Telangana TET Paper 1 Exam Model Paper 3

Concept:Simplify the square root of a decimal by converting it to a fraction or using perfect squares. Then isolate the square root term and solve for $x$ by squaring both sides.Explanation:First, find $\sqrt{14.44}$. Since $38^2 = 1444$, we have $\sqrt{14.44} = 3.8$.Substitute into the equation: $3.8 + \sqrt{9+x^2} = 8.8$.Subtract $3.8$ from both sides: $\sqrt{9+x^2} = 5$.Square both sides: $9 + x^2 = 25$.Subtract 9: $x^2 = 16$.Take the positive square root: $x = 4$.This matches the standard (3,4,5) Pythagorean triplet. Answer:The value of $x$ is 4, which corresponds to option B.