history of mathematics - National STEM Centre

3 a You get the answer 2 tou and 5 sheng.
b 1.32
Activity 5.8, page 65
1 x 2 +10;c = 39 so x 2 + 10;c + 25 = 64, leading to
(jc + 5) =64. Remembering that only positive numbers
were allowed at the time, .x + 5 = 8, so x = 3.
area =5*
area =5*
The total area is (x 2 + 10*) + 25 = 39 + 25 = 64 so the
side of the largest square is 8, leading to x = 3.
3 When you express the rule in symbols you obtain
x(n-x) + (^n-x) =(i«) . When you multiply out the
brackets, all the terms cancel.
4 No answer is given to this question.
5 Abu Kamil uses a rigorous Euclidean approach which
is difficult to follow geometrically. Al-Khwarizmi's
solution is clearer geometrically, and less rigorous.
Activity 5.9, page 68
20 -x
1 a By drawing the graphs of y = x 2 and y =
the point of intersection is at x = 2.59. x
20
b By drawing the graphs of y = x and y =
x-l
find x = 3.09.
you
Activity 5.10, page 69
1 The function takes its maximum value in the range
0 < x < a when x = ^a.
2 Al-Tusi may have known that the arithmetic mean
(AM) of positive numbers is always greater than or equal
to the geometric mean (GM), that is, that
i
+ a-, a n , \- VT ,
- > (a { a2 ...an j" . Now choose the three
numbers a - x , j x and \ x . Their AM is -y a , and their
the conclusion.
14 Answers
^
Then }a> (\x 2 (a- xty leads to
3 Look at the graphs of y = x 2 (a — x) and y = c and see
where they intersect.
Activity 6.2, page 74
1 Use compasses with a radius greater than ^ AB and
draw arcs with centres A and B, intersecting at points C
and D. Then join CD.
2 Suppose that X is any point on CD.
o
Then AX 2 = ACT + OX 2
= AC 2 -CO 2 +OX 2
D
= BC 2 -CO 2 +OX 2
= BO 2 +OX 2
= BX 2
Suppose now that X does not lie on CD and that the foot
of the perpendicular from X to AB is N so that AN # BN .
Then AX 2 = AN 2 + NX 2
*BN 2 +NX 2
The combination of these two arguments shows that only
points on CD satisfy the condition.
3 The answer is in the text following Activity 6.2.
4 The answer is in the text following Activity 6.2.
Activity 6.4, page 77
1 a = 2
2 a = 3. 46
3 The positive solution of a 3 + 6a 2 + 12a = 90 which is
about a = 2.61.
777