In a right triangle, one leg is 6 cm longer than the other, and the hypotenuse is 15 cm. What is the length of the shorter leg?

["# Finding the Shorter Leg in a Right Triangle: Side Lengths and Experimental Setup", "If you’ve ever worked with right triangles, you know how powerful algebra and the Pythagorean theorem can be—especially when one leg is longer than the other by a fixed amount, and the hypotenuse is known. In this article, we’ll solve a classic geometry problem: In a right triangle, one leg is 6 cm longer than the other, and the hypotenuse is 15 cm. What is the length of the shorter leg?", "We’ll also explore how this problem illustrates systematic mathematical reasoning—ideal for students, educators, and anyone interested in geometry and algebra.", "---", "## Understanding the Problem", "We’re given:", "- A right triangle (angle C is 90°)\n- One leg is 6 cm longer than the other\n- Hypotenuse = 15 cm\n- Goal: Find the length of the shorter leg", "Let’s define variables:", "Let ( x ) = length of the shorter leg (in cm)\nThen the longer leg = ( x + 6 ) cm\nHypotenuse = 15 cm", "By the Pythagorean theorem:", "[\n\ ext{(shorter leg)}^2 + \ ext{(longer leg)}^2 = \ ext{(hypotenuse)}^2\n]", "Substitute:", "[\nx^2 + (x + 6)^2 = 15^2\n]", "---", "## Solving the Equation Step-by-Step", "Start by expanding and simplifying:", "[\nx^2 + (x^2 + 12x + 36) = 225\n]", "[\n2x^2 + 12x + 36 = 225\n]", "Subtract 225 from both sides:", "[\n2x^2 + 12x + 36 - 225 = 0\n]", "[\n2x^2 + 12x - 189 = 0\n]", "Divide entire equation by 3 to simplify:", "[\n\frac{2}{3}x^2 + 4x - 63 = 0 \quad \ ext{(not necessary to divide, simplify directly below)}\n]", "Actually, divide by 1 (easier to work with original):", "[\n2x^2 + 12x - 189 = 0\n]", "Use the quadratic formula:", "[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "Where ( a = 2 ), ( b = 12 ), ( c = -189 )", "[\nx = \frac{-12 \pm \sqrt{12^2 - 4(2)(-189)}}{2(2)}\n]", "[\nx = \frac{-12 \pm \sqrt{144 + 1512}}{4}\n]", "[\nx = \frac{-12 \pm \sqrt{1656}}{4}\n]", "Now simplify ( \sqrt{1656} ):", "Factor 1656:\n( 1656 = 4 \ imes 414 = 4 \ imes 6 \ imes 69 = 4 \ imes 6 \ imes 3 \ imes 23 = 2^2 \ imes 2 \ imes 3 \ imes 3 \ imes 23 )\nSo, ( \sqrt{1656} = 2\sqrt{414} ), but better to calculate decimal or simplify:", "Wait — double-check:", "( 1656 = 4 \ imes 414 ), and ( 414 = 9 \ imes 46 = 9 \ imes 2 \ imes 23 ), so\n( \sqrt{1656} = \sqrt{4 \cdot 9 \cdot 2 \cdot 23} = 2 \cdot 3 \cdot \sqrt{46} = 6\sqrt{46} )", "Thus:", "[\nx = \frac{-12 \pm 6\sqrt{46}}{4} = \frac{-6 \pm 3\sqrt{46}}{2}\n]", "Only the positive root makes sense (length can’t be negative):", "[\nx = \frac{-6 + 3\sqrt{46}}{2}\n]", "But this is exact—but can we find a whole number?", "Let’s test integer values near expected range.", "---", "## Estimating and Testing Integer Solutions", "Since hypotenuse = 15, legs must be less than 15. Suppose shorter leg = ( x ), longer = ( x + 6 )", "Try ( x = 9 ): longer leg = 15 → check:\n( 9^2 + 15^2 = 81 + 225 = 306 <br/>\neq 225 ) → too big", "Try ( x = 8 ): longer = 14\n( 8^2 + 14^2 = 64 + 196 = 260 ) → too big", "Try ( x = 7 ): longer = 13\n( 7^2 + 13^2 = 49 + 169 = 218 ) → still > 225? No, 218 < 225", "Wait — 218 < 225 → too small", "Try ( x = 7.5 ): longer = 13.5\n( 7.5^2 = 56.25 ), ( 13.5^2 = 182.25 ), sum = 238.5 → still too big", "Wait — we need sum = 225", "Try ( x = 6 ): longer = 12\n( 6^2 + 12^2 = 36 + 144 = 180 ) → too small", "Try ( x = 7 ): 49 + 169 = 218", "Try ( x = 7.2 ): longer = 13.2\n( 7.2^2 = 51.84 ), ( 13.2^2 = 174.24 ), sum = 226.08", "Close — slightly over", "Try ( x = 7.1 ): 50.41 + 177.61 = 228.02 → too high", "Wait — earlier at ( x=7 ): 218", "At ( x=7.3 ): longer = 13.3 → ( 7.3^2 = 53.29 ), ( 13.3^2 = 176.89 ), sum = 230.18", "Wait — inconsistency.", "Actually: ( 7^2 + 13^2 = 49 + 169 = 218 )\n( 8^2 + 14^2 = 64 + 196 = 260 )\nWe need sum = 225", "So try ( x = 7.6 ): longer = 13.6\n( 7.6^2 = 57.76 ), ( 13.6^2 = 184.96 ), sum = 242.72 → no", "Wait — something’s off. Let’s use precise algebra.", "From earlier:", "[\nx^2 + (x + 6)^2 = 225\n]\n[\nx^2 + x^2 + 12x + 36 = 225\n]\n[\n2x^2 + 12x - 189 = 0\n]", "Divide by 3:", "[\n\frac{2}{3}x^2 + 4x - 63 = 0 \quad \ ext{(messy)}\n]", "Better: use", "[\n2x^2 + 12x - 189 = 0\n]", "Use quadratic formula:", "[\nx = \frac{-12 \pm \sqrt{144 + 4 \cdot 2 \cdot 189}}{4} = \frac{-12 \pm \sqrt{144 + 1512}}{4} = \frac{-12 \pm \sqrt{1656}}{4}\n]", "Now ( \sqrt{1656} \approx ? )", "( 40^2 = 1600 ), ( 41^2 = 1681 ), so ( \sqrt{1656} \approx 40.7 )", "Then:", "[\nx = \frac{-12 + 40.7}{4} = \frac{28.7}{4} \approx 7.175\n]", "So shorter leg ≈ 7.18 cm", "But is this exact?", "Try factoring original equation:", "( 2x^2 + 12x - 189 = 0 )", "Check discriminant: ( D = 144 + 1512 = 1656 )", "Is 1656 a perfect square? No.", "But wait — perhaps we made a setup error?", "Wait — hypotenuse = 15, legs: ( x ), ( x+6 )", "Try ( x = 9 ): 9 and 15 → 9² + 15² = 81 + 225 = 306 → no\n( x = 6 ): 6 and 12 → 36 + 144 = 180\n( x = 7 ): 49 + 169 = 218\n( x = 7.2 ): 51.84 + 182.4 = 234.24\nWait — still increasing", "Wait — 218 at x=7 — we need 225", "Try ( x = 7.5 ): 56.25 + 182.25 = 238.5 — too big", "Wait — perhaps no integer solution?", "But maybe it's designed for exact solution.", "Wait — try solving:", "Back to:", "[\nx^2 + (x+6)^2 = 225\n]\n[\n2x^2 + 12x + 36 = 225\n]\n[\n2x^2 + 12x - 189 = 0\n]", "Divide by GCD (3):", "[\n\frac{2}{3}x^2 + 4x - 63 = 0\n]", "Multiply all by 3:", "[\n2x^2 + 12x - 189 = 0\n]", "Now factor:", "Try: ( (2x - a)(x + b) = 0 )", "Expand: ( 2x^2 + 2b x - a x - ab = 2x^2 + (2b - a)x - ab )", "Set:", "( 2b - a = 12 )\n( ab = 189 )", "Try ( a = 21 ), then ( 2b = 33 ) → ( b = 16.5 ), not integer", "Try ( a = 27 ), then ( 2b = 39 ), ( b = 19.5 )", "Try ( a = 9 ), ( 2b = 21 ), ( b = 10.5 )", "Not working.", "But note:", "From earlier:", "[\nx = \frac{-12 + \sqrt{1656}}{4}\n]", "But ( \sqrt{1656} = \sqrt{4 \cdot 414} = 2\sqrt{414} ), and ( \sqrt{414} = \sqrt{9 \cdot 46} = 3\sqrt{46} ), so:", "[\n\sqrt{1656} = 2 \cdot 3 \cdot \sqrt{46} = 6\sqrt{46}\n]", "So:", "[\nx = \frac{-12 + 6\sqrt{46}}{4} = \frac{6(\sqrt{46} - 2)}{4} = \frac{3(\sqrt{46} - 2)}{2}\n]", "Thus, exact length of shorter leg is ( \frac{3(\sqrt{46} - 2)}{2} ) cm.", "Approximate: ( \sqrt{46} \approx 6.782 ), so:", "[\nx \approx \frac{3(6.782 - 2)}{2} = \frac{3(4.782)}{2} = \frac{14.346}{2} = 7.173 , \ ext{cm}\n]", "So about 7.17 cm — not an integer, but perfectly valid.", "---", "## Why This Problem Matters", "This problem is more than just calculating side lengths — it’s about:", "- Applying the Pythagorean theorem in real-world contexts\n- Setting up equations from word descriptions\n- Solving quadratic equations arising from geometry\n- Understanding when exact solutions involve radicals", "Whether you’re a student learning math or a teacher designing lessons, this type of problem reinforces logical reasoning, algebraic manipulation, and spatial understanding.", "---", "## Final Answer", "After careful setup and calculation, the length of the shorter leg is:", "[\n\boxed{\frac{3(\sqrt{46} - 2)}{2} \ ext{ cm}} \quad \ ext{or approximately } \boxed{7.17 \ ext{ cm}}\n]", "For exact form, keep:\n[\nx = \frac{-12 + 6\sqrt{46}}{4} = \frac{3\sqrt{46} - 6}{2} = \frac{3}{2}(\sqrt{46} - 2)\n]", "---", "## Further Learning Tips", "- Always verify solutions by plugging back into ( a^2 + b^2 = c^2 )\n- Practice factoring or using quadratic formula with different signs\n- Use graphing tools to visualize right triangles and derive relationships", "If you're teaching or studying, try recreating this problem with different differences (e.g., 4 cm instead of 6 cm) and observe how solutions change.", "---", "Keywords for SEO:\nright triangle leg difference 6 cm, hypotenuse 15 cm, solve right triangle algebra, find shorter leg given leg difference, Pythagorean theorem application, quadratic equation right triangle, geometry problem solving, solve for x in triangle sides, exact and approximate leg length", "---", "Explore more geometry insights at your next math step."]