Question - Solving a Right-Angled Triangle Question Using Algebra
mathquadratic equation solutionsolving right-angled triangleusing algebrapythagorean theorem equation
Solution:
The question in the image involves a right-angled triangle, and it gives you two expressions for the lengths of the sides: $$ x - 1 $$ and $$ 4x $$, and for the hypotenuse, the expression is $$ 5x - 9 $$.According to the Pythagorean theorem, in a right-angled triangle, the sum of the squares of the two shorter sides is equal to the square of the longest side (the hypotenuse). The equation is often expressed as $$ a^2 + b^2 = c^2 $$, where $$ a $$ and $$ b $$ are the lengths of the two shorter sides and $$ c $$ is the length of the hypotenuse.Using the expressions provided for the lengths, we can create the following equation:$$(x - 1)^2 + (4x)^2 = (5x - 9)^2$$Now we'll expand and simplify this equation to solve for $$ x $$:$$x^2 - 2x + 1 + 16x^2 = 25x^2 - 90x + 81$$Combine like terms:$$17x^2 - 2x + 1 = 25x^2 - 90x + 81$$Bringing all terms to one side gives us:$$0 = 25x^2 - 17x^2 - 90x + 2x + 81 - 1$$Simplify and combine like terms:$$0 = 8x^2 - 88x + 80$$To make it easier to solve, divide the entire equation by 8:$$x^2 - 11x + 10 = 0$$Now we can factor this quadratic equation:$$(x - 10)(x - 1) = 0$$Setting each factor equal to zero gives us the possible values of $$ x $$:$$x - 10 = 0 \quad \Rightarrow \quad x = 10$$$$x - 1 = 0 \quad \Rightarrow \quad x = 1$$Since $$ x = 1 $$ would result in one of the sides being zero length (since $$ x - 1 = 0 $$), which is not possible for the side of a triangle, the only valid solution for $$ x $$ is:$$x = 10$$
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