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Ent^red^ according to Act irf'the Parliament of Canada, in the year one
thouaand eight hundred and nineky^ght. b^ Amdrbw Doylb, at
tlie Department of Agrjkutturr.
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> VERY IHTEPSTING SELECTION
4)F
WITH SOLUTIONS
Designed as an Appendik or Supplement
TO
JlFitl^metie and JWensuration.
^>^m*-
mi
4 BY
^. IDO"5rXjE,
OTTAWA?
Ottawa Printing Co., (Llfmted), 3 and ^
1898.
rove Street.
^immmmmmmmmm-
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PREFACE.
■K-',.
Tt is valuable variety of useful exercises, is
istined to inspire students with an ardent desire
\r more extended mathematical attainments than
^ose acquired from a limited study of abridged
smentary schooKbooks. With a view of pro-
)ting intellectual progress, I have given many
^orems of great utility, witlfi the greatest
fssibte variety of useful problems and their
flutions, in a limited space. No two are alike,
id from each, ? rule or formula may be deduced
»r the working of similar questions. Those
[ho have acquired a knowledge of algebra and
sometry, will find these exercises really attractive
a source of profitable recreation.
This little work, containing elaborate solutions
all its exercises^ comprehends more proposi-
pns than the first four books of Euclid. It must
ridoubtedly secure a wide circulation and meritori-
is success. The principal propositions have
ien contributed by the distinguished mathe-
iatical correspondents of the Canadian Almanac
id Journal of Education ; selected and solved
their Math^mntical Editor, the Author.
I.
2.
5- «
6. I
i y* ; -« > ;M . ;rq-;ji«* i'MjI^it'^J '
" * g"" ' . "
*"
IMPORTANT MATHEMATICAL TlfOBLEMS.
1. Find the area of a triangle whose sides are,
J^ /J^T J^T Ans.'j^. . ^
2. The square inscribed in a circle : square in a
simicircle */. 5:2.
3. The square inscribed in a semicircle : square
in quadrant !'. 8 : 5.
4. If an isosceles triangle inscribed in a circle
* have each of its sides double of the base, the
squares described upon the radius of the
f circle and one of the sides of the triangle,
shall be to each otheir as 4 : 1 5. *
5. If r denote the radius of a circle, the side of
the inscribed square will be r rJ^t and the
side of the circumscribed square will be 2 r.
6. If a denote the side of a given square, rad^ of
inscribed circle shall be }a, and radius of the
circumscribed circle will be j/^J''
The rectangle under two sides of any triangle is
eqiial to the rectangle under the perpendicular
mmem9i^:\ifimtyhf.^'ys;iim,,iR-.?-im
***'<f*!9f^W!
■VMh
■fpfiMMi
m^^
to the bftse and diameter of the circumsci'ft-
ing" circle
7. Circle."
The rectangle under the
hypothenuse and ± of a right angled triangle
is equal to the rectangle under the sides. ^
8. The perimeter miTltiplied by half the radius pf
the Inscribed, circle is = area of triangk,
radius of inscribed circle rzr 2« -r P, a :=r:- arpa
and P =:r perimete"^. if '
9. The continued product of the 3 sides o? a
triangle = 2 area x diameter of the circuin-
. , f -. a b c ' V
scrioing circle .
. diam. = — ^-^ I
Z area |
10. The square on the diameter of a globe =4 3
times side of the inscribed cube. I
i
11. If r denote radius of a circle, side of the inscrib-
ed regular decagon = J^ r (f^T — i). |
12. U r denote the rad. of circle, the side ; of
the inscribed regular pentagon will be § r
Jio-2 jr I
I J. If a denote a side ot a given regular pentagon,
rad. of circumscribed circle will be = -^ a
N/50+10 ^5r~ t
14. The sum of the sides of right ai^gled triangle
divided by 5 =; X from the r ^
mMMm
*1-'-
ir-
'i^'^mm
15. The square on the side of an equilateral tri-
angle inscribed in a circle =: ^r.^
«
16. The J. of an equilateral triangle is equal 3 times
the radius of the inscrib. circle.
17. The side of a square inscribed in an equilateral
triangle is = the excess of 4 times the X
height of the triangle above its perimeter.
18. If the line bisecting the vertical angle of a
triangle be divided into two parts which are
to each other as the base to the sum of the
sides, the point of division is the center of the
inscribed circle- ,
19. Given the base, the area and line bisecting the
base of a triangle, to determine the remaining
parts. — LetAB=:i6;^
bisecting line C E =
II, area =i 82 ; then]
82^8=ioi=rl DC.I
99218 = DE ; then
8— 3-992 18 = 4*00782= A D •.•
is/A D^ -f D O 1 1.006 = a C
In like manner BC may be found.
20. Given the area of right angled* triangle 48 and
^ difference of the base and 1, 4, to find the
i mmnuw in
■iHii
8
sides. Let x denote the base, then x -f 4=
X i ii— =48 /, x^ -\-^r=g6; heL-ze x=zS
and ^ + 4cr 1 2=:the 1 A ^r5^~+8^ = ^i^
= 14.4222 = hypothenuse.
2T. Find the leng^th of a straight line bisecting a
given triangle from a given point in one of its
sides.
Let A C= 14; BC~
13 and A 6=15 ;
arear=84 and A E:==
2. Let B Y^=-x^ then
15 X 13 : 13 X : ; 2 : I
;. A:r=7i=B F, and
CF==Sf AG=:
GB=6t ;.
^* -^|»c=i CG=iTi
~7i=-9=
^^ » is/.9' + ii.2*=::C D=ii.236io2> Area
of triangle EFB=;42; 15 — 2=i3=BE;.-.42-r
6J=:1 F M— 6.46154 ; GB^6.6As 11.2:
6.6 : : 6.46154 : BM : 3.80769; 13—3.80769
=9.19231=? EM. Then
^^9.192312 -f 6.46T542 =EF= 1 1,23615
,-f r ■
length of the bisector.
\22. Given one side of a triangle and the lines
drawn from the angles at it, bisecting the
other two sides, to find the sides.
Let C E = i6; A E = i2 and C F=i5.
j^CF = CO=io ;. OF = 5; ?^ AE =
EO = 8, '. A = 4. On C E describe the
triangle CEO and produce E O to A=i.2,
' and CO to F
' -r^-Jf**;^
=15. Join A
C, F E and A
F ; ' then A F
is II C E and
=i of it, =^8.
In triangle A
OF, 8 . 9 : : I
: i}i diff. ot
segts. A M and M F ; Z'4^^^ F, and
^j-^e = MA. ^42 -3^Ty2 == MO = 2.04538.
Again, 16 : 18 : : 2 : 2j^ === diff. of segts,
CH and HE; then CH-= g}iy and HE =
6^- 1J102 —g}i2 = HO = 4.09076 + 2.
04538 = 6.13614 = HM = AN. CH — AM
= 9i — 3A = CN = sH' ^JHM2 — CN2
js/6. 136142 +5.68752 = AC = 8.36653x2
: .mnM i v iftmmtmmfi
lO
= 16.73316= CD. MF :-. 4fV; HE
MF
23-
= 6^ — 4t^ = PE = 2tV JPE2 +HM2
= »j3A^'+67r36r4^= 6.55743 = EF;;. E
Px2 ' 13.11486 = ED.
A pole 16 feet high is
broken by the wind in D;
in falling DC touches the
ground 2}4 feet from base
of the pole ] or the base
of right angled triangle
and sum of the hypotenuse
and perpendicular are
given, to find the sides.
Letx denote AD ; AC =
16 ; AB = 2}4 ' a; AC
= b ; DB :: b — x. Then x2 +a^ = b^ — 2
b2 '-a2 ;
A AD r
. 2 dx + «2 = ^2 , and X r
162 +25^2
2 b
= 7.8046875 ; DB or
2 X 16
DC = 8.1953125.
24. If a line bisecting the vertical angle of a iri-
angie, be divided into part?? wh»ch are to each
other as the base to the sum of the sides, the
point of division is the center of the inscribed
circle.
••T ">«««"«
w^
?*■
4
.,%:
1:
3t
II
25. In any triangle, it is required to inscribe a
rectangle whose sides shall bear a given ratio
to each other.—
Let ABC be the
giv^^ triangle ;|
AB=b;lcD=A,
and side of recti
< ± to base=rjic,
a,nCj adjacent
side = nXf n
denoting the given ratio. Then ^ ; ^ [] nx : k
:= G H, one Side of the
- X
X
:-:-X.-
h-^n h
required rectangle^
When side of the inscribed square is required. As h \h
. . . i • ^ ^
\ • X \ ft — X X,zzz ~ J— ■ '
rh + b.
26. A farmer has a triangular field, the distances
from whose three angles to the middle of the
opposite sides are : no, 140, and 160 yards^
respectively. Required the area of the field ?
Put AD :=d(iio), E C=i6o (^ ; B C = x^
AB =:v ; AC = ^.Then^ ^ +j/^ = 2^- -f JMr
X2 ;jy2 ^^2 =z2{i' +^ Z^ ; Z^ -{-X^ =22'^
-{-% y^ The sum of these gives 2x^ +
2y^ + 2z^ — 2 b^ •¥ 2c^ '\' 2 d^ + %
m^^
mmm^tmmmm
12
x2 +^jy2 + J^ z^\ andic2 4. ^2 4.^,2 «-.|
^2 -f I ^2 +1 rf* ; ivQm this subt. the first
equation ; then jt^ = ^ rf^— tH^-^-^ c^—
yi x^ , or gx^ = 8^2 _|. 8^'2 __-. ^2 J ^ ^ 2
= 81J2 -f 8^2 --. 4^2 ; whence x = ji
J 8c^ + 8d^ — 46^ =C B = 186.547 +
' ' y = H JSi^ + Sd' — 4^2 . 2: '= J^
fj 8^"+ 8^2 >_ 4^2 • 186.547, 15748,
129.615 = Sides.
27. A stone is weighed in a pair of scales which
are known to be incorrect ; when placed in
one scale, it weighs 714 lbs ; but being pu'
in the other, it only weighs 37 J lbs ; required
its true weight.
tJ7^\^37i = fJ 2700 = 51.9615. (nean pro-
portional.
i
S¥*waw« .TT^'^i'«'%^f*^m4»<>''
13
28. The base of a triangle is 80, and sides includ-
ing the vertical angle are 65 and 55 perches
respectively ; required the length .of a line
drawn from a point within the triangle, 8.53.
perches from the side AB, so as to cut off 4
of the area. CB = 65 = f ; BF = x ; ratio
5:7; AB = 55 = d ; w + w = J = 1 2 ; AC
== 80 = //; BE = 9I = g ; EP = 58.67
= / ; AP = 8.53 : CG = 59.71 ; BH = 8.
53 ; EH = 3.67. By similar triangle s, we
px
have : g + x : p :
I X BF : BA X BC
g + x
m : m + n.
= BI. Then B
p X2
^+x
: ^/
: : m
SpX2 *
J. Then bmf = -, and S ^ x2
^x X ' ^
= bmfx4-bmfg .'. x = 32.617:5 = BF .'. Fl
is easily found.
. ■Aiuw^itdmSai
^il^tiifi^»^4m&
59. In a given triangle, the base AC = iod; A
B = 70 ; BC = 90 ; uhat is the length of a
- line (i) drawn 11 to the base. (2), x to the
base (3), inclined to the base at a given <
15*^, so as to cut off t\ of the area ?—
ag.
i). Let X denote the length of the line 1 1
base AC : then (xix.vi. cr) 1002 ; x2 : 1 3059*
41 : yV of 3059.41. From this we get x =
7977 = QR-
(II.) Area ol triangle B D C = 61.1882
X V = 2019 2106. Let X c required L, H L
then, as B D» : x« :: B D C : y\ x 3059.4-
3744 : x« : : 2019 2106 : 1946.891^. From
this proportion we obtain the value ofx = X
to base. A C = 60.08258 r H L.
<II1.) Tlfe < A = 6o«^ 56' 28" : < ACE = 150,
;. AE is found = 26.68, a^id EC = 90. 1 1 ; tri-
■mmmimmMm!mf'm»'t-x*nf»'<''-<»m^Mmtim
15
angle A E C = 1 166.03, and C B E = 1893.38.
From triangle A E C, cut off ^ of area of ^
whole by a line 1 1 C E ; remainder will be
•j\ of the whole.
E O : X" : : 1 166 08 : ^j of 3059.41, .'. x = •
88:0176; then 81 19.812 : X* : : 1166.083 :
iiia.152 ; .*. X r 88.0176, the line required.
(1 1 1 1.) Bisect the triangle by a line whose
length is 49.32 perches.
ACxBC = 2 KCxH C, or 2a6xx
C H, and C H-7-;;. By similar tri-
2,X
a b a b d
= CL; KL =
angles b: d::
KC— CL = X . .
2 6 X 2 6 X
4 3« r» x» —4 ^ x» — «» b^ d* -v^ab^ d x"
2 X * 2 ^*x
abd 2bTii^^abd
a'^lf
b'd'
4 A« x«
: ;. 4 *«x4
4 b* c^ x« —
4 d* X"
4 hi/ x« =1 — a» b' ,\ X =70. V
30. There is a room in shape of a rhomboid
whose adjacent sides are 12 and 7 yards re-
spectively ; the shortest diagonal is 1 1 , find
the length of the other. Sum the squares of
the diagonals=sum of the squares of. the
sides ;. (122 +72 )x 2 = 386 : then (386—
1 1* ) = 265 then ^Ji65-=i6.28.
i6
31. The base of an isosceles triangle is 30, and
a segment of one of
the equal sides made
by a perpendicular
from one of the base
angles, on the oppo-
site side is 10. Let
A D = x : jr+io = A B ; ix-\-io)x 10=
T- :. ^ = 35» and A B or A € = 45—
^^45^ —15' = 42.4264 = 1x15 = 636.-
3961, the area. ^|i
32. The radius of a circle Ts 10, finJ the sides of
an isosceles tri-
angle inscribed in
it having the base
equal one half
each of the other
sides. Let x de-
note the side.
The square on
the radius : that upon one of the sides : :
4 • 15 .'. 15 ^=4 »'<^^ . r^ : x^ :: 4: 15, .'.
m^
33-
■^fe
34-
17
The three sides of a triangle are 15, 14 and
13, how far beyoiid the base must the sides
(14 and 13) be produced so as to form a
trapiziu m
cpntaining-
an area
equal 2 ^
times that
o f th e
given tri»
angfe. Area of triangle AB €^=84 x 2^=
2 10 = trapezium B D E C + 84=294
B C^ : D H2 : : 84- :'' 294 : whence
D E = 28.0624/' *( 1 5 + 28.0624) -r 2 = }i
sum of n sides, B C . and D E =
2^.5312, 210 ~ 21.5312 = 9.7523 = X
A slick of fimber is 4 inches wide at one end;
and 8 inches at the other, and 12 feet long,
where must it be cut, so that one half may be
at each sit'e of the cut? ad = 4 : bc = 8 : ae
or PN = 144 ; AP = EN = BE = 2 inches.
When BA and cd are produced to meet in
V, if is easily shewn that ba = Av, and pv
.:t.3;Tt':"
"p' '^
^
s PN ; A NV= 288.
Then 4 x 288 = 1152
r area of the triangle
VBC, and 11 52 -4- 4 =
288 - triangle avd.
NV = 288.1152-288 =
, 864; then 432 4- 288 =
720 :; triangle vgh
1 152 : 720 ;; '288* •
/ VL^^;8 : 5 ;; 8^944
' and vL r 227-684 +
227.684-144 = 83.684]
r PL .'. LN - 60.316,
Bc' : GH* " 1152 : 720 : from this proportion
we find the dividing line gh = 6.3245$ inches. ,
35. Given the differences between the diagonal
and side of a square to find the side. Rule. —
Square the diiF., double if, extract the fj~^
and add the diff. to it When sum of diagonal
and side is given to find the side. Rule. —
gquare the sum, double it, extract the ,^~~"
and subtract the sum fram the la^t result.
36. Given the sum of the diagonal and longer sid#
of a r < </ parrallogram, to describe it, whei^
/.-■*.
»9
the square of the diagonal is equal (» + i )
times the square of the shorter side. Let
AB = sum. From B draw BC 1 to AB r , and
make BC**= n AB*. Join AC, and make CD
= CB. From D erect IDE, meeting A B^ in
E ; join E C. :^hen E B^ ^BC» = ED« * EXC*.
.•. EB8 = ED«, and EB = ED.Then AE is the
1 1 "™ required. Let AB* = 25 ; then BG* = 125;
n being denoted by 5. ^25*125 = ^Ji^Q
AC = 12.24744, and JJ^ , i I.I8o34.^725
= 5. 12.24744 — I 1.1803 r 1.0671 r AD. Now
in similar triangles ADE and ABC, we have
11.18034:5:: ED: 1,0671, .*. 5 ED =
• 11.18034 X 1.0671 = 11.930540814 ; then ED
= 11.930540814-^5 = 2.386^082 r 5.693511 ;
1.06712x5=5.693512.
37. The sides of a triangle are. 26, 28, and 3o>
what must be the sides of a similar triangle
Ur
20
contatning 3^ times its area? Area of giv^n
triangle = 336 : 336 x 31^ or as i : 3^ : : 26*;
X* . *. X =46.87216. In like manneri we find
AB r 50.4777 and BC r 54.0832*9.
38. The perimeter of a r < <^ triangle, is 74.4* and
^rom the r < on hypotheiiuse is 14,88 ; jSnd
0t,
'^ne sides. ^^ — Let P = Perimeter ; CD = a; AC
= X ; BC = y.
"* ThenAB = p.-(y + y),
'and AC2 4. BC2 ?
AB ; whence x^- -f
+ x2 + 2 xy + y^
transpose and divide by 2 and P (x + y)~J^ P2
r xy (i). By ^similar triangle s, AB : BC : :
AC : CD .-. ABxAC, or a P-^a (x + y) = xy.
(11). By^ Substitution (a-hP)x (x + y) =^ a P
+ JP^ ; whence x-f y r
^ P (a + ^P)
or
P(fl + JP)
. « + P "' T '
— X. Substitute these vatues for
a + P
(x + y) and y in Eq. (11.); the result when
simplified and reduced, givesi {a 4- P) x^ — P
(rt + iP) x= -i flP2 . From last Eq and value
P(« + :,P) +
of y abpve is found, x or AC
2 {» + P)
21
P.
/ rw^ iJ(a-hP)^ -2a^' and if the result of
2(a+P) ^ * '
the two sides be taken from P, the result
t3
will given AB«P p-(x+y)«-72 — •. which
a(<i+p)
expressions are == the values of the 3 sides of
the triangle, which may be fpund to be AB-
31 ; BC = i8.6; AC = 24.8.
39, Given the difference between the side and 1 of
an equilateral triangle to find the side, Let
the side AB = x ; AD=y ; and x-'y= 1.071797;
then, ^^2x ry, or x-^Jxl ^ i.o7i797»^*-.
^ ^^ ~hJ 3 =^1.071797.-. 2 x-rJ3~=?.i43594
2X- 1.73305 ; X =2.143594, . ;. .26795 X =
2.143594
,26795
■an 'J X =8.
40. The inside dimensions of a school house are :
the inside length is 3 feet more than 3 times
the height ; the inside breath is 4 ft greater
" than twice the height ; and the inside surface
of the walls is 508 less than the surface of
the floor and ceiling together. Required the
dimensions ? Let y = height ; then 3 y + 3 =
, length ; 2 y-H4 = breadth. C y^ x C y = area
.Ji.lM>V*iii
11
22
of the sides ; 4 y* + 8 y r area of ends ;
12 y' + 36 y + 24 = area of floor and ceiling.
12 y* + 36 y + 24 - 508 = 10 y* + 14 y ; and
y* + II y = 24a. Solved gives y = 11, the
height ; 3 x x 1 1 + 3 r 36, the length ;
2 X 1 1 + 4 = 26, the breadth. W. D.
41. The wheel of a carriagfe is 5 feet in diameter,
required the length of the successive
cycloidal arcs generated by a nail in the
circun\f<*rence of the wheel, in a distance of
24 miles ? By means of the calculus, we
find the height of arc of a cycloid as follows ;
J= 1^2 r x-^ -f vers — 'x, the equation
pf the curve ; -4^ =
dx
t-x
^2 r x-x
■2 s
_'J
2 r z - X
^J2 r X -ui? 4-
*. ds :?= d X
X
dx S
dx
X
J~ .'. S= 2 7-^9
a;-* dx = 2 fJYrx^ C. When a;=0,
*
S = 0, /. C = O, and when a; = 2 r, then
the semicycloidal arc = 2 ^2 ^ ="4 ^
and the whole length of the cydoid is
8 r-4 times the diameter of the gener-.
lii
^l_
23
ating circle. . ' . ^^^ ^ 30-55767 miles
3.1410
travelled by the nail. -
42. There are two circles inscribed in an is Obceles
triangle touching each other and the sides of
the tri.
angle ;
• their dia-
■ *
meters
I are 4 and
6 respec-
tively.
Find the
' sides.
From the
centers
DartdF
draw D
G and F '
H p e r-
pendicu-
lars to BC, and FO n CB : draw CFDA.
Put DG = r:,FH=5: DO = y- — 5 = ^:
FD^r-^s^-d. Then FO = Jd»— c»
^d; CB = a. By similar triangles D
■MHIMIl
MiiiMiMiaiiUHiiiii
'^('^'■ii
' ''*.•,''> -■■••,".
•i,;V-,.:>^i.^.v:;:;
A -■'■ \ 7-. ■' ■-
' ""-..- ^ ■ . - ■
■:■■-. ! , ..-■. -
5 ■
».„— ■*■ *c
F O and B C A, we have b\ d\\ a :
1 1
I
-V = A C ; and c : b
o -
r: — = CD, then
¥ r
a —
+
^r
c d d '
43-
■^- ':
Find the ladius of a circle whose centre being
taken in the circumference of another circle
containing two acres, shall cut off one half
its area ? Radius of first circle:=20. 18501 18.
Now, suppose A O = 11.7 (nearly) ; t'hen
20.1850118 X =
(11.7)^ ~ 136.89
;. X = 6.7817615
= height of the
lower segment.
Then 11 .7 —
6 . 7 8 I 7 6 I 5 =
4. 9 182385 =
height of upper
- segment. Then the area of upper segment
A C E is found = 65.7318402, and area of
lower segment A O E :: 94- 3903972 : sum of
segments = 160.1222374, too much. Again,
suppose 11.69, ^"^ proceed in like manner
we have : * . I ^
■■.■'• ^- y .' ' v'm'"*', ■ -■ ,-'
- '^s-rWi;' 5-^5.-1;. ':«^14'2^.
- ;* . 1''
f. ■ ' - '' ' .''-»,-■'.
l.v
65.7299442268 = area of upper segment.
94.1699210470 - area of lower seqfment.
159.8998652738, too little; error = . 1061347262
and .122237465. Using these errors by the
rule of trta/ and error y the radius approx-
. " imates'to 11.6944789, and simi of segments
= i6o,ooo,ooo + a decimal. f
44. A line 16 inches long is a common tangent to
^ -;. 4 circles touching one another. It is divided
in the points of contact A B C D, such that
A B» = 3 B C« = 5 D . The difference be-
tween the diameters of the ist and 4th
circles is 7.311521963. Reqd. areas of the
4 circles. '
■•ilfl'l
"I "•
I'\
Hi
26
A B^ =3 B O , and B O =5 D C^ .';
3 B O = 15 D C^ , hence A B^ = 15
D O ; BO =5 C D^ , then A B =
^rsxDC: BC^^ixDC lAJTl^
J^^ I) X D C = 16 and D C = j-^n
-f I = c. Then A B = ^^15 x z =»
8.71673739. B C = Jl X C D =
5.03261066992 .'. C D = 2.2506. Let
^jc. y, z, and u represent the diameters
^respectively : then x y^a^ =3^ =liii^^j8r
=5 f^z /. sy-=5 ^» and j=*f ; x^3z .*.
xi^=! c^ or u^ -f du = 3 c*' This Eq.
gives w = >i fJd2 4. 1 2C2 = — - 1 .6^8478-
2
037;'S^=-~r''3-oa»i36j;:K=— =8.44»S5-
1418: x = -~ =0 then
y
3. 0000 1 36 1 'x. 7854 = . . 7.q686 ~ areaof 3«'dclrcIe-
I.688478*x.7854 r 2.2'i9i = area of 4th circle.
9*x .7854 63.6174 = area of i**^ circle.
8.442351 'x. 7854 = . . . .55.978 = area of 2nd circle.
■fai
45-
27
An aged man two daughters had,
And they were very fair ;
To each he gave a i^iece of land,
A (^ircie and a square,
At twenty shillings an acre just
The land its value had ;
The money that enclosed the whole
Just for the land was paid.
If every shilling be an inch.
As it is very near, -^
Required the acres in each piece
The circle and the square ?
For the circle. — Let x denote the circumference in
inches ; then o^ x J^d by =0795775 = area in
square inches, and 6272640 square inches in
an acre .*. — ' — ;; = price oc all.
6272640
' X t. 3941214.54 r circumference .*. 394214.
W X .0795775-6272646 = 197060 • 7338 =
area of the circle.
Forthe square.— 3941214.54-7-20 = 197060.727
= area as before for circle. Let x denote the
side of square, then x^ = area in inches.
x*-
6272640
= 4^
X r 1254528 inches.
mmttnk
28
12545^^^ ^ 20
4 X
X r 1254528.
6272640-^
1254528 > 4 -i- 20 r 250905.6 = area of the
square. *
46. A tre6 standing* in the water is just 15 feet
; above the surface. When the wind is blowing
the tree is bent over and touches the surface 20
: feet from where it stood. Find the length of
the pole or tree.
Let the line A B represent tlie water ; C P
the pole or tree. When P touches the water
at B, C B
must be
equal C P
.*. triangle
B C P is
isosceles —
AB r 20 feet
: AP = 15 .-.
^202+152
= 25 =PB .-. PD or DB r 12^^ ; 1 AE = 12,
and PE rQ; then 9 : 12 : : i2j^ : 16^ = CD.
Then ^CD2~7 PD2 = ^jr^>^^M6^ = ^P
= 20J, length CP, as required.
29
47-
■A
■ 1*','
48.
..■;«■.
4Q.
50-
AD r 24 ; BC = 18 ; DE = EC ; required ED,
AB, AE, EB and DC. It is. easily proved
that triangles
ADE and BC
E are equian-
gular, and D
E = E C .-.
they are = in
every respect*
and AE = 18 : BRTiif; ,-. M^^f^^
30; DC =42.4264. The 1 from M, middle
pt. of DC. finds pt. E. ^ '^
The parallel sides of a trapezium are 20 and
26 ; the area is 996 ; find th« ± height. —
Area ^ by Ji sum of s'des = x height.w^!^^^r
To find the area a rhombus. — Multiply yi of
one diagonal -by the ^other. -!^^|t' PW^r0f •
To inscribe a circle in a rhombus.— Intersec-
tion of diagonals is the' center.
■■ .'■■■■■»*f#*-'' '■'.''V'-gK. ..-"••tw: '^
.*t;f,ii -•.' ..->v.- .,'- -*^' -^ . ;- .-'1 •'.^•ft'-
|i. To find the X Jp* ^*^J from the ?- < on the
- hypoteituse, when the sides are represented
r^by 3, 4, 5,- or n time^y^ese number, sum of 3
sides??- 5 r x
ft:
'n^'^^r^f
'5 V
30
52. Find the diagonal of a cube, the length of
whose side being 18 inches.
T^,ei|Sqi«araillK>t 0^3 fftlMthe square of side
A='t'e vft«rnm diagdnaL ; .
'he length of a rootn iis'^S.feet ; breath i3>i,
ind io}4 feet high ; recttiired the distance
from any angle of foor
)f the ceiling ? ^7
e farthest corner
= 22. 5 ;
the required diag-
onal.
54. Find the size of the largest square stick that
can be cut from a cylindrical piece of wood,
5 j4 teet in circumference and 1 2 yi feet high.
Inscribe a square in circle representing the
base, and Inultiply its area by the height.
55. The sides of a
trapezium in-
scribed in a
circle are 40^
60, 80, and
67, to find the
angles. Tri-
angles ADO
and B CO are
similar.*. AD :
* —
?l
BC : : OD : OC ; for a similar reason, AB :
DC : : AO"i OD : OB : OC ; hence we have
the proportional lengths of AO, OB, OC,
and OD that is AO = i.79tV ; OB = .37^^ ;
OC = I, and OD = 2. The relative lengths
taken two by two, give the rates AC*:«BD,
and AC >^' BD== AB>^ CD + AD " BC, calling ^c,
bd, the hypothetical values of the diagonals,
we have bd : ac, :: 7220 : AC^ , or 2.89f| :
2.79y\ :: 72120 : AC^ .-. AC = 83.42 + , and
7220 — 83.42 = 86.54. We have now the 3 sides
of each triangle to find he <s .*. ABC =
' »ir 29'42* " ; DAB =^74" 50' 17" ; then ADC
=68" 30' i9>i% and BCD^ios" 9' 43".
j6. Given the diagonal of cube, 20.78461 feet ;
find the length of the side or edge. The
' square of the diagonal of a cube is equal '3
times the square of the side.
57. A cone 28 inches high is bisected by a circle ||
base ; how far from the vertex is this circle ?
Let *= height of upper cone, cut off. Then
28* • Jt* :: 2 : / .-. 2jic3 = 288 ; x^ ~ io^y5 ^.^
X = 1^^ fJT =2242 in.
^59. Z]^ slfipt'heigfht otil cone is 12, and its solid
ity 521.1537408 r required the height?
mni
3a
2 1563 - 46r2224
4612224 .-. x* = jj^^-^
Let y deflate the hilg>it, aod i^ the diameter
of the base ; then fjiz^-x^'^y^ »"^ 4 ^ ^ ^
.7854X? ^521.1537408- 3i4»6^2y ^,563.
•. 144 — ■
3.14107 '■'''-:::■}/■
1S63 4612224 ^ 2, and y3- i44y --497.
. ,3.Hi6y
664, solved, y = 9 • 6. the height.
59. A^piece of square timber is 12 feet long ; each
side of the greater base is 1 1 inches and that
of the less, 5 inches. What length must be
cut off from the less end, so as to contain a
solid foot ?. . . . AD = II .-. AE ' S'M BC
is 5 ; BO = 2i. ThYoughBdrawBF ii HE •
then AF r 3, and FE = 2^ ; AF : FB :: AE t
EH. or, 3 : 144 :: si - 264 inches = EH ; but
• EG = 144.*. OH
= 120 inches. 5*
»< 40 = lopo cubic
inches ; iocKy+
1728= 2728^ = soli-
dity of pyramid
KHL; u2 X H^
= 10648^ solidity
ADH. LetKLt
X ; then 11' ix^
'\r.
r.i): ' ;*.'!;. v^'^;.\>'V-. *... -^
:.»-i.
■•■ ■■ ,:jt.i7'.-.-. -^■■iiiiiJ- ,j , r.l.^i' ;,^i_
33
:: 10648:2728; whence x3 = 341 .-. x = 6.9863,
length of the dividing line As 2J : 120 ::
3.4932 : 167.6736 = Pn ; .-. 167.6736- 120 =
47,6736 = PO. ^
60. A bubble of air having a diameter of 4 inches,
passes from ♦^he bottom of a lake to ti.' (op
120 fathoms. Required the diameter of the
bubble on reaching the surface.
120X6X f2
— g — x62i =3121^ = pressure per square
inch on the bubble. 42 x 3. 1416 = 50.2656 r
surface of the Jdubble. 312J X 50.2656 =
15708 = pressure on bubble at the bottom.
>tV%^ 5^*2^5^ = 3*636 : pressure of = size at
•the top. As 3.636 : 1570S : : i : 4320 ;
1^74320 — 3. 1416 - II* 13-
61. Two men purchase a circular race course, i
mile in diameter, and divide it by a line //
diameter' ; required the length of the dividing
y fence, so that one may have fi of the area,
and the other ys ?
The area of a circle whose diameter is i mile,
is .7854 mile, yi of which is .2618 of a mile.
The versed line answering to this area is
•36753395 which taken from the diameter i.
l^mmtMM
M.
r
II l.llllli MW
«U
m
34 .
* /^- •'
*;i>f/-
4;>v leaves .63246605, remainder of diam. Then i§^
semichord is a mean proportional bet\veen 4S^
,;■•••. -^^/ci-,- >.. }«., .■.;.'.,',:•.■
' the segments of the diarn. 5?v we have li^^J;
2 ^^[63246605 X .36753395 = .96426162 mile
^ -. #7697 yards. ,. ,- ;.^:,-^,::.n^-i,i.rr'\:yy^'"-->>-
62. A can mow a field in 15 days by getting 7
„n days helph from B ; and B can mow it in 24
days by getting 2^ days help from A. In
what time could both working together mow
.:^,::..,^'-,:,;..^*.,.>^^^^^ Asi ^4 : 7 :: I : 5
A + B B ;yj> V, .'.I ofA=i|4 of BAs I : 2Jj^ : i || : 4
• o^ ■, K I i^X rrtimereg-.rfl"
' a ■.■ .'Vrt O U -■ •.-r'.- ■ ■ '.,.1. - V" ■ ■ - -■ .. .- >■-(■'■;.«: .-.
■''■/■";■- v« ,. - - .•- ■ ' - ^ ■'■ •
V • ■■■' > *■* J* ' ■-st"'' '^'i"'-.,' - -V *■■-" ■.
63, Find the length of a band to surround two
wheels, the distance between, whose centers '^
being 14 feet, and diameters 10 and 6 feet
respectively. C N = 14 ; C P = 5 ; Q N = 3.
When tangents are on alternate sides of the
circles, on the iine joining centers, discribe a
semicircle, and make C L^ = sum of the radii,
r 8. Then ^""(14^ -82 > = 1 1.489125. DN.The
»*1
^^^'h:]
35
^i':^r%:<-
!j"' ^ii'^'..''i''."^i'
;1|;| triangles C D N and O H N are similar ; .
S*-^^*^ 8 : 1 1. 4891 25 ;: 3 14.308422 r HO, r B O ;
*. II. 489125 -4.308422 = 7.180703. : A O,
^... = G O. ^~ (A O'^ + A C2 ) = C O r r
" (7.18070324 52) = 8.75. Then, in triangle
CAO, 8.75 : 7.180703 + 5 :: 7.180703-5 :
3'^3S7^3 - <^^ff- of segments = O S - C S ;
hence C S = 2.857144 ; and S O ^ 5^892856.
V~(5' --2.857144' ) -- SA . SG r 4. 10325.^8=
^j^^~( 2. 1428552 + 4. 103252 ) r 4.6291 = P A or
P G r chord of i the arc A P G, and 2 A S r
8.206512 5 chord of the whole arc A P G ;
hence the arc A P G = 9.608762. Circunife-
-■'■/ ,
36
.,- .^. -_
rence of the whole circle = 31.415926,-9.608762
= arc ATG, = 21.807164 + 2 X 7.180703 =
36. 16857, length of band O A T G O. Again,
AH-GO = OB or HO = 4.30842^. NB or NH
= 3 ; 14-8.75 = ON =5.25. Then, in triangle
OBN ; the base : sum of sides &c &c .*. 5.25
: 7.308422 :: 1.308422 : 1.82 1428 = diff. of
thesegts. .*. OV =3.535714 ; NV r 1.714286;
QV = 1.285714. ^~(BN2-VN2)= BV =
2.461955. (B V2 + Q V2 ) i = B Q = 2.7774^ =
chord of i arc B Q H. Then, chord of whole
arc = 2X2.461955 =4.92391. Hence, th^ arc
B Q H may be found to be = 5.765225 .6 X
3.1415926 = 18.8495556. -5.76525 = 13.0843 r
BZH, +2X4.308422 =21.70114 =belt OBZH,
+36.16857 = 57.86971, the entire length of the
band. - >
64. When the tangents are not on alternate sides :
CN =14; CP =5; NO =3; PO =6; CD =2.
Then, ^"^{14^- 22).= 13.855406= DN =
AB = EF.^(DN2+DA2 ) = AN = 14.177446.
In triangle CAN, we have the base to sum
bf sides &c. &c. .•.14 .'19.177446 :: 9. 177446 :
12.571427 = N S - C S, and the sum = 14, ;••
N S = 13.285713, and G S =.714287. Then
r-^.
g.
"fli^", w<i'ii,'« uA»i,ji .1^ .1 .,.^«i j,i iiwi n««pf^;i^^jwiii^)pMpit^ippBaiai|MP^ppMfp(pHpqi|p
mm
37
^""(52 -.7142872) rSA =4.94876, X2 =AE
r 9.897432, chord of the whole arc APE.^
{AS2 + PS2 ) = AP = 6.550353 r chord of i arc
APE. Hence, length of whole arc APE =
14.168464. 3.1415926 X 10 — 14.168464 =
17.246462 rare ATE. PT beiniJf diameter;
10 : 6 :: 14.168464 : 8.5010784 = arc B Z F.
ATE r 17.246462 ; BZF =8.501078; 2AB =
27.712812, .-. 53.46135 = length of the whole
bandATEFZBA.
m
65. A promissory note is offered for sale, on which
is due $24,3225 iaterest for one year, at P per
cent, simple interest. The owner of the note
agrees to cancel the interest, and sell it for
$23 less than the principal, which is at a
38
discount of the above per cent. Required the
amount of note? •
Let X r the amount ; P - rate W unit ; n equal
number of years. Then x P « = int. for 'n
years at rate. Also i + P « = amt. of $i for i
xP n
year. i + Pw:i ::xPw:
X P ^^
I + P «
= discount.
„ r 2-1 *? question ; x P « = 2^+2 P n :
X P = 23 4- 23 P .*. « - I ; but X P « = 24.3225 ;
.*. 23+23 P=24.3i25, and 23 P= 1.3225 .*. P=
.0575 ; but X P «=xX.oi75 x 1 = 24.3225 .*, x=
24.3225 -f ^575=amt. reqd r 423.
66. A boatman rows 6J miles down a river, and
up again in 182 mmutes, the stream having a
uniform current of 2;^ miles an hour : find at
what speed he can row in still water. Let x
denote the rate in still water, and X+2J = rate
downwards ; then 6 J — (x - 2^) = 26 - (4X - 9)
= time of ascent ; but 182 minutes = 3^^ =
91 -=-30 = whole time ; .*. (26-r (4X + 9) + 26-r
(4X-9) = 91 4^30 ; x=5^, thc! answ.
67. The interest on a certain sum of money for i
year, at simple interest, is 317.0465 and the
discount on the same sum for the same time,
at the same rate, is $297. Find the sum. Let
i
" i.^,^111. JU«>i««!«nniP>^n>^pa(;ii;i«|HVi(PHP|>Maf^p|^^
PiPi"PI
39
X denote the rate; then 317.0465 X
100
• 31704.65 -r X P. 1 00+x : X :: 31 704.65 -rx : 297.
From this proportion, x = 6.7496633 .*.
31704.65 -^ 6.7496633 = 4697, the principle
required. . .
68. -Two. railroad trains 109 and iii feet longf
respectively, are moving on parallel rails;
when they move in opposite directions, they
pass each other in 2j^ seconds; but when
they move in the same direction, the faster
train passes the other in 15 sec. Find the
speed of the trains. Let x = ft. travelled by
the faster per second ; a >d y =^ ft. per second
travelled by the slower train. Then, — ~ 4-
x + y
III 220
or = 2}^ = combined speed of
x+y x+y '^
both; and 2^ x-f 2^ y = 220 .*. 5 x-f 5 y =
440, aj\d X H- y = 88. Again, — -^ H or =
x-y x-y ,
220 , /
— « = 15 .-. i5x-i5y = 220, and x-y= 14^,
We have now the sum and diff. .'. x ~ 51 ^»
and y = 36^. .*. speed of faster per hour =
35 m, speed of the slower 25 miles.
mm
MiaaiMiM
40
69- A person changed a fifty dollar gold piece for
31 pieces of foreign coin ; some of which
were worth $2.26 each; others $3.01, and
the rest 77 cents each ; how many did he get
of each sort ?
Let x+y-f z = 3i ; 226X-I-301 y+77z = 5cxx)
77 x+ 77 y + 77 2 = 2387
I49x-h224y * -2613
and 224 y = 2613 — 149 x = a whole num-
ber;
x-y
224
= a
whole number = P. If
P = 0, x=i ; y=ii and z = 19. /. the
required numbers are i, 11, and 19.
70. A cubic foot of gold weighs 1 1 cwt ic4 lbs,
and a grain can be beaten out so thin as to
form a leaf of 60.25 square inches; how
many of these leaves will be required to form
an inch in thickness ? , j .
144 : 175 :: iiio^ : 1350 Troy = 7776000
grains -r 1728 =• 4500 grains in a cubic inch ;
but one grain covers 60.25 •*• 60.25X4500 =
271 125 leaves form an inch in thickness, and
the thickness of a leaf is i -r 271 125.
j7i. a vessel [A] contains 20 gallons of wine;
another [B] 8 gallons ol water. How many
gallons must be taken from each and poured
■l^MliHy
Uilti^^i^Mltlka
iidMiyb
■m'
41
into the other, so that after repeating the
process any number t)f times, the quantity of
wine in each vessel may be the same as after
the firtt operation ?
WINB WATER '
20, and 8 ; Let x denote the quantity
WATER
reqd. x + 20-v, and 8-x+x =first*remain-
, _, 19 x , 380 -r iQ X , 54-7 X
ders. Then -^— and ^— and ^^-^
20, 20, o,
7X 8 - X , X X 20 - X 20-x
8 •
8-x ^ X
and ;r- . . . .
20, 8
20
26
20
-180 - 19 X . X ^ ^ « .
^^ ^— + o ; or, 8oo~4 X = 760-38 X +
20-0
5 X . • . 7 X = 40, and X = 54.
. Otherwise. — Let X gallons to be taken from
[A] ; y = number to be taken from [B] ; then
2o~x : y : : 20 : 8 ; but x = y . • . ^ =54.
72. An Arab tent is composed of canvas k yard
wide and ^ inch thick ; and when not pitched
is wound on a pole 2 inches in diameter,
forming 115 rounds of cloth ; how many yards
in the tent ?
2.04x3.1416- 6.403864
- 4.56x3.1416=^23^^
20.73456 XII5
= 2384.4744 in. = 66.2354 yards.
MttcMiiiaiMlHUilHMIMMilliMM
If^-
42
Otherwise. — ns^^z - 4.6 inches = thickness
of cloth ; then 2 x 4.6 + 2 = 11.2 = diameter of
pole and cloth ; .*. (11.2)2 ..22 = 121. 44X. 7854=
95.378976 inches, surface of end of cloth ;
then 95.378976^36 = 2.6494i6-T-^*y=66.23i4»
73. A grain merchant having a quantity of barley^
sold I of it at a certain gain per cent ; fi at
twice that gain, and the remainder at 3 times
the gain on the first lot. He gained upon the
whole 30%. What was the^gain on each lot ?
Let P denote the required price ; r = rate per
cent.
Then | x f^^^ = f ^^ = profit on the first lot.
I - X T?F ~ l?f == profit on the second. .
' TT X \h = l?f = profit on the third.
Their sum is fy^f = profit on the whole ;
29Pr _ 30 p .
r =
29 r = 450 .•
1 5VI = profit % on the first.
31 j\ = profit % on the second.
46VI = profit % on the third.
74. If A had travelled -^ mile an hour faster, he
would have finished his journey in JJ of the
time ; but if he had travelled ^ mile an hour
slower he would have been iJJ hours longer
43
on the road. How many miles did he travel ?
Let X = miles travelled ; y = miles per
hour. Then — = time he couldfinish;y -h
_2
II
1 1 X _ 37 X
or, 420 X y =
II y + 2 39 y, '^ ^ "^
407 X y H- 74 X, or, 22 X y = 74 X .*. 1 1 y =
37, and y = 3^^ miles. Again, y - tt : x
II X X , 31 . ^ ^ 2 r^
* :: I : = - -f — . . 35x = 363y2 -66y
J I y-2 y 35 "^^ o oj J
= 3885 .*. »= III miles.
75. A broker has two kinds of money. It takes
m pieces of the first to make a dollar, and n
pieces of the second, te make the same sum.
Seme one offers him a dollar for r pieces ;
how many of each sort shall he take ?
z Let X = pieces of the first ; and y = piece^
of the second. Then niY. -ny— i.oo. If
m=Jy x= 100 ; ;;^ = 2 ;x = 5o ; m = 4, x =
25;. ;;^ = 5»x = 2o; ;;^=io, x=io; m =
100, x= I. Again, if ;^= ico, y=i ; n =
20,y = 5; ;/=io, y=io; ^^ = 5, y = 20.
U r=:Sf then we have^ ;;^ = 5, and ;^ = 20.
.*. 4x20+4x5 = 100; .*. he takes 4
44
twenty cent pieces, and 4 five cent
pieces.
76. A, B, C, are 3 equal vessels ; the first con-
tains water ; the second, wine and the third
contains wine and water. If the contents of
B and C be put together, it is found that the
mixture is 1 1 J^ times as strong as if the con-
tents of A and C had been treated in like
manner. Find the proportion of wine to watpr
in C. y'■•'>■■'^■-
Let X denote de wine in C ; then C - x =
-water in C. Then
1st mixture.
Bjfx
C -X
= strength of
\\r-__ = strength of sec-
ond mixture ; ^'^^ = "^ ; .*. a + x-x^
I - X 2 - X
-ii}i x-iij^ x^, and x^ - x ?= - jy •'• x
= .73425 = wine in C ; I - .73425 = .26575
• = water in C.
77. The discount on P dollars due x years hence
: $ P :: m : n, find rate of int.
Let /'ssrate, and x = time ; then loo-f rx
P'>* X
-——J = discount on $ P for
^^,
%
-■Mi
''if;
'■i>
: rx : : P:
*^-
45
X years ; ; : r : : m : n, or^
: I :: m : n: r x : loo+r x : :
r X
loo+r X
m : ft] this proportion gives x =
TOO
(£)
or r=
100 / m \
X I n-m. I
78. A steamboat started from Hamilton, and
sailed down Lake Ontario. Owing to rough-
ness of the lake, during the first four hours,
she only sailed 23 miles ; of this, the third
hour's sailing was double the first ; the
fourth was i ^ miles less than sum, and the
second was yi of the distance =18 miles
more than twice the third hour's sailing.
Continuing this progression, how far was the
boat from Hamilton, at the end of the tenth
hour ?
Let x^Ist hour's sailing; 2x = 3rd;
==4th;4^i±iK=2nd.
^ 3
22 X =
7i>i ; x = 3J<. Then 3J<+4?<+6>^ +
8>^ + 10^ + i3J<-h 16+19+ 22K + 25^
= 130 miles.
i
■teiita
IM
46
79. A local superintendent divides $360 between
four schools in G. P., and each receives as
many dollars as there were pupils attending
it on an average. N". i receiving the smal-
lest share. It was then found that the.amount
received by N" 3 exceeded J^ of that received
by N" I, by $90. Required the 4sum appor-
tioned to each ?
Let x = first term, and r the rate ; then
x-f rxH-r^x-f ^^x = 36o, and r^x-| = 90
.'. 4 r^x-x = 36o ; and (4x^~i)x36o,
^ndx = -^Then 36o . 360 .
4/^—1 4 x
2 _
4r^— I.
= 1 .•. /^ +;^ -f-r-f 1=4 /^ -I, and r^ -
3 ^ -!_;.= — 2, solved, x = 24 /! r^2\
hence the respective values or shares,
are : 24, 48, 96, and 192 = 360. W. D.
So. A, B, and C commence bunsiness with an
aggregate capital of $1200 ; C*s capital
exceeds A's by $100. A sells out in 10
months, B in 7^2 months, and C in 54 months,
and each receives in stock and profit $700.
Find each per.son*s stock.
iiL.u^
,JmA^
..vOj^^,--^,,:.^-. ■.■...■>^!..:...:^...\..*.i--..;.^ ..■>.■.
^H^^d^i^iMiiriitaHiMiMMIiillMliH
■T *** S? ' ^ *,"^ f ■« ■ fT'"*^ "."^ f^
47
Let x = A's capital ; ioo4-x = C's ; then
IIOO-2 X=B's. XX io + (iioo-2 x) 7^2
+(ioo+x)x5«=5J^'^^50.
tly, 12^+79250.
; consequen-
16200 X
10 X
900
9 .. , ' * x+ 15850
= x .*. 162000 x = x''^ 4- 15850 X .*. x = 35o
= A's profit. Then 700 — 350=350==
A's stock. 1100—700 = 400= B's stock ;
x4-ioo = 45o = G's stock. J.C.
81. A miller mixes flower which cost him $5 a
, barrel, with some which cost him only $3 a
^ barrel, and sells the mixture at $5.40 per
barrel, making 42.5 per cent. Required the
proportions of the mixture ?
Let X = first ; y = second ; then 5 x 4- 3 y
= first cost ; 5I (x 4* y) = selling price ; or
5I x4-5l y = (5 >^ + 3 y/ = l X + 2I y = pro-
fit. As 5 x4-3 y : | x4-|y : : 100 : 42J.
Whence, x= 15, y = 23, and the ratio is
15 : 23 or 23 : 15.
82. When two metals are mixed in equal volu-
mes, they form a compound of specific
gravity 9. When mixed in equal weights.
jk
x.r
jjj^tij^tMiijMim
■lii
'J.iS?'?. ■■;■
^^^"
they form a compound of specific gravity 8^.
Find the specific gravities of the metals.
Let W, w, and w' denote the weights
of the compound ; w, w' being equal.
S, s, and s' = specific gravities. (S — s')S:
-r(s-s')SxW-w;and^4^^§^ xW
(s-s ) X S
= w. Then ^-^^P^-~— X 10 = 5. From
(s-s ) , r ,
this 4o(s4-s')=9 ss'. Again ^ 4-
5000 S _ 50D3 S 4- 5000 S . g _j_ g'^jg"
s' 9
By substraction, transposition &c., 8ic,J
we obtain s = 10 and s' = 8, the specific
gravities reqd.-
83. Two men, A and B, take shares in a Petrolia
oil well to the aniount of $1850. They sell
out at par ; A, at the end of 2^ years ; B at
the end of 7^, and each receives in capital
and profit $1400. How much did each em*
bark ? ^
Letx = A's capital; 1850 — x~B*s; xx
2>^ 4- (1850 -x)x7>i = 13875 -5 x =
sum of products. 13875-5X : 950::
49
2jx :
. 4?5x _
2775
A's profit; 13^75-$^ '■
95o::i3875-7ix: ^^36250-i425x ^
^*^ O /J /» 2775-X
B's profit. As 950 : 13875 - ^ x : : 1406
_ ^ , 3885000-4175 x+x^ ^ ^^ j^^j ^f
190 . '^
A^s capital and time .'. we have,'
(3885000-4175 x-|-x2)-r475 = A's capi-
tal =x. From this equation, we obtain
x= 1091.863755 ; and 1850- 1091.863755
B's capital.
84. A renner had a tank full ot alcohol, contain-
ing 177 147 gallons, from which he drew a
certain vessel full, and filled up the tank with
water. He repeated this process 1 1 times,
when he found only 2048 gls of alcohol left
in the tank. Find the capacity of the vessel ?
- The quantity left after the nth draw is
(17714.7 -~ xV'
fouad to be ^ '' ^' ^— ' 2048 ; .'. 177 147
177147/° ^ /# T/
-X = 1771471^X2 : and 177147- x= 1 18098
. • . X = 59049 - Yz tank.
85. There are two quantities, m*a.nd n whose
. arithmetial mean is x ; . the geometrical
■* *•>
■JJ.kjiik.>M I I I'l III"-'
so
.-1
mean is y ; and tjie harmonic mean is sf;
if x^y=3l, and X — z=«4twi» find «» and
^"^^ ^. r — *. ^^^
n.
z.
Putx=*«^; y=;«. Then
3I; and
m-^n 2 m»
2
— .1 76
--4T¥Y'
fJmu
Then
X - y = 3I and x - z = 4t^|t. x + y * 2jxy
•f 7V, and x4-y= ^^+9T?T.*.wehave
4xy
4xy
• 2 fJxy H- H-
^^FyHhyr'^"^^^""^' ^^^^' ""^
fractions, ^ ^— ^ 1 + 252 -« 250 Wx y
.V500 xy-h 504 iJxy + 18141 = 500 xy+
1800 ^^i 296 ,J5^=*i8i4| .•.5iN/xy='7»
and,J^««li,aad xy-y ; x+y«=2V+'
Tiswio. Now, we have the sum, and
product given, hence x«9t, m^g\;
y«V •*• «*•• -
86. The dtffcretiir between the quotient add
divisor is 81339, atid sum of the divisor and
m.
■ 7t
«•»'
51
dividekid is 3007249* It is required to Biid
:e&cn« '9i^ii^m^
' Let X « divisor ; then 81 239 4^ x =* quo-
tient; but(8i2394-x)xx*=dividefid,and
3007249— x« dividend ; /. x^ +81239X =
3006249 - x; whence x^ + 81240 x =
3G07249. Then X =* 37.
87. The time/ rate» principal, and gain at com^
pound interest are all equal. Required the
time?
Let X denote each ; p R^ ^ s ; per ques-
. tion,^=x; r^-^^ R=i+-^; t = x;
/ X \x
2 X .•. X X j i X — - — I = 2x; divide
\ 100 J
by x;
X Vx
1 4- = 2 ; by the nature
100
Wgs, we have x x
(
\
1 +
100
.3pro300.xx \^r--^
ipo 20000 3000000
X M =
&c
3010300 x^ 3^
M
.X-
&c.,
100 aoooo 3000000
6g|i 7j by reversion,x ij« 8.49824, the answ.
fW^-^^-" f'^^'''fyT'^y''<^yr
5* •
#■ ' -
88. In what time could $j 5 amount to the same,
if placed at 6 per cent simple, and ^ per cent
comp0ond interest.
By a few trials, the time is found to be bet-
ween 43 and 44 years ; then by the general
rule of **Trial and Error," the answer is 43)4.
89. A merchant bought a cask of spirits ^£4^^
and sold ^ quantity ^exceeding three fourths
of the whole by 9 gallons, at a profit of t5%.
He afterwards sold the remainder at siich a
price as to clear 667;. ,on the whole transac-
tion ; and had he sold the whole quantity at
the latter price, he would have gained 175 per
cent. Find contents of cask.
' Note.— Henri Mondeau. the Shepherd of
Touraine, a wonderful French youth of
extraordinary powers of mental calculation,
having visited Jersey in order to csitibit these
powers, and when asked the foregoing
question, answered it almost instantaneously
as fottlows :
He sells the^ first portion at a profit of 35
per cent, and the last at i75 P«* ^®"*» ^^^
gahied 60 per cent on the whole. The first
profit is hssBi Aan the meitti profit Jg?^ 3^ per
I '»>,.
S3
cet3t» and theisecond ts greater by ii^ per
cent I he has .*. sold 115 parts ot the, first
against 35 of the second, that is, the first
portion sold w^s \H of what the whole cost ;
and the last -^ i but the first portion was |^
of cask and 2 gls. more ; and the difference
between {^^ or }J and ^, is ^V • *• 2 gls. ,s
^ of cask . *. lao gls = whole.
90, Which is the best interest, at 6 per cent com-
pounded annually on ^1000 for 20 years ; 5%
compounded every instant ?
Let X = 00 all the instants in 20 years ;
^=1000; thenA = « ^j -I- ^^ x = ^
(I^^y^^ x(x-I) ^r^ ^ X(X->I)X(X^2)
X 1*2 X2 I '2 '3
X ~ &c., hence, since x = infini ty, x (x — i)
I, and the series is. (I-f■I^-J^-
&c.>== 1000 X 2.7828= icoox,
X
1.2 1.2.3.4
Naperion base of logs = 27i8.28, at 6%
it is 3207.18 ; diff. = 448.90 in favor of
6%. J. Ireland.
91. A cylindrical piece of wood is 12 feet long,
and ^j4 ft. in diameter.^ Find the solidity
ijMet ■.
54
of the greatest paralleloptpedon that am be
cut out of it. Find the area of a square
i whose dtagonat is 3j^, and multiply this area
by the given length.
92. A farmer uses a roller, 4 feet 8 inches wide,
and 2 ft. 8 in. in diameter. How many
revolutions does it make over 7A 3R 25P ?
8809.32, the answer.
93. A cow is tithered with a rope sp as to graze
over I A 35 P of pasture ; but the grass being
insufficient to teed her, what additional length
of rope will allow her the use of another acfe?
Answer 16 yds i fooi.
94. Required the dimensions of an upright cylin-
drical vessel, capable of containing 16 gal-
lons, when the depth is equal diameter of the
base? .7854x2 X X = 277.264 X 16. This
equation gives x = 17.809 = height, or diam.
of the base.
95. If into a cylindrical vessel whose inner diam-
eter is 3 inches, we put as many wires of
^ inch diameter, as possible ; how much
water can be afterwards poured in, allowing
the height of the vessel to be 12 feet ?
96.
• 55
Put radius of the cylindriciir vessel suf* and
small radius = 4 ; thep the number of small
circles that can touch the original <y* and
one another, in one rounds is found as fol*
lows : Let the required number of (circles = x ;
then sin. (-^eo-rax) = s, and iis-7-(s+i) =
i ; .-. X will be found = 128. There are 21
terms in the progression, or the rad. must be
divided into 21 = parts, each =diam. of the
small circles. Then by the foregoing formula, *
we. have 21 concentric circles to be described
whose diameters are : l}^, lu, lu, Iuj ^u»
&c., &c., and the number of small circles of
1 inch in diameter that can b^ described
between each pair of circumferences, in suc-
cession are : 128, 122, 116, no, 106, 98, 92,
86, 8o» 74. 68, 62, 56, 50, 44, 38. 32, 26, 20,
14, 8, 2 ; their sum = 1420, total n^ of wires.
Area of end of each ^[^f x .7854 =004007
X 1420 = 5.7 square inches covered by the-
wires. 3^ x .7854 = 7-o689--5.7 = 1.3686,
area of empty spaces at the bottom 1.3686 x
12x12 = 197.0784 cubic inches of water the
vessel can hold betv^een the wires.
There is a ppint in an equilatorial triangle,
&'M.'>.
■■ip
S6 .
from which 3 lines are drawn to the angftes,
measure 2 J, 2, and 1} respectively, construct
the triangle and find length of side.
^he 3 given lines are 1^, 2, and 2^ tespec-
tively equal half 3, 4, and 5 .*. they form a
r < ^triangle. Now draw a line A B = i J ;
from A erect j.* A C = 2 ; join B C = 2 .** Oil
A C des-
cribe an
equilate -
ral trian-
gle ACD.
From D
let fall the
X DEon
BA pro-
duced ;
and from
C, draw C F at / <s to A D, bisecting A I>
in F. Join E F ; join also D B, and on D B
describe an equilateral triangle DBG. Then
DBG is the triangle required.
It is easily proved* by the principles of
equilateral triangles, that triangle D C G and
D A B are equiangular .'. C G = A B = ij ;
C D = 2, and CB=2^; .'. C is the point
^^
f
57
within the triangle G D B, from which the
, lines C B, C D, C G, drawn to the <s at
B D G are = 2J, 2, ij. Triangle D E F is
easily proved to be equilateral -*. D F = i ;
A F • I .'. Jj^ - I- = fjj = 1.73205 =
C F. ,J(C F + CG)« "^1 = is/11.4461472025
= 3.383214, side of required triangle. —
D. E. Scott.
97. Prize Problem. — If a rifleman man can
plant 1 1 per cent of his bullets within a circle
of 1 f pot in diameter, at the distance of 100
yards ; find the diameter of that circular
target which he might make an even venture
to hit the first, shot.
■""""" r
Formula. — a-r^l Log- 2
V
V
Log. 2.
Log. m. 1# Log.(i — H).
Let /*= J^ foot : H=tWi: ^^i^* whose
log. is 0.050610, and log. 2=0.30103, then
« = J^\/?5^ = 1.2195 + feet, which doub-
^M 6061
bled, gives 2 feet 5.2656 inch, diameter
required. — A.D.
98. Another Prize Pr6blem.— A sledge loves
down an inclined snow-plane, 140 feet long,
and 30® inclination ; and when arrived at the
bottom, procejds along a horizontal one
.■..-■:.,,.-gf..-A:-j^. u.^ j.-l..v.^---. -.-r... .^-.^^\i: :i^:iie:::t.^^i-^.^.j^^^L.L.LJ2L.^,-....i^^
iuiiiiiiil
y '
. -s., ^
S8
until friction brings it to rest.. If the coeffi-
cient of friction between the snow »nd
sledge be taken equal .05, what space will
the sledge describe along the horizontal
plane, neglecting resistance of the air ?
A B « 70 ; B C »» 121.2435 ; A C « 140.
Bisect A C in H, and draw the x H G. Then
|>y similar tri-
angles H G «=
BC; but HG
IS the COS. a
(30*), or COS. a
X coeff. of fric-
tion « friction
.'. I2I.243$X
.05= 6.062175
units of work destroyed. Or, the Normal
reaction multiplied by the coefficient ot fric
tion is the friction. . • . .866025 x .05=043301 ^5
X 140=6.062175, units of work destroyed .'.
70 w- 6.062175 « 63.937825 = units of work
accumulated in the sledge at C .-. 63.937825
-.05=1278.7565, th^ answer.
Formula W (h - base x coef. frict ^^)
(70-6.062175^11) === 1278.7565. The
followincr is another solution :
f--
59
The pressures : « :: base : 140.'. 140^^=
131.2435565
121.2435565 w, and jf
140
pres-
sure.
work of friction.
121.2435565 ^ I _ 121.2435565
i4<^^ 20 2800
70 w _ 121.2435565
140 280P
work of gravity over friction. Then, V**
r 64t >^ 1278.7564435 wx i4o -| yi ^ . ^^
units of work corresponding to this velocity
w
«63,937822i75w^ — =" 127S . 756435435*
A. D -
9^. The frustum of a right cone is 24 inches in
diameter at one end, and 16 inches at the
other end, and 60 inches in height. At what
' height will half the solidity of the whole be
found ? Answer : The segments of the line
representing the height are : height of lower
frustum = 24. 17 1 2847, and height of upper
frustum = 35.8287 16. Limited space prevents
an entire solution.
100. In a railroad excavation there is a rock to
be cut, % mile long, 40 feet wide at the top,
and 25 feet d6ep. A contractor agrees to
6o
complete the work for $59888!, or $.175 per
cubic yard. How wide must he leave the
cut at the bottom'? Answer 16 feet.
Let X denote breadth at the bottom. Then
40'f-jc 1000+22^
2--— X25-
iooo+25JKr
X 1320 =
1320000 + 33000 Jf
2 X 27
cubic yardsofexcavation=» if x
660000 -^ 1 6500^
27
000 ft 660000+ 1 6500 AT ,^
= 59888I . •. = 34222} ;from
27
this jf=i6.
101. The height of a cone is 41.088 ft., and
diameter of the base 8.56 feet? Find the
length of a straight line || to base, and diam-
eter of a circle that will bisect the cone.
Let 41.088 = A, and :r-the bisector. ; then
8.56«'x .7854x^— ^ = 788. 192274= solidity
of the cone, and its J^ = 394.096137. Then
8.563 : jc* :: 2 : i ; hence Ji; = 6.78575' x 7854
= 36. 1 7550475 = bisecting circle . •. 394.096137
■^3^' 1 7550475^ — ; whence ^'=32. 68.
3
■^fl
6i
103. Any part may be cut off, if instead of the
ratio 2:1, we use the ratio of the whole
solidity : the part to b^ cut off.
103. Find the side of the largest cube that can be
cut from a sphere 10 feet in diameter ? The
solidity of the greatest cylinder that can be
inscribed in a given sphere, is the revolution
of the greatest inscribed cube, and the side
• of the latter, is
1 10'
/T
5-7735-
104. Three men stand on a plane, in the same
straight line ; the first is 6 f t 2 ; second 6 ft ;
and the third is 5 ft 9 inches high. The
distance between i*' and 2"** is 10 ft. Find
the distance between the 2^ and y^^ whei^
'■ /
the tops of their heads are in the same straight
line.
A C=74 inches ; E N=72 ; and B 0=^69.
74 -: 69=5=-L C ; 72 - 6g=3=3==^S N or L P
.*. 2 : 10 :: 3 : 15 = 8 D or E B.
■J
^ .i
h-
:■ '<&.':
'•:-■. ■*■
HAXIMA AND fllNIMA.
105. Find the greatest rectangle that Can be
inscribed in a given triangle. The greatest
inscribed rectangle is when its height — j^ A of
given triangle.
106. To find the shortest, line that can be drawn
through a given point between two other
^ lines forming a right angle.
Let TB and BV be the two indefinite
straight lines, B the right angle, and P the
given point. From P draw P M x hase T B,
and draw PA || T B ; then P M i^ given = ^,
. and P A==a=^M B. Denote T M by jf ; then
X is found « ^fja ^ ; the twa points T and
B being then giyen» we determine th^ direc-
tidn and lenght of TV, '
107. To find the greatesit trapezium that can be
inscribed in a given semicircle. When the radius
==a, and height of trapezium =^ ; then x=
X
~,<H< , ,1^
^ .. .
]. ■ • ■■ . . . ■■ ■
1 08. Of ail the cylinders th^t can be inscribed in
a right cone, determine that which has the
greatest solidity. — Let height S C of the cone
=a ; rad. A C of the base =d ; then let S D,
height of upper section of the cone=ji[;, .*.
the lower section D C=a-x ; then the solu-
tion gives a-Xf height of the required
•. height of greatest
jnsC, cyV. =r j4 height of cone.
By this result, we can find the sideof great-
est inscribed cube or globe^
^09. AB represents the front of a street lot ;
^ re«tuir«dthe minimum length of the three
^remaining lines bounding the lot which con-
80
tains 80 perches. Let ^ r A B ; A C s — x
. X
., . 2 a a
cylinder = /J = -
a + 2 a; = sum of the 4 sides r
160
2 AT r
i6o + ajif*
min. solved gives jk: p 4 ^c
j[xt>. To find the greatest rectangle that can be
inscribed in a given semicircle. The greiUest
rectangle that can be inscribed in a semitclrcle
is, when its height is equal the radius divided
by the square root of a.
r
h\t^^ ..
!
1 i
64
I CI. The equal sides Qf an isosceles triangle are
opened from the vertex ; what must be the
length of a line joining their evtremities, so
that the quadrilateral thus form d, may be
the greatest possible ?
Answer, 4 ^ = ^^a + 8 *» 7 ^'
£'.■■• ^
. V- ■ - . ■ (
ii2. If the given triangle is equilateral, substitute
ing a for b, the result is 4 at = ^^a ^. g «■
a •
-a; or ^ a-a '2 a ,\ X - — and the maxi-
mum trapezoid equals 3 times area of the
' original triangle.
113. Two cofds, one of 6, the other of 8 feet ia
length are attached to a weight 100 lbs, and
fastened at their extremities to hooks in the
ceiling, 10 feet apart. Required the strian on
each cord.
The sum of the sides represents the whole
strain on both sides ; and the strains on the
sides are in the inverse ratio of the length of
the sides. Then, as 10 : 14 :: xoo M40, the
total strain ; .'. 14 : 140 :; 8 : 80 = strain on
€ ft cord. As 14 : 140 :: 6 : 60=^ strain on
8 ft cord.
i<i%
ItittMb
6s
1 14. A bar of wrought iron 150 ft long, and^ inch
square in section, lengthens .289 inch under
a certain strain ; what must be the additional
strain necessary to produce rupture ?
L = 50 ; strain 2240 lbs gives . 289 ; / :
L ::
2240
~7r
9
I 29.000000, or modulus of elasti-
city. .• . 290 / = 84, and / = .289. The strain
sufficient to produce rupture is ^j x 6720
tenacity = 2688 - 2240 r 448 lbs the additional
■ strain. '- .. ,:,■.;;, . : - v> \-..- -..j,
115. An iron wedge whose angle is 14", is driven
into a rntlass of oak by a force of 125 R)s.
What force is necessary to extract it?
' (3>'5o +7")
W
sin. 38^ 50' ^ 627057- ^' - -66976x135
:S4.72.Ibs, the force required.
116. A beam of oak i foot square, has its end
firmly embedded in Masonry, from which it
|^rojectS9feet ; to what height could a wall of
brickwork, 2 feet thick, and resting on the
4)eam, be carried, without producing rupture ?
A cubic foot of brickwork is equal to 1 12 lbs*
n
66
^..
Let a = natural length of the beam ; b its depth,
and c its breath. W = — x ; s being the
. 3 «'
"Inoduius of elasticity. Then, w - — x
3
12x144 42 . 1. r
^-^ r ^4""«~ ~ pressure for every inch of
42
length of beam. Then 246 -^ x 108 = 26618
89
86
— — 5)s, sufficient to produce rupture. 9x2x i x
86
112 =20i6.*. 26618-^ -r 2016 = 13.203 feet
high.
117. The rafters of a house are each 18 feet long
and tied by a wrought iron rod 30 feet long,
and section ^ square inch.
What weight must be sus-
pended from the vertical an-
gle so as to break the rod ?
Wx 30^-4 AD = horizontal
pressure ; A D = 0,95 .'.
WX30 67200
39-8 " 4
.*. Wx;Jt) >
i; il
39.8x16800 .*. W = 39.8x^60 : 32288 K>S.
118.
•i
I 19.
120.
IJ^vi
■jat^tiu^glitfiiliiiim^ttiitgbuuittit
-WpUUfiV'^R^
67
118. A bar of wrought iron suspended vertically
breaks by its own weight ; what is its length ?
The tenacity of wrought iron = 67200 lbs. Let x
■*' = length of ihe bar and n the area of its section;
67200 n = breaking weight of the bar. Specific
gravity of wrought iron = 7.788 .*.
7788
r X
144x16
• u. i-u 7-788
nx = weight of bar ; ,\^—^ — x , « at = 672O0 n
2304
V ,% 7.788 X =67200 X 2304 ; X = 19880 feet,
length of bar.
1 19. If the traction power of 97 lbs is required to
draw the fore wheel of a carriage over an
obistacle 6 inches high, what power will be
required to draw the hind wheel over it ; the
diameters of the wheels being 3)^ and ^%
feet respectively ?
As the required power varies ih tersely as
the radii of the wheels, we have 4^4 : 3^ ::
97 lbs : 75.4 =the required power.
120. To what depth may an empty glass vessel,
capdble of bearing a {> v^ssure of 216 lbs to
the square inch, be sunk in watei' before it
breaks? ,
.03616 lbs avoir du poids £ weight of one
cubic inch of water, at a temperature of 60^ ;
68
then 2i6-f. 03616 =5973-45 inches.-. 5973.45
■7-12 =497.7875 = the required depth.
121. If team of horses draw 3500 lbs, and the
center pin is removed one inch from the
center, how much will each horse draw ? If
the center pin is moved one inch from the
center to to the right or left, the horse draw*
ing on the short end wiU pull about ^ more
>. ' than the other. 3500 x-jV = ^75 lbs, differ-
ence. 175 + 1750 = 1925 ; 1750 - 17s = 1575 ;
hence the diff. of dratt = 350 lbs. ' --
122. A owes B $455, payable in 14 years, viz, at
the end of every two years $65. But he agrees
,, to pay him in 7 years by equal payments each
^ year, which B agrees to, and at the rate of
6% compound interest. What must be the
annual payment? First, find the present
- worth of the seven payments, which were at
first to be made, which is found to be
% $293.2583. Then find what annuity to con-
tinue 7 yrs at the given rate $293.2583 v/ill
purchase, which you will find to be $52.5^,
the answer required. /
123; If a body weighs 60 lbs at; a distance of
3000 miles above the earth's surface, what
^m
i"l"iH' I
, r-. ,"v •
69
will it weigfh at 3000 miles below the surface ?
Find the weight at the surface. Bodies weigh
directly as the tnasis^, and inversely as the
square of distance above the earth ; but they
weigh directly as the mas^^^below the surface.
Then the body is 7000 miles from the center,
what will it weigh at the surface? less .•.
7000^ : 4000* ;: 60 lbs : 19ft. If a body
weighs 19!^ lbs at 4000 miles from the cen-
ter, what will it weigh at 1000 miles from the
center ? less. ^
As 4000 : TOGO :: i9ff : 4f^ lbs, the answer.
124. A belt is 16 inches
long, and drawn
round a circle whose
diameter is 4 feet,
until it reaches a
point (P) in the dia-
meter produced.
Find the distance
FP, BP. Also, if
the circumference is
equal to the length
of the belt, BPmust
be equal to the arc
BF, 4ndBP + B'P
70
= the arc B F B'. Let belt PBDCEB' = i6 ;
3.1416x4 r 12. 5664 -f 2 r arc DCE = 6.2832.
Then 2^ + 6.2832 = BDCB';2^ + 2JC =9.7168,
aLndv-\-x = 4.8584 ; which is an indeterminate
equation. , ^
If xzi; zy z 3.8584 ; and^3.8584» +2«
= 4.34590 = A P, .-. FP = 2.34594. FP + 4 =
- C P = 6.34594 . •. ^6.34594 X 2.34594 = y =
. 3.8584; 2j/ =7.7168; BDCEB' =2 + 6.2832
= 8.2832+7.7168 = 16.
125. Two men engage to excavate 100 square
«
yards, divided by its diagonal into equal
parts of sand and rock. They work frorn
opposite sides, each, at sand and rock, until
they meet in a line parallel to the sides at
which they com-
menced.* Each
must receive
equal parts of the
whole cost which
is $100 ; the sand
at 75c., and rock
at $1.25 per yard.
How much rock
and sand must
each escavate and ^^^
where will they ^^
7J
2'
fc^'-A
'■^
••■■'*■
meetr^LetBE =jerAE = io-:»;; then EB = EO,
and AE=CF=FO. Side of Mpare = lo ;
A B C is sand and B C D is rock. Then
lO-f
fy (io-a:) xH +
lO
300 - 3JC' 500 - iooa: + SJC** ^ 300- 3^'
8 "^ 8 '\'> 8~'
. 500- 100^+ icjc^ 2x' . 100 ^-5 a;*,
_|-2 ^ 5 — =^-— H Q-^~^ ;hen-
*•■•'■, ,.\r
i'-Tf^/C'-Lin:
ce we whave x^ - ^o x ^ - 200 : solved gives
^=4.38448; 10- a; r 5.61552.
126. If E and D be the points of trisection of the
sides A B, AC of a triangle (nearer to A),
and F the point ot intersection of C D and
B E ; prove that the triangle B F C is half
the triangle ABC, and the quadrilateral
A D F E is equal to eithej ot the triangles
C F E or B D F. On A B, side of triangle
A B'C, make A D = ^^^^ A B, and make A E
r ^ A C. Join BE, CD, E D and A F.
The triangle B C F will be J^ of triangle
ABC, and the triangles E F C, D F B, and
the quadrilateral A D F E will be equal to
i one another. For in similar triangles A D E
and A B C, E D = J^ B C, and in similar tri-
angles DEF and BCF, side EF = ^ FB ; but
l|||fi|«fi^^l!i^»iP'w-i^i^'j *|i .u -r^^!^'-^^/ .'-r-'. '■
7»
triangle C E B, by construction, = ?^ triangle
A B C or # A B C ; but th^ base E F - }{
BE.-, triangle E F C = J^ A B C and B CF
r I or yi ABC. Triangles BCE and
BCD on Same base are equal » take away
common triangle B C F, and EFC and DFB
will be left equal. It can be readily seen that
the two triangles forming quadrilateral =
E F C or D P B.—D. Scott, C.E.
127. The girth of a heifer is 6)4 feet, and length
from the shoulder blade to the tail bone 5. 25 ;
6J^« =42.25, and 5x5.25 t 26.25'; multiply-
ing this together, and dividing 1.5, gives
739.375 lbs the approximate weight of the
heifer when dressed. A shorter method i%^
to multiply the square of the girth (baqk of
the fore shoulder) by the length ; then multf-
.,,..^
• ■ (, ,«.,'.
t of barrels^ casks.
. 73
ply that result by 7 and divide the product
'by 2^ ", '';f^ ^.rf^'. vr
laS. To find how many bricks in a watl or buil-
ding, multiply the length, height and thick-
ness ^ in feet by 20. A brick 8 x 4 X 2 = 64
inches.
Jo find the Ci __
juare one hJ& the sum of the bung and
head diameters in inches, and multiply by
the height In inches ; then_multiply by 8, and
cut off the I ight hand figure ; this gives the
cubic Inches which divided by 277 Ji( gives
.the number of gallons, and divided by 2150.4
gives the number of bushels.
130! Required the contents of a barrel whose
^ iniddle or bung diameter is 22 inches, and
diameter 18 inches, and 30 inches high? 22 4-
J, j tS 4-2 =20, the average diam. 20 x 20 x 30 X
fiB =9600 -T- 277 J<Q54^ gris-
131.- How many gallons in a round tank, 6 feet
in diameter and 6 feett high ? 6x6x8-
288.288 X 6 = 1728 gls or 1440 Canadian gls.
A cistern is 5 feet in diameter and 8 feet
deep, how many barrels will it hold ? 5 x 5 X
8 = 200-5-5 :^ 40 barrels,
132.
V t\
liliililiiilMiall
mmmmfm»
133. From the top of a mountain, A miles high
the visible horizon appeared depressed a
degrees ; it is required to show that, if d be
the distance of the boundary of the visible
horizon, and D the diameter of the earthy
d^hcot }i a; D + h =cot? }i «. Let A B
be the height of
the mountain, BC
the diameter of
the earth ; < P AF
the dip, and AF
the distance of
the horizon. Joiii
CF and FB, and
produce them to
meet a line
through A at r<s
to ABC in D and
E. Now the > D
is common to the
r<d triangles D F E and D E C ; . •. <f C =
<E;but<AFE=<C; .'. A F E =E ;
hence A F^E and A E F are each = J^ D A F
r i^ «, and A F r A E. Again < D is the
complement of <; E, arid A F D the comp.
6( A F E ; hence < A F £> r D and A F =
is '*S
■MHIilMIIIHlii
'34
75
•
A D. The r < flf triangle ABE gives A E =
A B cot. E ; hence d^ h cot. ^ a* Agait>
r < d triangle CAD gives AC = A D cot C i
hence X^-Vh - d cot. ^ a -h cot? y% a.
Example. — From the top of a mountain 3
miles in height, the visible horizon apperired
depressed 2^-/»i 3' 27" ;. Find the diameter of
the earth and^ the distance of the boundary of.
the visible horizon.
Log.
Cot.
%. h = 0.4771 21 \
t. }i fi - 1 1. 71 1941 j
sum = 2.189062 ; d =
= i54.:^4; + ^=7961.3.901003.
h =
.>
• .*, D = 7958 miles the diameter of
the earth.
135. To construct a horizontal sundial. — In any
straight line A B take a point O, and from O
\, erect a x -0 T ; take O as center withe the
chord of 60^ as radius and describe a circle
F T S D. F S will represent the 6 o'clock
line, and O T the meredian or 12 o'clock.
From T on the quadrant arc, lay off T H =
the latitude and join O H ; from T draw the
X T P which will be the sine of the latitude.
Make T G = T P, . and from G as center with
iHilittilili
y6
G T as radius describe the quadrant G T L
and divide it into 6 = parts. Through each of
these parts draw the lines G U, G V, &c.,
meeting- a tangent from T, indefinitely produ-
ced in the points U, V, W, X, and Y.
Through these points respectively, draw the
lines O U, O V, O W, O X, and O Y, which
will be the hour lines from 12 to 6, p.m. Make
the arcs on the quadrant FT r respectively to
{•/"'■'-..^i''-: :.
ff-
■w
fT^^
^
*.
'^i"
:f
'Ui'
-i'/
%^
. ■«' '
, fe.
!^'-
;!,'
,\
.'?iv
\,
'i-S
rf
•^'■.-
"■■ii-.-
-%
■■H-.
these, and we have the hour Unes for the fore-,
noon. 7 (a.m.) produced gives the hour liiie
for 7^ (p.m.) and 5 (a.m.) gives 5 (p.m.) &c.
To find half hour lines, divide the quadrant
GTL into 12 r parts and proceed similarly.
The quarter hours will be sufficiently appro-
ximate by bisecting the half hours arcs. ^
The angle of the gnoTion or style, standing'
j^" ly on the meridran Hn^. O T, must exactly
touch FS at O, and the meridian line must
? ; coincide with the breath of the gnomon. The
* < of gnomon must be equal to latitude pf
the place for which the dial is constructed*
To describe a south inclining vertical diaU
we use the complement of the latitude ; then
the construction is the same as the foregoing,
lo describe a dial for the equator. Divide a
circle into 24 equal parts and place a x style
in the center. This placed on an inclined
plane, having an angle equal the latitude of
a given place, will show correct solar time.
A dial engraved for a given lat., will show
correct time by placing it on a wedge having
an < = the diff. of the two latitudes.
136. The three sides of triangle are 18, 12, and
10. It is required to bisect it by the shortest
78
line possible. Describe an isosceles triangle
A F G - }4 triangle A C B, having a common
<^ C A B, by the VI. 15, and the base F G
is the line re-
quired.
Bisect A B
in E ; then
A E
r 10.3^2304 = A G or A F. Area of triangle
A B C = 56.56854, hence A F G = 28.28427.
Denote^ F G by x^ then we have the follo-
. wing equation ; f^a' - x^ x ^ = s = 28.28427:
this equation gives jc = 2.828427 x 2 = 5.656854
r F G, the required minimum line,
137. A, B, and C in partnership gain $1800. If
we takeC'stimv^ from the sum of A*s and B's,
7 times the remainder will be equal to 1 1
times the sum of A*sand C'sdimished Hy B*s
C's stock is to the sum of A'sand B*s stocks;
as A*s time is to 6 times B*s time ; the sum
4Qi all their times divided by the sum of B's
and C*s minus A's, equals 19 ; and 3 times
the difference between the stocks of A and B,
is equal to twice C's stock. Requlired each
person *s gain, by simple proportion.
79
7 (A's + B's-Cs) time r n (A's + C's-B's):
hence 7 :ii :: A*s + C's-B's ; A*s + B's-Cs;
then f8 \ \ :: 2 A's : 2 B*s - 2 C's ; .*. 36
B's - 36 C's = 8 A's or 9 B's - 9 C's r 2 A's
also,9 B's-1-9 C's : 10 A's, hence 18 C's =8 A
.•• C's tkne = % A's time, and 18 B's = 12 A's
.•. B's =9X I2-T-I8 =6 ; then A's =9, B's =6,
and C's =4. Again, C's stock : A's + B's
stocks ;; 9 : 36; from this proportion, we
find C's = (A's + B's)-- 4 = (3 A's - 3 B's)-r 2 ;
this gives B's stock = \ A's.
Let A's stock rix9=9 \^5« ^^^o ;: 9 : 1080 =
A's g:ain.
** B's •* =4x6 =44/ 15; 1800;; 44:5144 =
B'9 gain.
** C's *' =fx4 = 14/ »5 J 1800;; 14:2054 =
Csgr'n
sum of products of
S & T - 15
■>.
$1800 =
sum of gains.
138. A and B are candidates at an election when
680 persons vote, and A is defeated. The
same electors vote the following, year, when
A and B are again candidates, and A is
successful, having carried his election by i|
times as many votes as he h-jfore lost by, and
his majority ; B's the year before ;: 9 : 5 ;
IBMdHi
gUMaM
mm
m -,,
So
how many electors changed their minds
during the year ?
Let I lo denote B*s gain and A^s loss the
first year ; then i lox i| r 198 = A*s gain the
second year. Then 198 : no :: 9 : 5, and
198 - no =8l5 who have changed there minds
at the second election. This question admits
36 solutions clear of fractions, giving as
many sets of answers ; and the majorities
change 36 times from 10 to 360 included.
139. A owes B $1000, and agrees to pay him in
ten equal annual instalments, at a rate per
cent, simple interest, equal to the TRUE
equated time for all the payments : how much
must B receive annually ?
Let ^ =th# rate = time ; then ^' = int. on
each payment, and 100 + ^' = each annual
payment. The most correct method of find-
ing the equated time is, when the interest of
the i*ums payable before the equated time,
from the times when they are due till that
time, should be equal to the discount of the|
sums payable nfter the equated time for the'
intervals between that time and the times at
which they are due. Then, when x is the
xs.
:"«&
8 1
equated time, the times for intefesntre -
•^ -- 1, .f -- 2, X - 3, .1? - 4, and ^ -5 years. The
times for discount are : 6 - *, 7 -- ^, 8 ~ •^%
9- *,andio~^ years. ^^ (100+ ^* ) X (^ - i) =
AT* - A^* + 100 x"" - 100 Xi X X (100 + ,ir* ) X
(at -- 2) r ;,>> - 2 .t3 + 100 AT* - 200 /r ; .tr X
(100+ a:' ) X (a: -3) = A^ -3 x^ + IGO at* -300.^;
:r X (lOO + Jic:* ) X (^ - 4) = AT* - 4 Ar3 + 100 JC^ --
400 ^ ; :v X (1^0 + -^* ) X (^ - 5) ~ ^* -- 5^^ +
100 X' - 500 X ; The sum of these products =
(5^ - 15 ^ + 5°o^' -: -Soo^} ^ Interest.
As tco-\r6x-x^ : 6;itr-rjic2 ;: 100 -♦•a:^ : \
6x3 u. 6oojir - a4 - - I ooji;^
As 100 4- 7a; -:v^ : yx-x' :; ioo + a:'^ :
7x^ 4- 700X - x4 - lOOX'
IOv?4-7X--X»
As ioo + 8x--x* : 8x--x* :: ioo-{-x^ ;
8x3 + 800X -X4~ lOOX^
ioo + 8x--x*
As 100 + 9X -X* ; 9x-x» :: 1004.x* :
9x3 4 900X - x4 - ioox2
ioo + 9X"x2
As loo+iox-x' : iox--x2 ;: 100+ x^ ;
iox3 + looox-x*-- ioox2
100+ I ox -X"
CA
•^^'T*** '
,'%.!"■ '
'^- fiv^^ :■<} 'i,'^ ,'"( 's'"'...,-
' """.■:*? :/!'wi,^i;ii!!.jM iiv:;',i",i!
8a
We now have —
6x^ ^ 600 - x3 -'^loox 7x « ♦ 700 - x^ - loox
ioo+7x-x2
ioo-f»6x - x^
8x2 . 800
100 + 8x -x^
x3 - loox .gx^
900 - X3- 100 X
IOO-1-9X -x*
- i5x*2 + 500X,- i|oo
' ox^ " 1000 "X>^ "jgg^^^y^
100 r lOX - X* 100
This equation solved, gives x - 5.29484 - rate
.r time, and x« = 28.03533062565, the annual
payment therefore, = 1 28.0353
i40.The sum of the squares of two numbers minus
their sum is 14 ; and their product added to
their sum r 14 ; find the numbers.
xa +y2^(x4-y) r 14, and XY+ (x + y) ►
14. Denote x + y by u ; then x« 4-y2 = u +
14 ; and XY ri4 -^ u. Then 2 XY = 28 -- 2u ;
X2-f2XY + y« =u-f I4H-28-2U = 42 - u,
.-. x + y = J^T^, oru = ^42 - u • "* =
42 -u, .-. u2 -f u =42 : this equation gives
u =6 =xH-y. .'. x2 -fy* z 20, and a xy = 28
-12 =16; .-. X2 - X Y + y* = 20-l6:£ 4
.-. x-y=2; andx4-y=6, hence xr4 and
y = 2. ■ _^ .
143, Solve x^ ^^Ji ' i8,by a simple equation.
142. Find the compound interest of $80 for one
month, at 6 per cent per annum ; for 2 months,
3 months, etc.
3i,-
J^^S^W^S
•^^>/