Square root spiral

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.

The next hypotenuse will be (\sqrt{64}), and (\sqrt{64}=8). At a perfect square, an exact whole number is obtained.

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 (1) unit perpendicular and right angle allow Pythagoras theorem to apply correctly. This forms successive square roots.

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.”

Since \((\sqrt{109})^2+1^2=110\). Therefore \(\sqrt{109}\) is the correct previous hypotenuse.

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.

A square root spiral is a stepwise construction based on Pythagoras theorem. Each new 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.