polygon" is itself kon-
日五月三年八六九一曆公年七十五國民藏中
真二第張四第
日七初月二年申戊麿
WAH KIU YAT PO
#
B
1968
英文中學會자문
Since ABCD --- MA 52.
in- siden regular polygon
數學科
(+)
歐陽鐙交
MATHEMATICS (19)
And
24
LESSON 12
Int. & Ext. as of a polygon Parallelogram
$! Important theorems for reference on:"
A, Ext. + inta of a polygon
1a: The int as sum of a
2 4.4)
2": Ext. < of a equals the sum of int opp <s). 30: Int. Is sum of n-sided polygon (20=4 se}
Sum of n-sided polygon (4.4.45)
* Ext. 45
8, Purallel lines
<*: Two fundamental Axiams
(a) Two coplanar st. fines
parallel.
are either intersect
connot
\((b) Two intersecting copanar St. haes
both be parallel to a third st. line
Play fair's Axiom)
If either (a) alt. Is equal, of 163 car as equal.
<< (c) int. as supp. se. Lines are parallet.
a st. Line
Lives
cuts two o
then (a) att. «.
rep B: Two Y Sewe
the
int. 4. Supp.
a+nat 2 was]{ int a," ABIIMH){
cox int. a sum of ABCD ..... MN=32=223=lap 2 dari(2)
Eg(1) and Egra) are identical ( same" int. 4 Sam)
77
(7-2)(2-1) 4 » '77 (2-1) (72-22-#42) † 2 to 72-7
1+
X =
which is meaningful only when n is even; in other no parallel aidés tov an odd-sided regular робудет
words.
Example 3 As in fig. ABCD is
then the famo
Plotz
.*.
mare parallel aquat (b) com. 2. equal,
st lines which ar parallel to same st. lines,
am parallel to one another.
"9": St. lines which the perpendicular to the same coplanar vt. Live, are parallel to one another.
c) In
a parallelogram ›
sides, #gram
(are equals
4s, gra
I cil
qual I
12*
dling
#gram & bisect each other. ding of ligram bisects the ligram into two, congruent triangles.
a squan
APB =★
that (1)
CXQ #AP, DQHBPX. Prove PX = AP - PB
AP/QC
ره کرد
< BXC = ¿ 6PA =H.2
b+b= a+ 5 m slmut Sam
... b, a
In a BCX, ABP?
- BXC=APB = Aa ̧ proved proved
tod
↓ BL = AB
· ABUR ♣ ABP
... BX = AP
Sides of $8.
AAS
cor sides of mon
b
PX = BX - BP
= AP
Simila
we find
JSP 2 ABCY
CBQ (AAS)
quadrilateral to be a parallelogram:
opp sides & equal.
ding
equal.
ETV
bisect each other`,
e pair of opp. cides (= &
Themems on other properties I's guad
18 Rectangle, square and rhombus are // gram
s. Alder for a Mect.: the ding, are equals and
lebs for atsporth
`bisect ""each other.
the ding are equal and
· bisect each others at the.
(C) for authombus clothe ching, & care treguet
but bisset each other at t
#424 Insangises, trapezium ((a) the ding are equals (b) the base as are quat. [20*1{};a" kitc` = One of the diag. bisects the other at #. As (26) ¡fectp=yligian" +"La
*
مهر
„Square #s\rect + adj. sides" equal. Rhombacsmil/gmm + aq İsides qual
"Ja"Etamines
*
Example Lys In T&ABC, prove that the angle formed" \the #attitude with the angle bisector
at 4, equalezhalf the difference_of
GARE CABCTerith ADL BC'
AE bisects _BAC $.0,+Q,#Q ̧!
ABC (f. 187.C)
*«BAC{~ € 90*-'«B]
= $(/80--8--C)~ (90°-3«8 >
Conversely, if B <_C" then È lies betwan.
ffence,
faqjarticular
Q1 = ±(< C - «8) Mishen «B ≈ «C, -4?. ««Aặc is G06] wa= 1 − 8 - «8) ̧ = −";
Consequently, BAB=1/48240]
[may ̄u3toF[«8-26} {{{6}]=BLIC
4
lo dangte, the difference
ku (0)}{zB • «8) mfluent thelabsolute valut
of the i difference?
p 1 Put all
•
Pax is isosceli <QPX = 45°
ande 4. HBP. ACE are equilateral as
of ABC, DAEF is a gram outside
Prove that FBC is also an equi...
[Figen)
SireniŢA, ABD, ACE an equilateral :
ADFE A
a #gram. To Prove FBC is an equi, anz
Proof Ket
then,
<BAC = cm
« DAE ≈60° + 60° + œ
b
= 120° +
2 DAF = 240° – »<? <FDA 180° – « DAE
(As in fig. a)
= 180′′- (zao*+ œi = 60* - a
- FDA-
¿FDB = <
in A, ABC, DBF 1
AC = DF (AE)
KONG PER
means the "difference" be turen <B>_C_{ ts, the greatte subhach the smaći
Examak 2・ AB is a side of an n-sided regular polygon Find the condition for an other side of the polygon parallel to
48
B
Solution Denote the polygon by ABCD--MA---
Let MN be the xth side
5.4.
MAILAB #
Join AN.
***
ABC)--MN is therefore a (X+1)-Sided D
polygon
int. ds sum of ABCD---MN is
[*cx+1)=4}** = (32-23 14
M
Similarly,
« BAC¤ < BDF ( = d
BC= CF' AFRO is aquilationl
(in both
I
SAS
in fig. ali
Com vidus 7243
2 MB CR GEFC ̧
Example 5. The diagonals of rect. ABCD cut at K, and
Ak is greater than AB. The circle, centre A. radius AK, cute 48 produced at £- £f ZARB = 4.8kk Find <BAC.
·~: Rect. ABC D
AKAE
*
To find: a
Solution!
HENCE
ABK
ARRE
b+
(KA-KB
{Q+ b + k‚=180*, +6+48=1800
= 673*
HINTS ANs. To Ex, 18.
Given
Į base maisos. 4"
stiff
{sing of rest!
'baza as, isos, mi
< sum of a
I Let A be the reg'd maon. Then 2, 4, 7Fare`in”AP)
the comman Wiff. = 'g-AmA - X
NOTE: The arithmatic mean of 2
quantikes is half their sum.
? Let a be the 1st term "and" d be the cammon dif.
Then, the 300 term = 044d # ?2~$7
and
the 21 dem = a + aod = 232-407
The need series: 33 47-29 52-44,
He c. p. ('5% + C) - (34420) m ab-c.; the_{# tem# 36+20
-- the 1*tcm= {3b+a¢3+ (n=1) {~b¬6) ⇒ 17b-5c
一日料理
(#)
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$ = #Laa+cn=v9d] is 3/2 =#{(2)(+2)+(N¬d(+3)} n'- sqm + 208 m (1-43) (2−16)=0
Both values Satisfy the condition of the equ. For, if you' Write down the 14th, 15th and 16th terms, you waakt dl, find that they are 3, a. -3; the sum of which is
zero. Thus, the sum of first 16 terms = sum of 13 terms, NOTE: If the equ. give a root which is tractional or
negative, then this value would be rejected as in compatible with the conditions of the case. no. of terms = x+2
1st term = X
精
total
last term = * + [xx+2) = 1}  %==
sum of (2 + 2) terms filx+3x)=2x(2*% sum of avitt, means = 2x (2+2) = (2432)= 12'
there will be (n+2) terms in G P or Including 2 and
of which
"I is the 1st from, y is the last. Gir I take the C.R. {nta)th term =x+/+ xy, y=m/ The ragút means are:_*(*)** ***
where his the nth term of £, £,
↑ 1
Similary, we find · A =(4) » Hence,
T
and of which.
The regi summ$, +5, = { 1 - £) ++C!~ $2.
8, The amounts in each successive birthday "
$x..
Exa
which is equivalent to
un # P. of which a•ž, α-§. If to yo. be the regid age of the són. Then ; s= Hearin÷(d]=(3x20
2*> Each a has,
a
perimetr of tulf - length. preceding p
~~ ** •} + { + ! Sump is p÷90-£) = 22
ANS. 16 YL
As the
vir Eacht. Thas an area of of the preceding
“Sume S.
Exercise 19
1}{if the vertices of ligram EFGH fie çok ilgram ABCD,
then the four diagonals· are'con cument.
bisectors
2, EFGH is a quadrilateral formed by the amale
of “quadritateme ABCD. Prove that (a) the sum lof opposite angles of EFGH is a \do】if ABC,iskasparatlilogram, then EFCH iskaj
rectangle;
(c) if ABCD is a rectangle, then Ef bif'isa*square? 3,ABCD is a square,
* is a point on CA producat such that the parallelogram) DAPQ is a rhombus If QC cuts PD at R. find the angles of land prove that RP - RC3
As shown in fig., Qis fany point on the diagonal "AC" of #gram ABCD. Prove that, the areas of ligram XQRD] land PBYQuae equal,
- DRQ