Daily Ohio statesman. [volume] (Columbus, Ohio) 1855-1870, 1867-11-30
;THE ruzzLEw,s OoLUnr.
ColIrNsv1LLr. O., Nov. is, 1857.


teman's family. We can eat the pig and potatoes
and still have few bivalves before leaving town
e.nra place. s and Il vii average the distance. @ Bs
!i,s this time. The weigh. q authority is against
Pub;ishe.. solution. the printer put for flour
Guern.ey and Delta are about right on 2s!; Otiose
my answers: minor skis = sO.s1s: area oes .
square feet and exontricity ,.6a575. My ans
to as1 c n Is a Sc S. ss sEsn
as diameter 7957X mi es and .oOJ circumference
dVas ths and this VIII 81 give We
hight of segment=haf the radius. Hence angle
and So and
radius by ene-half gives hight from center ami this
minus radig. gives siSs, at TIE man must be elevated
No. a1s-sl4 inches u. That is merely the crust
No a75@Medeira q. so: Port sa ao Sherry q. 10.
No. a7s-42 feet 8 inches X. s. x. give the rule.
siass. ee Not %efy ''fd'
No. 2sa-Hight $40 feet and 100 feet. r
No. 2s3-$i02s 1853.
se.s Me- assm.issa, ss9 iss ny ,ss S"n-aS4 twice.
No. 256-l3 "A lbs nesrly. z
Thus endeth the chapter. SAMMY.
isues r. .+r7 ssao NArCHEz. o. Nov. 10Th, 1867.
Jewhilikens what a column greeted our eyes
last week. but we will consider it next week. Here
are answers 70 problems of Nov: o: -. 1''
No. I s7s.-dhe cost Of fencing is s4s.5<. IFe a,
No 81 The Madeira is worth g 50 per gallon.
the Port % 80 per gallon, and the Sherry @ 10 per
gallon. A. s. s. please bring few bottles Of each
r. No. a7s.9Beiouss to SIX: lis worked by resolv-
No. 27s. Soiution.-Letx= X number of first size.
and y=number of second size. By conditions of
Sy pas dvdsd as By from n1 winch %,;ase
q No. .2sO.-x=ss.1 ~ 20, and perpendicular =
NJ. a r..x l.:.ht
No. 2s2-The sides are $50 feet. 24o feet. and 300
this answer. viz Radius Of Inscribed circle IA
feet, and the sides of the triangle 1l025. 13 30,
and .st ~ y tt feet d rob. No on and
myself. The answer, should have been 1898 SIS
joking with E. b. concerning No. a7o The matter
was. lcnuld not it., SAMMY, gives sie as an
answer To- probem2 274 Iwould like to see his WORK,
: mast humby beg pardon Of Sammy. In looking
over my work discover that made very grevous
mistake in my problem No. I57 Tho problem should
have read thus The difference between s the east
daftans. Here are the numbers: .1a50oo. 2eiss6
29e296 I'd f,fnituns. and 57e703 7o37o3 ad iituna.
San hs a4d Of .
and will will i treat. How do you take yours9 Mine
sammy and gave the same answer ta Delta's loaf
Of brea It's said that we are not c,rrect-that
Guernses's GUESS is nearer the answer. Now,
ton. Ny problem No. as3 should read four mum
bers.n'. instead of " from numbers" Some of the
wonder Why tn y are not equally good in answering
problems. hope they will soon commence. AS
am unwe!l this week. don't feel very msthemati-
cally inclined. OT1osE..
- p GREENvILLB, o., Nov. 18th, 1867.
Mr.. EpIToR:-I am glad to see that the interest
in the Puzzler's Column is increasing. Here is my
msthod with 244: Assume the quadrilsteral B q D.
The an of the perpendicuars let fall on the dag
onal G. is equal to the diagonal D into ths sine
of the angle. formed by the intersection of the di
Is 5s half .ss sum into
~ ACXB sine so 200 sine so dez.=sls IS.
Is,nt wrong answers to problems 2s7 and 259 from
tie fact that had not time To review my solutions.
No. z53-I think Sammy could put q few more
cannon balls in that box. There are several ways
ting 81 in the LIST course. II in the a. 90 in the II
in the 4th. and 50 on. IL this way, he will get II
No. a54-Sammy says this problem is absurd. As
it is one of INV own make. think can show Sam
X equsl to into square root of 2. get equsl to
u into Square root of 10 minus square root of 2)
into % into Of
and YAK into square root Of 10 minus !, into
square TOO root - Of e nt 81 u iptc ito
hence XI,wXa,.Y3. Similarly. it can be proved
No. 1863 --2l5 s95S4 feet.
No 264.-Mnor axis eqnals sos7 feet. area equals
No. a7o -Otiose nisspprehended this problem.
Sammy is right. -sr
Brot.er Otiose is wrong on No. e72.
No. 276.-344 u cubic in. s.e-
No. ~. 1.-Madeira 50 go per gallon Port % 80; and
Sherry s2lO. 1H-
~ No. 27S.--426 81 feet. --.. f- -s-f.=
No. 279.-ls of the one size, and 81 Of the other.
No. 2S0-As circle cannot be inscribed in a rect
angle, unless it is square, l presume Guernsey
means that the circle is inscribed in one Of the two
equal triansles. nv solution is; The perimster Of
the perpendicular into the base. because each
duct is equal to the of the rectangle hence we
have, after reducing, 35X plus IS 5=Y2--69 ID;
from the right nng ed triangle have X2=af2 plus
az plus CO. Multiplying equation % by 2. aiding
the resulting equation to equation 20. then adding
JI55 to both numbers and extracting square root we
get plus 35=lY-34. Find the value of Y in this
equation in terms of Y, substitute in l2 tran pose
reduce. and quadrate. and we find =ll9 conse-
quently plus l=i2o. aud X l69.
No. 261-6l .97 teet.
No. 2ss-The sides about tho right angle are $50
feet and 240 feet. hypothenuse 300 feet So'ution
Let x2 -y2. and 2xy equal rides about the right
ANGLE then x2 plus ~ equal hyp,thenuse. First. |
the rerimeter 2X2 plus 2xy equals 720, X plus
equals $50 ID. Secondly, From similar tri-
x2-ya: 2xy IA whence KY (x2-y) equals 72 (xe
plus y2) (2l. Dividing equation (2) bv ID have,
af,er reduction, x2 plus eye equals Ky 49, combin
irg equations D and (3, so as to eliminate Ky. we
and az we find W equals 81, consequently x2-y2
egu us l50, and 2xy eq lals 240.
N. 2S3-Sl .O26 3D..
No. 2sf-Ist PARTLY applying algebra to this |
pro 5lem. obtain equation whose roots mega
tive which proves an Inconsistency.
No 2s4-2d part-The sides of the triangle are |
$685. 1926 and I6 .2o93 feet radius of inscribed
circle 2 .SOO4 feet. |
No. 2S5-S1230 81.
No. 2S6-l3 23S pounds. E.B.