / \ Q-------R \ / \ / \ / S ` Step 2: Locate the midpoint T and draw the path. 1. Mark the midpoint T on the edge QR. 2. The straight path from P to S, passing through T, is a straight line segment connecting P and S on this unfolded net. Step 3: Calculate the length of the path PS. 1. The line segment PT is the altitude (height) of the equilateral triangle PQR from vertex P to the base QR. The formula for the altitude h of an equilateral triangle with side length a is h = sqrt(3)2a. For PQR, a = 4 cm. So, the length of PT = sqrt(3)2 × 4 = 2sqrt(3) cm. 2. Similarly, the line segment ST is the altitude of the equilateral triangle SQR from vertex S to the base QR. For SQR, a = 4 cm. So, the length of ST = sqrt(3)2 × 4 = 2sqrt(3) cm. 3. On the unfolded net, P, T, and S are collinear. Therefore, the total length of the straight path PS is the sum of PT and ST. Length of path PS = PT + ST = 2sqrt(3) cm + 2sqrt(3) cm = 4sqrt(3) cm. 4. To get a numerical value, we use sqrt(3) ≈ 1.732. Length of path PS ≈ 4 × 1.732 = 6.928 cm. The length of the straight path of PS through the midpoint T over the edge QR is 4sqrt(3) cm or approximately 6.93 cm.