育教僑質三第張六第,日五十月二年酉己曆夏
WAH KIU YAT PO 郭日僑罩
文中學會考試題預習
數學科 (廿二) 歐陽文
MATHEMATICS (22)
LESSON 22 AREAS
IMPORTANT THEOREMA
Arsa theorems on triangle:
= < base x height>
Taron's formula
A, on the same base (or equal basesy and between
the same parallels are equal in area.
Area themems on quadrilateral.
Area of rect.
Length x breadth
Area of syuare = sguan. on one dio
Area of Kgram = base x height
ligram (ine. rect, and sp.) on the same base for equal
bases) and between the same parallels are spust in area
venage of two bases x height
= median x height
Area of trapezium =
Area of rhombus =
of the product of two ding
期星
日一月四年九六九一整公年八十五國民華中
EXAM
km bisects the
TRY IN ARBC, AS=5" BCES
« ABC=45° Calculate the area of JABC and length of AC. (h) ABCD is a guadrilateral in.
which eh=3=90° AB=6" BC= 10* AD=18" Prove that BD bisects & ADC".
SOLUTION: (0) Draw AD LBC
"ADB = 90%
Const.
bau cx} and blew ZAlfCx
A CX2 =d. CRA
of which, ACXY, 7*
By Subtraction
Common "to both
A AXY.
they are on the same pase (ZT) and between the some parallits (?¢ // ZY)
AAXY and A CZY=)
LAXY
Hence
<BAD
BD = AD
sides opp
+(AD)(BC).
HINTS & Ans. To Ex 21
Produce AM to E such that Mỹ anm ABÉC is then a ligram
AB > BE
Hence
x 10.6 (38.m) 丹吉
cas
DC= BC » BD = 6-
rom a EBCT BE +KACSBC From From Adding
CF+ GBA = CA.
AD +ICB > AG
Bythagoras,
ACTM= AÐ" + DC"
=(-5)+(6-=
18.57 HC=W18:57
vihagoras" thesem:
the area of the square on hypotenuse equals the sun of areas of squares on two adjacent sides
EXAMPLE 1 Find the area of a triangle whose sides are (p"-7*)\
*pg, (P+8") inches Grue Prasoo$.
Let a, b, c be the sides of the o such that
ADISC
Take a point x on AD such
JD E BC
XBCD & a #gram:
From H. △ABX
·⋅AX=18-18
AB=6
By Pythagoras.
the
mangled
aea = fb =(pg2p*8*
This problem can also be solved by using
ama formula.
Ascs-ases-b>cs-c)
where
AD+BE+OF Y÷CAB+BC + CA)
<by Produce AD to Ï such What DPAD
Join BP
From AABP:
Ssimilarly
AB+AC > 2 AD
BC+ BAD & BE
CATCBY 2 OF
Adding, AB+8C+ CA> AD+BET+ CF
Hence, in iguam XBCD • BY=8C=/0 (m.) XBCD is a rhombus BD=ding.
BD bisects LADC.
EXAMPLE'S;As shown, ABCD
is a ligram and
Prove that
PR HAÐ RSH AB
(a) & SPA+/65PQ=&SAD APRBA
ASCR=2. ABKD
(C) APER
门
PROOF.
(a) Join KD. PR
ASPD
= AKPD+AMPS +AKSD
às on the same base and beksaan.
the same ils are egust in area.
APR A
AKPS 13 'n
common.
AKPDT AKSD+AKPS=AAPS + 3K5Q+aKPS
ASPA + ASPA➡ SSAD
13. Join KA
A ASQ == AKSA+« KSQ ̈ + 4QKA
OKSA=AKSP #4 of from KSBP oksa = ± of Higrom KSCQ
*4 QKA = 4 QKR, mL of thram MD RK AASQ +▲APR
tal KS BP + KSCQ +ØDRK♣ APKR)
5 of ABCD
ABD
SHPX = DAPY (AAS)
PNBPNC (SAS)
APXBS APYC (RHS)
And
PX = AB - BX = AB-C
= AB-1AY-acy
= AB - AX FAC AX== CASTAC)
and excentre of △ABC
AOD 13 A st. line. DOA, FAE are the ut. ext. anglew blsectors of ch
DOA & FRE
an altitude of Að
Similarly, FOB, FOC are altitudes of ADEF
NOTS
AD, GE, SF intersect, at
the ortop centre
ADEF
AC
AB,AD, BD
1 bisectors of AB,
ACAD CD bisecta of AB
A4
both coïncede, with
In a circle
the I bisacker of
•passes through the cante, S. bisectors are concurrent af of the given circle )
chord all the
Prent Produce HA to K
Such that Ak=BC
4 BDC #4 ABK (SAS)
d=by
7. CDL BK Similarly BFIKO
KH, BF CD am alt. of a KBC AH, BF CD are concurre cat the artho contre of in KBC)
EXAMPLE 21 HD is an attitude of LABC; P Q are point
on AD produced such that DP DA = AC. Prere that BQ =cp
PROCE· IT AB>AC, then from the fig. as shown
ADPL BD C
480, 800, PDC, CIÀ GH IT. AS
by Pythagoras' theorem:
•From W. A BDA: BD" — BATM- AD"
From H4 6RDI BA?
BQ* - BA - A2,
From: vt a CDA
From it. acP DE
C.P2 = ACT
Fring (1) & cay.
DPZAB & AC=D&
CP*DQ AD + AB
線
CA
BQ
By adding the case for ABC AC
then completed
EXAMPLE: 3.
the proof.
shown in figure, ABP is a st.
and M is the mid- point of AP Prove that KM bisects the area of AABC
PRODE: Join PK
A, PMK, HMK ONE OF
equal bases (AMxMA)
and of the same alfitude
AKMP = AKMA
AKMPS AKMBT AKBP..
AŞ KBP, KBC am on the same bate(KB) and between the same /S (BK#PE)
KBP= AKBÓ
Hence.
akmas &KMB+ <KBC =KMBC
SKMP= ARMA
SKMAS KMBC
Proved
Join KC
· of ABCD
ABDKŁADEK FACBK ABDC = £ of #gram ABCD ADCK A ADAKT ARČK
of (DQKR + QCSK)
ACBKI A CSK + ASBK
of(cskQ + SBPK)
ABDK + 2(DQKR+QCSK)
<sko + SBPK) ABCD = 2-4BDK+DQKR+ 2 @CSK + SBP大
TE (DAKR + QC3k + SBPKYŁAPKR
BDK
APKR - QCSK
EXAMPLE 6 ABCD is a parallelogram
to BA cuts BC AC
AX Prove that ABCD is a Ugram
XYZ TA8
Given
To Proves -4AXY-ADYZ
Proof: Join Cz.
In Ai CX2,CKA;
They are on the same
my line pariet
As shown
point
on
Prove that
EXERCISE
diag, AC
ARD. =PQYB
P.Q. R. are points in the sides BC. CA, AB
Resp of a ABC such that POIFAG, ORII BC,
• ABP = 4 ACR
Prove that In A ABC
is the m
ABCD is a
LADC
ACB is a right angle,
-point of BC, prove that
AB = AD+3BD*
nadorlatival so that chec and each of 40" IF AP AQ
CB drawn parallel to CD
at P. Q resp., prove that
RA· AB = PA · AD
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