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.
This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
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: $\triangle PQR$ and $\triangle SQR$. 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 $h = \frac{\sqrt{3}}{2}a$.
For $\triangle PQR$, $a = 4$ cm.
So, the length of $PT = \frac{\sqrt{3}}{2} \times 4 = 2\sqrt{3}$ cm.
2. Similarly, the line segment ST is the altitude of the equilateral triangle SQR from vertex S to the base QR.
For $\triangle SQR$, $a = 4$ cm.
So, the length of $ST = \frac{\sqrt{3}}{2} \times 4 = 2\sqrt{3}$ 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 = $PT + ST = 2\sqrt{3} \text{ cm} + 2\sqrt{3} \text{ cm} = 4\sqrt{3} \text{ cm}$.
4. To get a numerical value, we use $\sqrt{3} \approx 1.732$.
Length of path PS $\approx 4 \times 1.732 = 6.928$ cm.
The length of the straight path of PS through the midpoint T over the edge QR is $\boxed{4\sqrt{3} \text{ cm}}$ or approximately $\boxed{6.93 \text{ cm}}$.
