[Enrich20170101SQT] Calculating Square Roots by Hand

Introduction

          Happy New Year to our readers!  I wish this year will be a fruitful one for everybody.

          Today, I will illustrate how to calculate square roots by hand, using  54 756  as an example.  It is similar to long division, but has some modifications.

Solution

          Starting from the right, pair up the digits.

          2×2 = 4  is the nearest perfect square to  5.  Subtract and bring down the next two digits, giving  147.

          Double the digit  2  to get  4.  Think:  ? × 4?  gives  147  or nearest possible value.  We have 3×43 = 129.

          Subtracting and bringing down the next two digits gives  1856.  Replicate the digit  4  on the left and double the digit  3,  giving  46.

          Now think:  ? × 46?  gives  1856  or nearest possible value.  It turns out that  4 × 464 gives exactly  1856.  We are done!  The square root of  54 756  is  234.

How does it work?

          This relies on the algebraic identity  (10a + b)² = 100a² + 20ab + b², the right-hand expression is equal to   100a² + (20a + b)b.  For example, at stage 4, we have  a = 23,  b = 4  and  (20a + b) = 464.

          Did you learn something today?

[Enrich20161103NRP] The Napkin Ring Problem

Problem

     Two rings are made by drilling a cylindrical hole through a small sphere and a hole through the large sphere, such that the resulting rings have the same height (2h).

     Which ring has the larger volume of remaining material?

Solution

     The answer is: both rings have the same volume.  How can we know?

     There is a way to show this using integration.  But calculus is not necessary.

     Let  r  be the radius of any chosen sphere and let  a  be the radius of the cylindrical hole.  By Pythagoras’ Theorem,  h² = r² – a².  Consider a cross-section of the ring sliced a distance  x  from the centre of the sphere, perpendicular to the axis of the cylindrical hole.  The outer radius of this cross section is the square root of  r² – x².  Hence the area of the material in the cross-section is

               p [(r² – x²) – a²]  =  p (r² – a² – x²)  =  p (h² – x²)

Note that  r  does not appear in the formula.  That means the cross-section does not depend on  r.  A bigger (or smaller) sphere would have the same cross-sectional area for each distance  x  away from the centre.  By Cavalieri'sPrinciple, the other sphere will have the same volume!

     By the way what is this volume?  It is the same as that of a sphere without hole  i.e.  where  a = 0  and  r = h.  This works out to be  ⁴/₃ p h³,  where  h  is half the height of the ring.

Heuristics Used

H02. Use a diagram / model

H09. Restate the problem in another way

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘A’ Levels H2 Mathematics (Number patterns, with algebra)

* revision for IB Mathematics HL & SL

* Advanced Placement (AP) Calculus AB & BC

* University / College Calculus

* other syllabuses that involve volumes and Pythagoras’ Theorem

* any learner who is interested