a. To find the angle QCP, we can start by finding the coordinates of the points P, Q, and C. Since P and Q are midpoints of the edges FG and GH, respectively, we can find their coordinates as follows:
P = (0, y, z) where y = (3+4)/2 = 3.5 and z = (0+4)/2 = 2
Q = (x, y, 0) where x = (3+4)/2 = 3.5 and y and z are the same as before.
Since C is the center of the cube, we can find its coordinates as:
C = (2, 2, 2)
Now we can use the dot product formula to find the angle QCP:
cos(QCP) = (PC · QC) / (|PC| · |QC|)
where PC and QC are the vectors from P to C and from Q to C, respectively.
PC = C - P = (2, 2 - 3.5, 2 - 2) = (2, -1.5, 0)
QC = C - Q = (2 - 3.5, 2 - 3.5, 0) = (-1.5, -1.5, 0)
|PC| = sqrt(2² + (-1.5)² + 0²) = sqrt(5.25)
|QC| = sqrt((-1.5)² + (-1.5)² + 0²) = sqrt(4.5)
PC · QC = 2(-1.5) + (-1.5)(-1.5) + 0 = -3 + 2.25 = -0.75
cos(QCP) = (-0.75) / (sqrt(5.25) * sqrt(4.5)) = -0.166666...
Therefore, the angle QCP is cos⁻¹(-0.166666...) ≈ 100.53° (rounded to two decimal places).
b. To find the total surface area of the solid remaining after pyramid PGQC is removed, we need to find the area of the four triangular faces and the square base of the pyramid and subtract them from the total surface area of the cube.
The square base of the pyramid has side length PQ = sqrt((3.5 - 3)² + (3.5 - 4)² + 2²) = sqrt(2.5) and area PQ² = 2.5 cm².
Each of the triangular faces has base PQ and height equal to the distance from P or Q to the plane containing the opposite vertex and the center of the cube. This distance is equal to the length of the perpendicular from the vertex to the opposite face, which is half the length of the diagonal of the cube.
The diagonal of the cube is sqrt(3² + 4² + 4²) = sqrt(41), so the height of each triangular face is sqrt(41)/2.
Therefore, the area of each triangular face is (1/2)PQ(sqrt(41)/2) = (sqrt(41)/4) cm².
The total surface area of the cube is 6s², where s is the length of a side of the cube. Since the volume of the cube is 1728 cm³, we have s³ = 1728, or s = 12 cm.
Therefore, the total surface area of the cube is 6(12²) = 864 cm².
The total surface area of the solid remaining after the pyramid is removed is:
864 - 4(sqrt