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history of mathematics - National STEM Centre

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Practice exercises<br />

Chapter 2, page 151<br />

1 a 42.42638889, 1.414212963<br />

b V2<br />

c The answer needs to be a diagram like Figure 2.11.<br />

d Input N, A<br />

repeat<br />

display A<br />

until satisfied<br />

Output The square root <strong>of</strong> N.<br />

e 2, 2.75, 2.647727273, 2.645752048, 2.645751311,<br />

2.645751311<br />

2 Babylonians used the result in specific cases.<br />

Pythagoras's theorem involves pro<strong>of</strong>, and there is no<br />

evidence that the Babylonians proved the result.<br />

3 1 ; 0,45 and 0; 59, 15,33,20<br />

4 a How much grain is left over if as many men as<br />

possible are given 7 sila from a silo <strong>of</strong> grain.<br />

b 40,0 times 8,0 to give the number <strong>of</strong> sila in a silo.<br />

c To divide by 7, multiply by the reciprocal <strong>of</strong> 7.<br />

d By multiplying<br />

e The reciprocal <strong>of</strong> 7 has been taken as 0; 8,33 instead<br />

<strong>of</strong> 0; 8,34, 17,8<br />

Chapter 3, page 152<br />

I a Let AB intersect / at the point K. Draw a circle<br />

centre K, radius KA, to cut / at C, and a circle centre K,<br />

radius KB, to cut / at D. Then KA + KB = KC + KD so<br />

AB = CD.<br />

b Let AB produced intersect / at the point K. Draw<br />

circles as for part a. Then KA - KB = KC - KD so<br />

AB = CD.<br />

c Take any point C on / and draw CA. Construct an<br />

equilateral triangle on CA, with the third point called E.<br />

Draw a circle centre A, radius AB, to cut EA at the point<br />

14 Answers<br />

F. Draw a circle centre E, radius EF, to cut CE at the<br />

point G. Draw a circle centre C, radius CG, to cut / at the<br />

point D. Then CD = AB.<br />

2 Construct an equilateral triangle ABC on AB. Draw a<br />

circle, centre B, with radius BP, to cut BC at D. Draw a<br />

circle, centre C, with radius CD to cut AC at E. Draw a<br />

circle, centre A, with radius AE, to cut AB at Q. Then<br />

AQ = BP.<br />

3 First construct a rectangle twice the area by<br />

constructing a rectangle with a common base and the<br />

same height as the triangle. Then bisect the base <strong>of</strong> the<br />

rectangle to give a new rectangle with the correct area.<br />

Call this rectangle GHIJ, with HI as one <strong>of</strong> the shorter<br />

sides. Draw an arc <strong>of</strong> a circle, centre H, with radius HI.<br />

Extend GH to intersect this arc outside GH at N. Find the<br />

mid-point <strong>of</strong> GN and construct a semi-circle with this<br />

point as centre. Construct a perpendicular to GN at H.<br />

Call the point <strong>of</strong> intersection <strong>of</strong> the perpendicular with<br />

the semi-circle K and construct the required square on<br />

HK.<br />

Chapter 4, page 153<br />

1 a Diagram <strong>of</strong> the following form.<br />

b c d<br />

b a(b + c + d) = ab + ac + ad<br />

area <strong>of</strong> large rectangle = sum <strong>of</strong> areas <strong>of</strong> small rectangles.<br />

2 Using the algorithm on a graphics calculator, the<br />

greatest common divisor <strong>of</strong> 2689 and 4001 is 1.<br />

3 If a measures b then ka = b for some integer k. Hence<br />

k 2 a 2 = b , so a 2 measures b 2 .<br />

4 Archimedes: spirals, measurement <strong>of</strong> a circle, or<br />

approximating pi, quadrature <strong>of</strong> a parabola, trisecting<br />

angles, measurement <strong>of</strong> area and volume<br />

Chapter 5, page 154<br />

1 a x 2 + IQx = 39<br />

185

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