history of mathematics - National STEM Centre

14 Answers
3 If AD intersects the line I in E, the circle with centre
A and radius AE is the circle required.
4 You are not allowed to draw this circle directly
because, although you know its centre, you don't know
its radius. Constructing the perpendicular to I through A
enables you to find its radius.
Activity 3.2, page 31
1 a Draw two equal circles with centres A and B,
intersecting at C and D. Then CD is the required
perpendicular bisector.
b Both constructions rely on the geometric properties of
a rhombus for their justification.
2 Draw a circle, centre A, to cut the lines at B and C.
Then, with the same radius and centres B, C, draw two
more arcs, cutting at D. Then AD bisects the angle at A.
3 The geometric properties of a rhombus explain all
these constructions.
4 These constructions are left to you.
Activity 3.3, page 32
1 a A line through D is drawn parallel to CE. BC is
produced to meet this line at F.
b Both regions consist of the quadrilateral ABCE and a
triangle, in the first case CDE, and in the second CFE.
These triangles have the same base, CE, and the same
height, the distance between the parallel lines.
c Bisect AG so GH = \ GA.
d It has the same height, and half the base.
If a : x = x '. b , then — = —, so x 2 = ab.
x b
RD CD
Triangle ABD is similar to BCD, so —— = ——.
5 DA DB
g Draw an arc of a circle, with centre H and radius HI.
Extend GH to intersect this circle outside GH at N. Then
GN is a line segment divided at H into lengths a and b.
Find the mid-point of GN, and construct a semi-circle
with this point as centre. To find the mean proportional,
construct a perpendicular to GN at H. The point of
intersection of the perpendicular with the semi-circle is K
and HK is the mean proportional of lengths a and b,
since triangle GKN is right-angled at K. Now construct
the required square on HK.
Activity 3.4, page 34
1 Antiphon's solution was to find a series of regular
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polygons whose areas get progressively closer to the area
of the circle.
Bryson's solution was to find the average of the inside
and outside squares shown in Figure 3.10a.
2 Aristotle believes that the methods of Antiphon and
Bryson are not scientific enough. He believes that
Antiphon's method assumes that at some point the last
polygon will coincide with the circle. Bryson's method
uses the false assumption that two things which satisfy
the same inequalities must be equal.
3 In neither case is a proper construction using Euclid's
rules given. In Bryson's case, the areas are not proved
equal. In Antiphon's case, the algorithm is not finite.
Activity 3.5, page 36
2 If AB is the given straight line, set the compasses to
have radius AB. Then, with centres A and B, draw arcs of
circles intersecting at C. Then triangle ABC is
equilateral.
3 Proposition 1.9 is the same as Activity 3.2, question
2. Proposition 1.4 also relies on the properties of a
rhombus, namely that the diagonals bisect each other.
4 Q. E. F. stands for Quod Erat Faceat. It means 'Which
was to be made'.
Activity 3.6, page 37
1 It is Pythagoras's theorem.
2
3 The proof is left to you.