A student says that \(\sqrt{13}\) cannot be represented on a square root spiral because 13 is not a perfect square. Which construction correctly disproves the student’s claim?
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Answer and explanation
Correct answer: The hypotenuse of a triangle with perpendicular sides \(\sqrt{12}\) and 1
In a square root spiral, drawing a perpendicular unit segment at \(\sqrt{12}\) gives \(\sqrt{(\sqrt{12})^2+1^2}=\sqrt{13}\). A number need not be a perfect square. In exams, apply Pythagoras’ theorem.
Frequently asked questions
What is the correct answer to this question?
The hypotenuse of a triangle with perpendicular sides \(\sqrt{12}\) and 1
Why is this the correct answer?
In a square root spiral, drawing a perpendicular unit segment at \(\sqrt{12}\) gives \(\sqrt{(\sqrt{12})^2+1^2}=\sqrt{13}\). A number need not be a perfect square. In exams, apply Pythagoras’ theorem.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Number Systems. Topic: Square root spiral.
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