Answer: a) To find the height of the pyramid, we can use the Pythagorean theorem. Let H be the height of the pyramid, and let M be the midpoint of QR. Then, PM is half of PQ, which is 15 cm. We can use the Pythagorean theorem to find H:
H^2 = PQ^2 - PM^2
H^2 = 30^2 - 15^2
H^2 = 675
H = sqrt(675) = 5 sqrt(3) cm
b) To find the angle that PS makes with the base QRST, we can use trigonometry. Let A be the foot of the perpendicular from P to the base QRST, and let B be the midpoint of PS. Then, we have:
tan(angle PSB) = AB / PB
Since triangle PAB is a right triangle, we can use the Pythagorean theorem to find AB:
AB^2 = AP^2 - PB^2
AB^2 = H^2 + (PQ/2)^2 - PB^2
AB^2 = (5 sqrt(3))^2 + (15/2)^2 - PB^2
AB^2 = 225/4 + 75 - PB^2
AB^2 = 375/4 - PB^2
Since triangle PBS is also a right triangle, we can use the Pythagorean theorem to find PB:
PB^2 = PS^2 - BS^2
PB^2 = (2 H)^2 - (PQ/2)^2
PB^2 = 4 (5 sqrt(3))^2 - (15/2)^2
PB^2 = 500 - 56.25
PB^2 = 443.75
Substituting these values into the equation for tan(angle PSB), we get:
tan(angle PSB) = sqrt(375/4 - 443.75) / sqrt(443.75)
tan(angle PSB) = -0.4385
Since angle PSB is in the second quadrant, we have:
angle PSB = 180 degrees + arctan(-0.4385) = 152.4 degrees (rounded to one decimal place)
c) To find the total surface area of the pyramid, including the base, we can divide the pyramid into four triangular faces and a square base. The area of each triangular face can be found using the formula:
area = (1/2) base * height
where the base is the length of one edge of the square base, and the height is the height of the pyramid. The area of the base is simply the area of the square QRST, which is (20 cm)^2 = 400 cm^2. Therefore, we have:
area of each triangular face = (1/2) (20 cm) (5 sqrt(3) cm) = 50 sqrt(3) cm^2
total surface area = 4 (50 sqrt(3) cm^2) + 400 cm^2 = 200 sqrt(3) cm^2 + 400 cm^2
total surface area = (200 + 200 sqrt(3)) cm^2 ≈ 532.4 cm^2 (rounded to one decimal place)
Explanation: