ANSWERS 749

26 a i ¼ 0 : 770 m^2 ii =2:704 592::::::¼ 2 : 705 m

iii ¼ 1 : 83 m^2 iv ¼ 2 : 60 m^2

b¼ 31 : 5 m^2 c ¼ 37 :4%

27 a t¼ 1 : 27 s bh¼ 0 : 994 m c h=

9 : 81 t^2

¼^2

diAbOB¼ 57 : 3 o ii 0 : 500 m^2

28 a i sides arexcm,(x¡3)cm,(x¡5)cm

) P=3x¡ 8

ii 16 cm iii CbBA¼ 43 o

bix^2 =(x¡3)^2 +(x¡5)^2 fPythagorasg

gives x^2 ¡ 16 x+34=0 on simplification

ii x¼ 13 : 48 or 2 : 52

iii AB¼ 13 : 5 m, AC¼ 10 : 5 m, BC¼ 8 : 5 m

(casex=2: 52 is impossible)

29 a i BbAD=BbCD=66o

ii ² BbAD=BbCD ffromig

² AbPB=CbPD fvertically oppositeg

) ¢s are similar fas they are equiangularg

iii PB=16cm, CD=8: 75 cm

iv area¢CPD=

n

4

cm^2

biBbTD=48o iidiameter is OT and OT¼ 23 cm

30 a i ¼ 39 : 3 cm^2 ii 25 cm^2 iii 7 : 13 cm^2 b 50 cm^2

CHAPTER 34

The University of Cambridge Local Examinations Syndicate bears no

responsibility for the example answers to questions taken from its

past question papers which are contained in this publication.

A INVESTIGATION QUESTIONS

1ak=6

n=2, LHS=5, RHS=

2 £ 3 £ 5

6

=5 X

n=3, LHS=14, RHS=^3 £^4 £^7

6

=14 X

b338 350 cim=50 ii171 700 d166 650

e¡ 5050

2a/b

The axes of symmetry bisect each other at right angles.

ca rhombus d^252

p

3 m^2 ¼ 21 : 7 m^2

eArea is a maximum of 25 m^2 when the angle is 90 o,

i.e., when the rhombus is a square. This is because when 60 o

is replaced byμ, area=

¡ 1

2 £^5

(^2) £sinμ¢£2 = 25 sinμ
which is a maximum of 25 m^2 whensinμ=1(a maximum).
3a5=2+3 is one example
bi16 = 3 + 13 = 5 + 11
ii 38 can only be written as7+31
c16 = 2 + 3 + 11
dFalse, as 11 cannot be written as the sum of two prime
numbers.
4a¼ 114 cm^3
b ¼ 132 cm^3
c ¼ 47 :6%,¼ 51 :4%
d box shape
(from above)
Area from above=¼r^2 + 3(2r^2 )+
p
3 r^2
=(¼+6+
p
3)£ 2 : 152
¼ 50 : 263 cm^2
) volume¼ 216 : 13 cm^3
) unfilled space¼ 91 : 2 cm^3
and%unfilled¼ 42 :2%
5aThey all have perimeter 24 units.
b 24
p
3 cm^2 ¼ 41 : 6 cm^2
cR: 24 cm^2 ,S: 30 cm^2 ,P: 32 cm^2 ,Q: 36 cm^2 ,
H: 41 : 6 cm^2 ,T:¼ 45 : 8 cm^2
d
6aL 123 4 5
B 125 8 13
W 024 8 12
T 1 4 9 16 25
bi 60 ii 41 cm
c T=L^2
diT=2B
iiT=2B¡ 1
7a1=1
1+2 =3
1+2+3 =6
1+2+3+4=10
13 =1
13 +2^3 =9
13 +2^3 +3^3 =36
13 +2^3 +3^3 +4^3 = 100
b 13 =1^2
13 +2^3 =9=(1+2)^2
13 +2^3 +3^3 = 36 = (1 + 2 + 3)^2
13 +2^3 +3^3 +4^3 = 100 = (1 + 2 + 3 + 4)^2
c 13 +2^3 +3^3 +4^3 +::::+9^3
=(1+2+3+4+::::+9)^2
=45^2
= 2025
d 3252 = 105 625
eIf n=1, LHS=1, RHS=a+b
) a+b=1
If n=2, LHS=3, RHS=4a+2b
) 4 a+2b=3
) a=^12 , b=^12
f984 390 625
8a i 2 n¡ 1 iin^2 iii n(n+1)
2
iv n(n¡1)
2
b
n(n+1)
2

• 
  

  n(n¡1)
  2

  
  

  n
  2
  (n+1+n¡1)

  
  

  n
  2
  £ 2 n
  =n^2
  9a iVA=9¼r^2 h, VB=3¼r^2 h, VC=27¼r^2 h
  ii 3:1:9
  60°
  5m
  5m
  5m
  5m
  axes of
  symmetry
  1cm 1mº
  r
  For figures with the same perimeter, as the figures get more
  regular the area gets larger with the circle providing max. area.
  IB MYP_3 ANS
  cyan magenta yellow black
  (^05255075950525507595)
  100 100
  (^05255075950525507595)
  100 100
  Y:\HAESE\IGCSE01\IG01_an\749IB_IGC1_an.CDR Thursday, 20 November 2008 3:33:07 PM PETER