history of mathematics - National STEM Centre
history of mathematics - National STEM Centre
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