A triangular prism has a base area of 15 cm² and a height of 10 cm. If the base area is increased by 50% and the height is doubled, what is the new volume?
["New Volume of a Triangular Prism: A Practical Calculation", "When studying geometry, understanding how changes in a prism’s dimensions affect its volume is essential. This article explains how altering the base area and height of a triangular prism changes its volume—using a real-world example involving a base area of 15 cm² and a height of 10 cm.", "### What is the Volume of a Triangular Prism?", "The volume ( V ) of a prism is calculated using the formula:", "[\nV = \ ext{Base Area} \ imes \ ext{Height}\n]", "For the initial prism:", "- Base area = 15 cm²\n- Height = 10 cm\n- Volume = ( 15 \ imes 10 = 150 ) cm³", "This establishes the starting point for our analysis.", "### How Changes Affect Volume", "Now, suppose we increase the base area by 50% and double the height:", "- New base area = ( 15 + (0.5 \ imes 15) = 15 + 7.5 = 22.5 ) cm²\n- New height = ( 10 \ imes 2 = 20 ) cm", "Using the volume formula again:", "[\n\ ext{New Volume} = 22.5 \ imes 20 = 450 \ ext{ cm}³\n]", "### Why This Matters", "The volume increased from 150 cm³ to 450 cm³—a threefold increase—even though only two of the three dimensions (base area and height) changed. This highlights a key geometric principle: volume scales directly with both base area and height in prisms.", "### Conclusion", "Increasing a triangular prism’s base area by 50% and doubling its height multiplies the volume by three. This insight helps in fields ranging from architecture to packaging design, where volume optimization is crucial. Understanding and calculating these changes ensures precision in both academic and practical applications.", "Summary:\nInitial volume: ( 150 ) cm³\nNew volume after changes: ( 450 ) cm³\nVolume increased by a factor of 3 due to proportional adjustments in base area and height.", "---", "If you want to quickly find the new volume from any original dimensions, use:\n[\n\ ext{New Volume} = (1.5 \ imes \ ext{Base Area}) \ imes (2 \ imes \ ext{Height}) = 3 \ imes (\ ext{Base Area} \ imes \ ext{Height})\n]", "This formula gives the new volume in just one step—perfect for quick geometry problems."]
