Square root spiral

Drawing a (1) unit perpendicular on (\sqrt{1799}) forms (\sqrt{1800}). The previous hypotenuse has one less number.

Since \(24^2=576\) and \(25^2=625\), and \(576<624<625\), we get \(24<\sqrt{624}<25\). On the other hand, \(729=27^2\), so \(\sqrt{729}=27\). Therefore, option A is correct. Option B is incorrect because 624 is one less than 625, so its square root cannot be 25; also, \(\sqrt{729}\) is a rational integer. Exam tip: Compare a number with consecutive perfect squares to locate its square root quickly.

The construction begins with a right triangle whose existing hypotenuse is \(\sqrt{16}\), and a new perpendicular side of length 1 is added. The square root spiral does not obtain the next length by simply adding 1 to the old length. Instead, it uses the Pythagorean theorem, because the old hypotenuse and the new perpendicular form the relevant sides of a right triangle.

Here, \((\sqrt{16})^2+1^2=16+1=17\). Therefore the new hypotenuse has length \(\sqrt{17}\). This proves that option B is correct. Option A is wrong because \(\sqrt{16}+1=5\), not \(\sqrt{17}\); option C confuses 17 with its square root, and option D subtracts instead of adding the squared perpendicular side.

In a square root spiral, successive hypotenuses are represented by \(\sqrt{n}\). Therefore, the hypotenuse after \(\sqrt{2115}\) is \(\sqrt{2116}\). Since \(2116=46^2\), we get \(\sqrt{2116}=46\), which is an integer value. \(\sqrt{2117}\) is the following hypotenuse, not the immediate next one. Exam tip: Identify nearby perfect squares to solve such questions quickly.

Since \(9^2=81<99<100=10^2\), we get \(9<\sqrt{99}<10\). Similarly, \(10^2=100<101<121=11^2\), so \(10<\sqrt{101}<11\). Therefore, on the square root spiral, \(\sqrt{99}\) is just before 10 and \(\sqrt{101}\) is just beyond 10. Option B is incorrect because \(\sqrt{101}>10\). Exam tip: Locate a square root by comparing the number with the nearest perfect squares.

In a square root spiral, the successive hypotenuses are \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Therefore, the hypotenuse immediately after \(\sqrt{3968}\) is \(\sqrt{3969}\). Since \(63^2=3969\), \(\sqrt{3969}=63\). \(\sqrt{3970}\) comes one step later, so it is not the required hypotenuse. Exam tip: Check whether the radicand is a perfect square before simplifying a square root.

Since \(54^2=2916\), \(55^2=3025\), and \(56^2=3136\), we have \(2916<3024<3025\). Hence, \(54<\sqrt{3024}<55\). Similarly, \(3025<3026<3136\), so \(55<\sqrt{3026}<56\). Option B is incorrect because only \(\sqrt{3025}=55\). Exam tip: locate a square root by comparing the number with nearby perfect squares.

In each new right triangle of a square root spiral, one leg is the previous hypotenuse and the other leg is \(1\). By Pythagoras’ theorem, the new hypotenuse is \(\sqrt{(\sqrt{132})^2+1^2}=\sqrt{132+1}=\sqrt{133}\). Option B incorrectly squares \(132\); it is the previous hypotenuse \(\sqrt{132}\) that must be squared. Exam tip: square the previous hypotenuse to get the radicand, then add \(1\).

In a square root spiral, successive hypotenuses are \(\sqrt{1},\sqrt{2},\sqrt{3}\), and so on. Therefore, the hypotenuse after \(\sqrt{4623}\) is \(\sqrt{4624}\). Since \(68^2=4624\), its exact value is \(68\). Option C is close, but \(4625\) is greater than \(68^2\), so its square root cannot be 68. Exam tip: Compare a number with nearby perfect squares to check whether its square root is an integer.

After (\sqrt{48}), (\sqrt{49}=7) is formed. Since (7^2<50<8^2), (\sqrt{50}) lies between (7) and (8).

(69^2=4761) and (70^2=4900). The number (4899) lies between them, so (\sqrt{4899}) lies between (69) and (70).

A unit perpendicular in the square root spiral creates a right triangle whose old hypotenuse is \(\sqrt{8099}\) and whose new perpendicular side is 1. By the Pythagorean theorem, the square of the new hypotenuse is obtained by adding the square of the old hypotenuse and the square of the new side.

Hence the new hypotenuse is \(\sqrt{(\sqrt{8099})^2+1^2}=\sqrt{8099+1}=\sqrt{8100}\). Since \(90^2=8100\), the exact value is 90. Thus option A is correct. Option B subtracts the added square, while option C doubles the original quantity. Option D has the wrong radicand: although 90 is written there, \(\sqrt{8101}\) is not equal to 90.

\(91^2=8281\) and \(92^2=8464\). Since \(8281<8463<8464\), taking square roots gives \(91<\sqrt{8463}<92\). It cannot be \(92\), because \(8463\) is 1 less than \(92^2=8464\). Exam tip: To locate a square root, compare the number with the nearest perfect squares on either side.

In a square root spiral, successive hypotenuses are \(\sqrt{1}, \sqrt{2}, \sqrt{3},\dots\). Since \(95^2=9025\), the new hypotenuse is \(\sqrt{9025}\). Therefore, the hypotenuse in the immediately previous step is \(\sqrt{9024}\). \(\sqrt{9025}\) represents the new hypotenuse itself, not the previous one. Exam tip: square the given perfect-square hypotenuse and subtract 1 to find the previous radicand.

Since \(99^2=9801\), \(9998\) is less than \(10000=100^2\), we have \(99^2<9998<100^2\). Hence, \(99<\sqrt{9998}<100\), while \(\sqrt{10000}=100\). Therefore, option A is correct. Option B may seem close, but \(9998\) is greater than \(99^2\), so its square root cannot be 99. Exam tip: Place a number between two consecutive perfect squares to locate its square root quickly.

The square root spiral changes the squared length by 1 whenever a perpendicular of length 1 is added. With an existing hypotenuse \(\sqrt{9603}\), the new right triangle has legs represented by \(\sqrt{9603}\) and 1. Pythagoras therefore requires addition of their squared lengths.

The result is \(\sqrt{(\sqrt{9603})^2+1^2}=\sqrt{9603+1}=\sqrt{9604}\). Since \(98^2=9604\), the exact new hypotenuse is 98. Therefore option A is correct. Subtracting 1 gives the wrong direction, doubling gives the wrong construction, and \(\sqrt{9605}\) cannot equal 98 because its radicand is not \(98^2\).

\(98^2=9604\) and \(99^2=9801\). Since \(9604<9800<9801\), taking square roots gives \(98<\sqrt{9800}<99\). Option C is incorrect because \(\sqrt{9800}\) is less than \(99\); in fact, \(99^2=9801\). Exam tip: To find the interval of a square root, compare the number with the nearest perfect squares on either side.

The square-root spiral is based on the Pythagorean theorem. When a perpendicular of length 1 is drawn to a hypotenuse √n, the new hypotenuse has square equal to n + 1: (√n)² + 1² = n + 1. Here n = 10403, so the new hypotenuse is √(10403 + 1) = √10404. The number 10404 is a perfect square because 102 × 102 = 10404. Therefore √10404 = 102 exactly. Option A subtracts one instead of adding one, option B doubles the radicand, and option D has the wrong radicand even though it incorrectly claims the same value. Thus option C is correct.

We have \(100^2=10000\) and \(101^2=10201\). Since \(10000<10199<10201\), taking positive square roots gives \(100<\sqrt{10199}<101\). Option C is incorrect because it would require \(10199\) to be greater than \(101^2=10201\). Exam tip: To locate a square root, compare the number with consecutive perfect squares.