Square root spiral

\(\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}\). Since \(\sqrt{2}\) is irrational, \(5\sqrt{2}\) is also irrational. In contrast, \(\sqrt{25}=5\), \(\sqrt{36}=6\), and \(\sqrt{49}=7\) are rational because 25, 36, and 49 are perfect squares. Exam tip: Check whether the number inside the square root is a perfect square; if it is not, its square root is generally irrational.

This relation matches Pythagoras theorem, so (OP) and (PQ) are perpendicular. This right-angle construction is essential in the square root spiral.

Taking (1=\sqrt{1}) makes the whole sequence clear. Then the inner number increases by (1) at each step.

The squares of the hypotenuses are (2,3,4), which increase by (1). The actual lengths do not have equal differences.

The exact value is (\sqrt{5}), while (2.236) is only an approximation. In exams, prefer the exact square root form.

(\sqrt{6}) is formed from the previous hypotenuse (\sqrt{5}) and the new perpendicular side (1). Therefore, ((\sqrt{5})^2+1^2=6) is correct.

(\sqrt{2}) is given by the first triangle, so up to (\sqrt{n}), the number of triangles is (n-1). Remember the starting term while writing a general formula.

The first hypotenuse is formed from (1) and (1), giving (\sqrt{2}). (\sqrt{4}) appears in a later step, not as the first hypotenuse.

Just before the final (\sqrt{32}), (\sqrt{31}) is formed. The spiral does not skip any number in order.

The square root spiral represents square roots like (\sqrt{2},\sqrt{3},\sqrt{5}) geometrically. It helps in understanding the number line and irrational numbers.

The square of the new hypotenuse is (35+1=36), so (OQ=\sqrt{36}). When the hypotenuse is asked, write the square root form.

In the previous step, the number under the root is (1) less. Therefore, (\sqrt{41}) comes before (\sqrt{42}).

The hypotenuse of the (r)th triangle is (\sqrt{r+1}). Therefore, the (18)th triangle has hypotenuse (\sqrt{19}).

(\sqrt{2}) is given by the first triangle, so up to (\sqrt{48}), the number of triangles is (48-1=47). Keep the initial (\sqrt{1}) separate in counting.

In consecutive hypotenuses, the numbers under the roots increase by (1). Therefore, (\sqrt{22},\sqrt{23},\sqrt{24}) is the correct order.

In a square root spiral, a unit side is drawn perpendicular to the preceding hypotenuse. By Pythagoras, \((\sqrt{n-1})^2+1^2=n\), so the new hypotenuse is \(\sqrt{n}\). Exam tip: identify both the unit side and the right angle.

((\sqrt{30})^2+1^2=31), so the perpendicular is drawn on (\sqrt{30}). For the next square root, identify the previous inner value.

((\sqrt{11})^2-(\sqrt{10})^2=1), but (\sqrt{11}-\sqrt{10}\neq1). In exams, understand the difference between length and square.

(\sqrt{72}=\sqrt{36\times2}=6\sqrt{2}). Taking out the perfect square factor is a good way to simplify.

(\sqrt{50}=5\sqrt{2}), and (\sqrt{2}) is irrational. Therefore, (\sqrt{50}) is an irrational number.