F
4.
Answer key for the examination
A.
Exnl,rplr oF
A
I
ke,lo Kt<s 5v o\l
coRREcr soLUTroN
""-\-:'_:_/
length of the altitude to the hypotenuse of a right triangle is the
CO
- OB Jfne
lgeometric mean between the lengths of the segments of the hypotenuse.
m nO _ m
m d6
m
Cii
mAO 10m
=10 m 20m
qo*H
mA6=5m
Since point A is to the left of point O, the coordinates of point A ar
*
"
;;;;;;;;;oBp
m ZQBP +m ZPBR = 180o
m ZOBP = 180o-m ZPBR
m ZOBP = 180o -52o
=128
MEASURE oF ANGLE OPB
The sum of the measures
m lep'+m ZOBp +m lpeg=
- 1gO. {Ltriangle is 180"'
m ,ZopB+12g. +23. =1g0"-
of the interior angles of a
m ZOPB = 29o
.,_-
mPB
mOB
(Law of sines applied to triangle OBP)
sin(m tPoB) sin(m zoPB)
J
mPA
20m
sin23o=-sin29o
m PE = 16.1189... m
HEIGHT oF THE TowN HALL
sin(m IPBR)=
sin52o
t t=I (Sine ratio applied to right triangle BRP)
mPB
m PR
- 16.1189...
m
m FF = 16.1189... mxsin52"
= 12.7019... m
-r
--r-\
\
rr
if',$
/\
o.It-
38 480 kg
2
\.-J
lh^
Mass of the frustum
Volume of the frustum
_ 2600 kg
Volume of the frustum
1m3
Volume of the frustum = 14.8 m3
Lb00
\
/'"/\
J
-l
r{F*
t{ ?,-*3
(\'
-=U€OSltVOlOfanlte
Ministere de l'Education, du Loisir et du Sport
565-410 - Mathematics - Science Option
Voluue oF THE PEoESTAL
Volume of the pedestal= Volume of the frustum =14.8m3, since equivalent solids are of the
same volume.
HEIGHT oF THE PEDESTAL
Volume of the pedestal = 14.8 m3
2mx2mxHeight of the pedestal= 14.8 m3
Height of the pedestal= 3.7 m
Hetcnr oF THE PYRAMTDToN
Since it is a right pyramid, a right triangle can be formed. The measure of one of the legs of this
triangle is equal to the height of the pyramid, and the other leg measures half the length of an
edge of the base.
h : height of the pyramidion, in metres
---) \
\\i
\-/
I
tan54o
=0.75h m
540
h = 1.0322... m
Height of the pyramidion =1.0322...m
i__
1.S
m
_l
0.75 m
Height of the obelisk = Height of the pedestal+ Height of the frustum + Height of the pyramidion
+4.8 m+1.0322...m
cousrnatnrs ro
BE TAKEN tNTo AccouNT wHEN DETERMTNTNG THE EeuATtoNs AssoctATED
THE TRAJECTORIES AND THE COORDINATES OF THE EXPLOSION POINTS
o
o
r
.
o
Both rockets must explode at the same height.
The difference between the y-coordinates of the vertices of the trajectories and the
y-coordinates of explosion points D and E must be equal to i.
The y-coordinates of explosion points D and E must be above 9.5322....
Explosion points D and E must be between 4 m and 6 m apart.
The vertices of the trajectories must be below 12.7019... m.
T
I
I
12.7019... m
A(-5,0) 'O(0,0)
Ministdre de I'Education, du Loisir et du Sport
565-410 - Mathematics - Sc,ence Option
B (20,
0)
wtrH
EouartoHs AssoctATED wtrH
THE TRAJEcroRtEs AND cooRDtNATEs oF THE
ExplosloN potNTs
FIRST PoSSIBILIw
Suppose that the rockets expiode.gt
vertices of the trajectories'q{]*_)t
Suppose that the x-coordinate of the vertex of the trajectory of the rocket that explodes at D
is
=f**
Equation associated with the trajectory of
the rocket that explodes at D:
Coordinates of explosion point D:
ro=f(x
+2)2 +11
o=a(-5 +2)2 +11
-t=!U
+z)2
-l
fi=$ *z)'
Y=
a(x +2)2
1
=
+ 11
a1-3;z
-11
-.8
=xt2
v11
I
v
=-!{r+2)2
+11
or
lo
or
=- .li- -2
Y11
x=-2.9045... or
x
/-o
^lL=x+2
Y11
x
fo
=.1!
-2
Y11
x=-1.0954...
Since the x-coordinate of point D is greater than
that of the vertex of the trajectory, the coordinates
of explosion point D to the nearest hundredth are
D(-1.10,10)
.
Suppose that the x-coordinate of the vertex of the trajectory ol_-![g_I9-c,Kel,"that e"Xplodes at E
is-F--
Equation associated with the trajectory of
the rocket that explodes at E:
Y
=a(x-9)2
+11
Coordinates of explosion point E:
_'l
10=-i(x-9)'+11
11'
'/
_'l
0=a(20-9)2+11
-1=--(x-9)'
11' /
-1 1 = a(11)2
11=(x-9)2
-1
-fi1=x-9 or Ji=x-9
x=-fi1+g or x=*l+g
x = 5.6833... or
x =12.3166...
11
_,|
Y
=V$-9)'+11
Since the x-coordinate of point E is less than that
of the vertex of the trajectory, the coordinates of
explosion point E to the nearest hundredth are
E(5.68, 10).
Distance between points D and E: 5.6833... - -1.0954... = 6.7788... m
The explosion points are too far apart.
Ministdre de I'Education, du Loisir et du Sport
565*410 - Malhematics - Screrce Option
Secoruo PossrBrLrrY
As we did for the first possibility, suppose that the rockets explode at a height of 10 m, that the
y-coordinate of the vertices of the trajectories is 11 and that the x-coordinate of the vertex of the
trajectory of the rocket that explodes at D is -2.
Equation associated with the trajectory of the rocket that explodes at D: y
,
={6
g\
+Z)2 +11
Coordinates of explosion point D, to the nearest hundredth: D (-l .10, 10)
Suppose that the x-coordinate of the vertex of the trajectory of the rocket that explodes at E
isL
Equation associated with the trajectory of
the rocket that explodes at E:
Coordinates of explosion point E:
Y=a(x-8)2+11
'to=.11(x-8)2+11
/
o = a(20
-1=-11(r-B)t
144', t
-8)2
144'
+ 11
144
11-(x\/ - 8)2
-11= a(1212
ffi=X-8 or ffi=x-8
,=ff *u or ,. =ffi*o
-11
144
-=a
v=-11(r-8)2+11
'144\/
x
=4.3818... or
x =11.6191...
Since the x-coordinate of point E is less than that
of the vertex of the trajectory, the coordinates of
explosion point E to the nearest hundredth are
E(4.38, 10).
Distance between points D and E: 4.3818...--1.0954. ..=5.4773...m
This answer meets allthe requirements.
Equation associated with the trajectory of the rocket that explodes at D: V
=-!6
Equation associated with the trajectory of the rocket that exptodes at E: v =
The coordinates of explosion point D: (-1
.1
0, 10)
The coordinates of explosion point E: (4.38, 10)
Minist€re de I'Education, du Loisir et du Sport
56$.410 - Mathematics - Scrence Option
+Z)z +11
ff08-
B)2 + 11
Note: There are several possible answers. Here are a few of them.
Equation associated with
the trajectory of the rocket
Equation associated with
the trajectory of the rocket
that
explodes at D
that
explodes at E
r =f {* +2)2 +11
r =frj{" -7)2 +11
y
(x +2)' +12
= -4.
3
Coordinates of
Coordinates of
explosion point D explosion point E
(-1.10,10)
(3.08,10)
(-1.13,
(4.54,11)
_.1
v
r
=fi(x-8)'
=ffi{*
+12
-7)z +12.6
1
1)
(-1.15,11.6)
(3.34, 11.6)
(-1.75,10)
(3.08,10)
v
=-fi{,
v
=-${*+2.5)2
+12
v
=-ffi{r
-7.s)2 +12
(-1.78,11)
(3.89,11)
v
=-ffU +3)2 +12.5
r
=ffi{r -t)2 +12.5
(-2.43,11.5)
(3.32, 11.5)
\'*
(l
: tsbr)tft
!,(-*,$
f3
t'*0,
', =-11(r-7\2
169\ /
+2.5)z +11
*'S
"zf+n
+11
,
J,} r* q,{tt,t}ZL
a'q" |JZLT,- . r"i.-