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Multi-Step 3D Geometry Problems - Math 2

  1. A right triangular prism has a right triangular base with legs of length 5 cm and 12 cm. The prism has a height of 8 cm. A plane cuts through the prism parallel to the triangular base at a height of 3 cm from the bottom. What is the volume of the smaller piece (the piece that contains the triangular base)?
    Answer: 90 cm³ The triangular base is a right triangle with legs 5 cm and 12 cm, so its area is A=12×5×12=30A = \frac{1}{2} \times 5 \times 12 = 30 cm². The original prism has volume Vtotal=A×h=30×8=240V_{total} = A \times h = 30 \times 8 = 240 cm³. When the plane cuts at height 3 cm from the bottom, it creates a smaller prism with the same triangular cross-section but height 3 cm. The volume of this smaller piece is Vsmall=A×3=30×3=90V_{small} = A \times 3 = 30 \times 3 = 90 cm³. Note that since the cutting plane is parallel to the base, the cross-sectional area remains constant throughout the height of the prism.
  2. A water tank in the shape of an inverted cone has a base radius of 6 meters and height of 9 meters. Water flows out of the tank through a hole at the bottom at a rate of 2π2\pi cubic meters per minute. When the water level is 6 meters high, how fast is the water level dropping?
    Answer: 18\frac{1}{8} meters per minute This is a related rates problem. Let hh be the height of water and rr be the radius of the water surface. Since the tank is an inverted cone with base radius 6 m and height 9 m, by similar triangles: rh=69=23\frac{r}{h} = \frac{6}{9} = \frac{2}{3}, so r=2h3r = \frac{2h}{3}. The volume of water is V=13πr2h=13π(2h3)2h=13π⋅4h29⋅h=4πh327V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{2h}{3}\right)^2 h = \frac{1}{3}\pi \cdot \frac{4h^2}{9} \cdot h = \frac{4\pi h^3}{27}. Taking the derivative: dVdt=4π⋅3h227⋅dhdt=4πh29⋅dhdt\frac{dV}{dt} = \frac{4\pi \cdot 3h^2}{27} \cdot \frac{dh}{dt} = \frac{4\pi h^2}{9} \cdot \frac{dh}{dt}. We're given that dVdt=−2π\frac{dV}{dt} = -2\pi (negative because water is flowing out). When h=6h = 6: −2π=4π(6)29⋅dhdt=4π⋅369⋅dhdt=16π⋅dhdt-2\pi = \frac{4\pi (6)^2}{9} \cdot \frac{dh}{dt} = \frac{4\pi \cdot 36}{9} \cdot \frac{dh}{dt} = 16\pi \cdot \frac{dh}{dt}. Solving: dhdt=−2π16π=−18\frac{dh}{dt} = \frac{-2\pi}{16\pi} = -\frac{1}{8} meters per minute. The water level is dropping at 18\frac{1}{8} meters per minute.
  3. A rectangular swimming pool has dimensions 20 m × 15 m × 2 m deep. The pool is being filled with water at a rate of 5 cubic meters per minute. Due to evaporation and small leaks, water is simultaneously lost at a rate of 0.5 cubic meters per minute. If the pool starts empty, how long will it take to fill the pool to 90% capacity?
    Answer: 120 minutes The pool's total volume is V=20×15×2=600V = 20 \times 15 \times 2 = 600 cubic meters. 90% capacity is 0.9×600=5400.9 \times 600 = 540 cubic meters. Water flows in at 5 m³/min but is lost at 0.5 m³/min, so the net rate of filling is 5−0.5=4.55 - 0.5 = 4.5 m³/min. Time to reach 90% capacity: t=5404.5=120t = \frac{540}{4.5} = 120 minutes.

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