\[ 150 = \frac{1}{3} \pi \times 25 \times h \]

["Title: Solving the Equation: 150 = (\frac{1}{3} \pi \ imes 25 \ imes h) — A Step-by-Step Guide", "---", "Meta Description:\nLearn how to solve the equation ( 150 = \frac{1}{3} \pi \ imes 25 \ imes h ) step-by-step. Discover applications in geometry, volume calculations, and real-world problem-solving using pi and linear equations.", "---", "### Understanding the Equation: ( 150 = \frac{1}{3} \pi \ imes 25 \ imes h )", "This equation plays a crucial role when solving for the unknown variable ( h ) (height) in problems involving geometric shapes—particularly when dealing with volumes of composite objects like cylinders, cones, or spheres. But more broadly, it’s a classic linear equation used in algebra and real-world applications such as physics, engineering, and architecture.", "---", "### Step-by-Step Solution to Find ( h )", "Step 1: Start with the given equation:\n[\n150 = \frac{1}{3} \pi \ imes 25 \ imes h\n]", "Step 2: Simplify the right-hand side:", "Multiply ( \frac{1}{3} \ imes 25 ):\n[\n\frac{25}{3} \pi \ imes h = 150\n]", "Step 3: Isolate ( h ) by dividing both sides by ( \frac{25}{3} \pi ):\n[\nh = \frac{150}{\frac{25}{3} \pi}\n]", "Step 4: Simplify the fraction:", "Divide 150 by ( \frac{25}{3} ):\n[\nh = 150 \ imes \frac{3}{25 \pi} = \frac{450}{25 \pi}\n]", "Step 5: Reduce the fraction:\n[\n\frac{450 \div 25}{25 \pi} = \frac{18}{\pi}\n]", "So, the height is:\n[\nh = \frac{18}{\pi}\n]", "---", "### Why This Equation Matters", "Equations like this arise in geometry when calculating volume. For example, if the formula involves the base area (here ( \frac{1}{3} \pi r^2 ) resembling a full cone volume formula without height) and you’re given a total value like 150, solving for height becomes essential.", "Additionally, understanding how to manipulate such equations strengthens algebra skills and prepares one for real-world scenarios where proportional relationships define dimensions—such as scaling architectural models or analyzing natural structures like tree trunks modeled as cylinders.", "---", "### Bonus: Visual Application — Cylindrical Volume", "Suppose you’re given a cylindrical container with radius 5 units and volume 150π cubic units. To find its height ( h ):", "- Volume of a cylinder: ( V = \pi r^2 h )", "Plug in known values:\n[\n150\pi = \pi (5^2) h = 25\pi h\n]", "Divide both sides by ( 25\pi ):\n[\nh = \frac{150\pi}{25\pi} = 6\n]", "This matches earlier methods—showing the power of algebraic manipulation in geometry.", "---", "### Final Thoughts", "Solving equations like ( 150 = \frac{1}{3} \pi \ imes 25 \ imes h ) isn’t just algebraic manipulation—it’s a gateway to applying math in design, engineering, and science. By isolating variables and simplifying expressions, you unlock practical solutions every day.", "---", "Keywords:\n( 150 = \frac{1}{3} \pi \ imes 25 \ imes h ), solve for ( h ), geometry algebra, volume calculation, linear equation, cylindrical volume, real-world math applications, math problem solving, algebraic manipulation", "---", "Ready to apply this equation? Use ( h = \frac{18}{\pi} ) or ( h = \frac{18}{\pi} ) units for your height calculations!", "---", "Keywords optimized for search intent around equation solving, geometry applications, and practical algebra use."]