Square root spiral

By Pythagoras’ theorem, \(h_{new}^2=h_{old}^2+1^2\). Thus, the square of the new hypotenuse increases by 1, giving \(\sqrt2,\sqrt3,\sqrt4\) in order. Exam tip: write the squares first to avoid adding lengths directly.

In a square root spiral, the previous hypotenuse becomes one leg of the next right triangle, and a unit segment is drawn perpendicular to it. Hence the new hypotenuse represents the next square root. Exam tip: look for the perpendicular unit side.

In a square root spiral, the sum of squares is taken by Pythagoras theorem. Directly adding lengths is wrong.

In the square root spiral, each new right triangle has the previous hypotenuse as one side and a new perpendicular side of length 1. If the current hypotenuse is \(\sqrt{224}\), then the next one has square equal to \((\sqrt{224})^2+1^2=224+1=225\). Therefore the next hypotenuse is \(\sqrt{225}\). This follows directly from the Pythagorean theorem and the fixed unit segment used in the spiral.

Since 225 is a perfect square, \(\sqrt{225}=15\), because \(15^2=225\). Thus option A gives both the correct next hypotenuse and its exact value. Option D has the right radical but the wrong value, since 14 squared is 196, not 225. Options B and C do not follow the successive construction rule. The supplied answer is therefore correct.

Neither \(2\) nor \(8\) is a perfect square. The square root of a positive integer is rational only when the integer is a perfect square. Hence \(\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\) are both irrational. Option C is incorrect because \(\sqrt{2}\) is not a whole number. Exam tip: First check whether the number under the square root is a perfect square.

By Pythagoras' theorem, if the square of the previous hypotenuse is \(n\) and the new perpendicular side is 1, the new hypotenuse has square \(n+1\). Hence it represents \(\sqrt{n+1}\). Thales' theorem is not used here. Exam tip: identify the right angle first.

To obtain \(\sqrt{5}\) in a square root spiral, the previous hypotenuse \(\sqrt{4}\) is taken perpendicular to a unit side. By the Pythagorean theorem, the square of the new hypotenuse is \((\sqrt{4})^2+1^2=4+1=5\), so the hypotenuse is \(\sqrt{5}\). In option A, the hypotenuse would be \(\sqrt{3+4}=\sqrt{7}\), not \(\sqrt{5}\). Exam tip: At each new step, use the previous square root and \(1\) as perpendicular sides.

In a square root spiral, the previous hypotenuse \(\sqrt{7}\) is retained as one leg and the other leg is always 1 unit. Hence the new hypotenuse is \(\sqrt{7+1}=\sqrt{8}\). Taking \(\sqrt{7}\) as the other leg gives \(\sqrt{14}\), not \(\sqrt{8}\). Exam tip: always check the unit leg.

Each new triangle in a square root spiral is right-angled, so the new 1-unit side is drawn perpendicular to the previous hypotenuse. If the old hypotenuse is \(\sqrt n\), the new one becomes \(\sqrt{n+1}\). Exam tip: perpendicular always means \(90^\circ\).

The governing idea of a square root spiral is the repeated use of a right triangle. If the existing hypotenuse has length √n, a new perpendicular segment of length 1 is drawn at a right angle to it. By the Pythagorean theorem, the new hypotenuse has squared length (√n)² + 1² = n + 1, so its length is √(n + 1). Repeating this construction produces √2, √3, √4 and so on. Thus option A correctly identifies both essential conditions. Option B is false because the hypotenuse generally increases, option C is false because most roots remain irrational, and option D is false because a right triangle is not equilateral.

In a square root spiral, each new right triangle adds a side of length 1. By Pythagoras, the new squared distance is \((n-1)+1=n\), so the radius is \(\sqrt{n}\). Exam tip: identify numbers by their distance from the origin, not by the angle.

Direct addition is not used in a square root spiral. The new hypotenuse is obtained from a right triangle and Pythagoras theorem.

Adding a (1) unit perpendicular to (\sqrt{223}) forms (\sqrt{224}). The number increases by (1) in the next hypotenuse.

The hypotenuse length of the square root to be marked is taken in the compass. Drawing an arc from the origin is the correct way.

In the spiral, each new hypotenuse is formed using the previous hypotenuse and a perpendicular unit side. \((\sqrt{9})^2+1^2=10\), so it is \(\sqrt{10}\). \(\sqrt{10}\) instead gives \(\sqrt{11}\). Tip: track the hypotenuse.

The first right triangle has two perpendicular sides of 1 unit each. Hence its hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). \(\sqrt{3}\) occurs in the next triangle. Exam tip: remember that the spiral begins with \(\sqrt{2}\).

Since \(4^2=16\), \(5^2=25\), and \(16<24<25\), we get \(4<\sqrt{24}<5\). Also, \(25=5^2\) is a perfect square, so \(\sqrt{25}=5\). Option B is wrong because \(\sqrt{24}\) is not equal to \(5\). Exam tip: compare a number with nearby perfect squares to locate its square root.

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.

The previous hypotenuse becomes one leg, and a perpendicular unit leg is added at its endpoint. By Pythagoras, if its square is n, the new hypotenuse has square n+1. A parallel line will not form the required right triangle. Exam tip: remember “perpendicular + 1 unit.”

In the standard square root spiral, each new right triangle has a perpendicular of length 1 unit. If the previous hypotenuse is √k, the Pythagorean theorem gives the next hypotenuse as √(k + 1), because (√k)² + 1² = k + 1. To obtain √110, we therefore need k + 1 = 110, so k = 109. The preceding hypotenuse must be √109, followed by a 1-unit perpendicular at a right angle. Option B is correct. Option A would produce √109 rather than √110, while options C and D use the wrong starting value or move in the wrong numerical direction.

In a square root spiral, a 1-unit perpendicular is drawn at an endpoint of the previous hypotenuse to form the next right triangle. By Pythagoras, the new hypotenuse becomes \(\sqrt{n+1}\). A parallel line would not create a right angle. Exam tip: each new outer leg is 1 unit.

In a square root spiral, hypotenuses are formed successively. In the usual construction, intermediate steps are not skipped.

The direct answer is A. A square-root spiral is a chain of right triangles. Begin with a suitable right triangle, and at each stage draw a new perpendicular side of length 1 unit to the previous hypotenuse. By Pythagoras, if the old hypotenuse is sqrt{n}, the new one has length sqrt{n+1}, because (sqrt{n})^2+1^2=n+1. Thus option A gives the essential construction. Option B is wrong because the spiral represents successive square roots, including non-perfect-square roots, not merely a list of perfect squares. Option C is wrong because square roots are not added directly; the Pythagorean relation is used. Option D is wrong because circles are not the defining construction. Memory cue: one new unit perpendicular plus the old hypotenuse gives the next square root.

In a square root spiral, a unit perpendicular at the endpoint of \(\sqrt{7}\) gives a new hypotenuse of \(\sqrt{7+1}=\sqrt{8}\). A number need not be a perfect square. Exam tip: each new hypotenuse represents \(\sqrt{n+1}\).

The new hypotenuse is (\sqrt{323+1}=\sqrt{324}), and (\sqrt{324}=18). When a perfect square appears, write the exact whole number.