3Question: In a right triangle, the inradius is $ r $ and one leg is $ a $. Express the hypotenuse $ c $ in terms of $ r $ and $ a $.
["Title: Expressing Hypotenuse $ c $ in a Right Triangle Given Inradius $ r $ and Leg $ a $", "---", "Introduction\nIn right triangles, understanding key measurements like leg lengths, hypotenuse, and inradius provides deeper insight into triangle geometry. One insightful relation involves expressing the hypotenuse $ c $ in terms of the inradius $ r $ and one leg $ a $. This article explains how to derive a precise formula connecting these values using fundamental triangle properties.", "---", "Understanding the Right Triangle and Inradius Formula\nConsider a right triangle with legs $ a $, $ b $, and hypotenuse $ c $, where $ a^2 + b^2 = c^2 $. The inradius $ r $ of a right triangle is known from geometric formulas:", "$$\nr = \frac{a + b - c}{2}\n$$", "This formula arises from the area-based derivation and the tangential circle property inside the triangle.", "Given that one leg is $ a $, and $ r $ and $ c $ are known in terms of inputs, we aim to express $ c $ explicitly.", "---", "Step-by-Step Derivation\nStart with the inradius formula:", "$$\nr = \frac{a + b - c}{2}\n$$", "Multiply both sides by 2:", "$$\n2r = a + b - c \quad \Rightarrow \quad c = a + b - 2r \quad \ ext{(Equation 1)}\n$$", "Also, from the Pythagorean theorem:", "$$\na^2 + b^2 = c^2 \quad \ ext{(Equation 2)}\n$$", "Now substitute Equation 1 into Equation 2. First, solve Equation 1 for $ b $:", "$$\nb = c - a + 2r\n$$", "Substitute $ b $ into Pythagoras’ identity:", "$$\na^2 + (c - a + 2r)^2 = c^2\n$$", "Expand the squared term:", "$$\na^2 + \left[(c - a) + 2r\right]^2 = c^2\n$$", "$$\na^2 + (c - a)^2 + 4r(c - a) + 4r^2 = c^2\n$$", "Expand $ (c - a)^2 = c^2 - 2ac + a^2 $:", "$$\na^2 + c^2 - 2ac + a^2 + 4r(c - a) + 4r^2 = c^2\n$$", "Combine like terms:", "$$\n2a^2 - 2ac + c^2 + 4r(c - a) + 4r^2 = c^2\n$$", "Cancel $ c^2 $ from both sides:", "$$\n2a^2 - 2ac + 4r(c - a) + 4r^2 = 0\n$$", "Distribute $ 4r $:", "$$\n2a^2 - 2ac + 4rc - 4ra + 4r^2 = 0\n$$", "Group terms with $ c $:", "$$\n(-2a + 4r)c + (2a^2 - 4ra + 4r^2) = 0\n$$", "Solve for $ c $:", "$$\nc = \frac{2a^2 - 4ra + 4r^2}{2a - 4r}\n$$", "Factor numerator and simplify:", "$$\nc = \frac{2(a^2 - 2ra + 2r^2)}{2(a - 2r)} = \frac{a^2 - 2ra + 2r^2}{a - 2r}\n$$", "---", "Final Expression\nThus, the hypotenuse $ c $ is expressed in terms of the inradius $ r $ and leg $ a $ as:", "$$\nc = \frac{a^2 - 2ar + 2r^2}{a - 2r}\n$$", "---", "Conclusion\nThis formula elegantly links the hypotenuse of a right triangle with inradius $ r $ and one leg $ a $, enabling precise geometric calculations. Whether solving for unknowns in geometry problems or exploring triangle properties, this relation serves as a practical algebraic tool grounded in geometric truth.", "Additionally, knowing $ c $ supports computation of the area, perimeter, and other characteristics of right triangles, making this expression valuable in mathematics, engineering, and physics applications involving triangular configurations.", "---", "Keywords:\ninradius in right triangle, hypotenuse formula, right triangle geometry, expresar hipotenusa r a, 3Question triangle, right triangle formulas, inradius and leg expression, solve triangle with r and a", "---", "Meta Description:\nLearn how to express the hypotenuse $ c $ of a right triangle in terms of inradius $ r $ and leg $ a $ using fundamental geometry. Formula derived algebraically with full steps. Ideal for math students and educators."]
