Square root spiral

A square root spiral uses a right triangle at every stage, adding a new perpendicular side of length 1 unit to the existing hypotenuse. If the existing hypotenuse is √6 and the new hypotenuse is h, the Pythagorean theorem gives h²=(√6)²+1²=6+1=7. Since h is a positive length, h=√7. Therefore option C is correct. √5 is the preceding member of the sequence, while √6 is the old hypotenuse already present, not the newly obtained one. √12 is incorrect because the added side contributes 1², not another 6 or a multiplication by 2. The construction therefore increases the squared hypotenuse from 6 to 7 and moves from √6 to √7.

Each new right triangle in a square root spiral adds a unit side. If the previous hypotenuse is \(\sqrt{4}\), Pythagoras gives the next hypotenuse as \(\sqrt{4+1}=\sqrt{5}\). \(\pi\) is not obtained in this sequence. Exam tip: remember the hypotenuse pattern \(\sqrt{2},\sqrt{3},\sqrt{4}\dots\).

In a square root spiral, a perpendicular side of length 1 unit is added to the previous hypotenuse. Thus, the new hypotenuses represent \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Exam tip: the added side is always 1 unit.

The square root spiral is generated step by step by adding a perpendicular segment of length 1 unit to the current hypotenuse. If the current hypotenuse is √14, then the next hypotenuse h satisfies h²=(√14)²+1²=14+1=15, so h=√15. Therefore the hypotenuse present immediately before constructing √15 is √14, making option B correct. √13 is one stage earlier in the sequence. √15 is the target that is about to be constructed, not the value already present. √16 is the following stage after √15. The essential relationship is that the squared value increases by exactly one at each construction step, so the predecessor of 15 must be 14.

In a square root spiral, each new right triangle uses the previous hypotenuse and a new perpendicular side of length 1. Hence the hypotenuses increase, not remain equal. Exam tip: remember \(h_{\text{new}}^2=h_{\text{old}}^2+1\).

The hypotenuse length of the required square root is taken in the compass. So the hypotenuse of (\sqrt{3}) is needed.

In a square root spiral, a unit perpendicular is added to the previous hypotenuse. By Pythagoras, the new hypotenuse has square

a^2+1

; hence successive roots are formed. A parallel segment would not ensure a right triangle. Exam tip: look for “unit perpendicular.”

While constructing a square root spiral, each new triangle is formed by drawing a right angle on the previous hypotenuse. A set square helps construct this right angle accurately, so option A is correct. The length of a hypotenuse is generally measured with a ruler; the main use of a set square is constructing angles. Exam tip: at each step of a square root spiral, draw a new unit-length side perpendicular to the previous hypotenuse.

At each step, a unit segment is drawn perpendicular to the previous hypotenuse at its endpoint. If the previous hypotenuse is \(\sqrt{n}\), Pythagoras gives the next one as \(\sqrt{n+1}\). In exams, check the right-angle mark.

The direct answer is B, (\sqrt{79}). In each step of the square root spiral, a 1-unit segment is drawn perpendicular to the previous hypotenuse. If that previous hypotenuse is (\sqrt{n}), then the new hypotenuse is (\sqrt{n+1}) because (\sqrt{n})^2+1^2=n+1. For the target (\sqrt{80}), solve (n+1=80); hence (n=79) and the previous hypotenuse is (\sqrt{79}). Option A, (\sqrt{78}), would give (\sqrt{79}) next, so it is one step short. Option B is correct because adding the unit perpendicular gives exactly (\sqrt{80}). Option C, (\sqrt{80}), is the required new hypotenuse, not the previous one. Option D, (\sqrt{81}), would produce (\sqrt{82}), one step beyond the target. The safe exam method is to subtract 1 from the number inside the target square root: (80-1=79).

By Pythagoras’ theorem, new hypotenuse² = \((\sqrt{10})^2+1^2=10+1=11\). Hence the new hypotenuse is \(\sqrt{11}\). \(\sqrt{20}\) results from squaring the previous hypotenuse incorrectly. Exam tip: square the radical first.

The direct answer is A: draw a 1-unit perpendicular at the end of (\sqrt{2}). The construction works by making a right triangle at every stage. Start with a unit-based triangle to obtain hypotenuse (\sqrt{2}). Next, keep (\sqrt{2}) as one side and draw a perpendicular side of 1 unit. Pythagoras gives the new hypotenuse (\sqrt{(\sqrt{2})^2+1^2}=\sqrt{3}). Option A is correct because this is exactly the next construction needed. Option B, erasing the segment, destroys the construction and cannot create the next square root. Option C, making a denominator zero, is unrelated to a geometric spiral and division by zero is not permitted. Option D, converting the triangle into a square, is not the method; the next value is created by a new right triangle, not by making a square. The word “perpendicular” means meeting at a right angle of 90 degrees. Memory cue: at every new endpoint, add one unit at a right angle; the hypotenuse then moves from (\sqrt{n}) to (\sqrt{n+1}).

Each added triangle must be right-angled. Drawing a unit side perpendicular to the previous hypotenuse gives the next hypotenuse using Pythagoras’ theorem. A parallel side does not form a right angle. Exam tip: mark the new unit side perpendicular first.

The first right triangle in a square root spiral has perpendicular sides of 1 unit and 1 unit. By Pythagoras, hypotenuse² = 1² + 1² = 2, so the hypotenuse is \(\sqrt{2}\). A square of side 2 has diagonal \(2\sqrt{2}\), not \(\sqrt{2}\). Exam tip: identify the triangle’s legs first.

In each new triangle, the previous hypotenuse is one leg and a unit segment is drawn perpendicular to it. By Pythagoras, the square of the new hypotenuse increases by 1. Exam tip: each added segment is perpendicular to the previous hypotenuse.

Since 7 is not a perfect square, \(\sqrt{7}\) is irrational and is shown as a distinct length on the spiral. In contrast, \(\sqrt{9}=3\), \(\sqrt{16}=4\), and \(\sqrt{25}=5\) are rational. Exam tip: square roots of perfect squares are integers.

The first hypotenuse is \(\sqrt{2}\). Adding a new 1-unit side gives hypotenuse squared as \(2+1=3\), so the next hypotenuse is \(\sqrt{3}\). Exam tip: each added triangle is right-angled.

By Pythagoras’ theorem, hypotenuse² = 1² + 1² = 2, so the hypotenuse is \(\sqrt{2}\) units. Option 2 is only the sum of the sides, not the hypotenuse. Exam tip: add squares, then take the square root.

In a square root spiral, each new right triangle uses the previous hypotenuse and a new side of 1 unit. Since \(\sqrt{16}^2+1^2=16+1=17\), the new hypotenuse is \(\sqrt{17}\). A side of 4 gives \(16+16=32\), not 17. Exam tip: check the sum of squares using Pythagoras' theorem.

Adding a (1) unit perpendicular to (\sqrt{56}) forms (\sqrt{57}). The previous hypotenuse number is always (1) less.

The square-root spiral advances by one under the radical at every stage. The construction uses the current hypotenuse and a new perpendicular segment of length 1. If the current hypotenuse is
sqrt{n}
, then the next hypotenuse has square length
n+1
, so its length is
sqrt{n+1}
. This is a direct application of the Pythagorean theorem.

Here the current value is
sqrt{43}
. Adding the square of the new unit side gives
(
sqrt{43}
)^2+1^2=43+1=44
. Taking the positive square root, as lengths are positive, gives
sqrt{44}
. Therefore option C is correct.
sqrt{42}
is the preceding stage, while
sqrt{43}
is the existing hypotenuse, not the new one. There is no doubling in this process.

A ruler is useful for taking the (1) unit length correctly. Wrong measurement can make the spiral lengths incorrect.

A square-root spiral is constructed through successive right triangles. A set square is a suitable geometrical instrument for drawing or checking a 90° angle, which is essential before applying the Pythagorean theorem. Thus option A correctly matches the tool with its use. A calculator performs numerical operations but cannot physically draw an arc. A balance measures mass, not a line segment or hypotenuse, so option C is inappropriate. A clock measures time and has no role in measuring a 1-unit length; a ruler or compass would be relevant for that purpose. The correct match is therefore the set square and right-angle construction.

Answer: option A, √1 → √2 → √3 → √4. The spiral begins with a unit length, represented as √1. Add a perpendicular unit segment: the new hypotenuse has square 1 + 1 = 2, so it is √2. Repeat the same construction: (√2)² + 1² = 2 + 1 = 3, giving √3. One more step gives (√3)² + 1² = 3 + 1 = 4, giving √4. Therefore the complete order is √1, √2, √3, √4. Option B skips the necessary √2 stage. Option C places the stages in an incorrect order, and option D reverses the construction. Each new hypotenuse depends on the immediately previous one, so no stage may be skipped. Memory cue: write the radicands in consecutive order: 1, 2, 3, 4.

The governing rule is the one-unit progression of the standard square-root spiral. If the present hypotenuse is √n, adding a perpendicular of length 1 gives a new hypotenuse h with h²=n+1. Therefore, after √63, the next hypotenuse is √64. Because 64=8², its principal square root is exactly 8, a whole number. Hence option A correctly states both the radical and its value. Option B goes backward to √62 rather than forward. Option C identifies √64 but evaluates it incorrectly as 7; 7²=49, whereas 8²=64. Option D doubles the radicand without justification and does not follow the spiral rule. The fact that 64 is a perfect square explains why this next length is an integer.