**Areas of Non-Similar Triangles

A(A) Non-Similar Triangles with a Common HEIGHT

~ABC, ~ACD and ~ABD have the same height, h.

Area of ~ABC = -3: -,< \ x hArea of MCD = -3:)("2 x hArea of ~ABD = ~~ 3:><.h 2 DB'"

:. ratio of these areas = 2 . 3 = ratio of +heIr base-- ----

(B) Non-Similar Triangles with a Common BASE c~ABC, MBD and MBE have the same base, b.

Area of ~ABC = ~ 7Z by. 5

Area of ~ABD = ~)< b"f...3

Area of ~ABE = -k7- b ~ \••21~~~~~~~---

1-------::L-----1f-----------=~

b

:. ratio of these areas = 5 3: = ratio of -Their he, hi-

C Non-Similar Trian les without common base nor commo

Two triangles' bases are in the ratio 1 : 5 while their heights are i the ratio 3 : 4.What is the ratio of their areas?

Area off A :: ~ 7< 1)< 3

A reo oV: B -= ~?< 5 ~ it

5

:. ratio of these areas = ~ : ~ :~ = ratio of ~Ir bQ.S<2. )<. he:0h1- .

5

- - f\:re'a ~NrLO cf - \0

b 50Arw..1 ~ QC. ::: 10)( ).Example: ::::'30 em ~

In the diagram, PQ is parallel to AC. Given that BQ = 4 em, BC = 10 em and area of MPQ = 8 em",Find the area of (b) ~PQC and ~ ~J3Q A C(a) MBC; __(b) MQC; a\X2.. ~mYYbr'\ ~'n+ --

(c) MQC. . Area o~ ~PGc. _ (, bcYY1

(0') .6 AI3CtV b. P e,Q A ceo o~ Ll f5Q - --;::- Q"'7 A.!$o o~AflBC = (Be:\). _, Area Cl~ AfQC.::: ~)( ~

A reo o~ A PE>Q g,Q, ) ). B::: 12cm *

Arw- of AABC. == ( ~ ):2. )<. 8::: 50c..m>"'~

In the figure, ABC is a straight line an L.DAC = L.BDC.(i) Name an angle equal to L.DB .Given further that AD = 5 em, BD = 4 em and DC = 7 em,(ii) calculate AC and BC,(iii) if the area of MCD = a em", [nd, in terms of a, the area of MBD.

c.i) L ADC;jX ~ = -~~ IiI) .6AOC r-: ~ DEe.

61) AC =:AQ.. Be A®4 L'lAOC _(~"{-PC ~B ~ ::: ~ Amoo;fADf3G - DB) 5Ac b .D (- J..

---==J = T ,6C ~ 5· b em ~ Area.or~ Art:.. =- -&:J )(ap '-'''''+-'----- b..

Exercise AC = <B-15 c.m~ ::: J5 0-:SIC, .b .:&-n;-

, . Areo q i1ABD ::: 11, Q - Q..

With reference to the diagram, find the value ofy. =1t~6.AOCf'-o'.6 EDc..

A

7

Q1 E

'3 GCAB -= Ac.

2_3-2'+ - 6(,

y C.A •• 36 D -..

Q2 In the diagram, L.LNM = L.LP . Name an angle equal to L.LNP and calculate x.4LNf'/\ f'..,.6. LPN

~L\....NP ::: LLMN

~ _ LM _ i!ltLLP - LN - NP.2.S- _ li-b - 4

In MBC, D is a point on the si e BC such that LABC = L.DAC, AC = 6 em, CD = 4 em and BD=xcm. 1 B(i) IfMCD~MCA,showth;tx=5. I,

(ii) Calculate the area of MC given that the area of MCA is 12 x'2

em . L.fx+' to = 3bCI ') Ac. AD rCD

Be. ::: BA::: CA 4x = :2-0

b X =5 * .sho~V" ':=. At> -:::.~ 6

x+Lt E>A b((', ) Arep ~ A~ z: (+):L

A('(Jo ~ R>CA, (.!±....\2. 2 .L).

A'V(o of .6AeJ);::' b)( \ =- 5 3 (,TV) ~

L 6 P M

Q3

Q4 In the diagram, ABCD is a quadrilateral with BA parallel t I CD. AC and BD meet at X where CX= 8 em andXA = 10 em.(a) Given that BD = 27 cm, find BX.(b) Find the ratio.~.rea of MXC : area of llAXD.

t;\ fj. A eX r-.., b CD)< (b ') .6)< - l.S... = .!2..llAJ XD - \2 It-

i?,x _ A X AJ'. -:::: ~ s-OX - C)< xc <6 z: LF B <::~f--L!-+---="":::'--..l;fl ;,r.--'-=---,-,-~

~ = ~ .: l.\BxC "" b. DXA (<;;,t\s)21- e,)<. 81~ Bx :::110 - \O~X ArE'oof ~B~ _ LI5 \2-

l%Bx ;:;.')70 l\reoo¥.6AXD -\\0)B)( :: \5 Ct11~ :::: ..:L

X@. - It-\':>u 3 ~1io IS ~; It

. In the diagram, CA = @, L.CAh? = L.BCD and CD = x cGI)hr Write down a second pair of equal lengths.el) W Name the triangle which is similar to. MCD.

Given that AC = ..J1O em, AD = 3 em,(iii) Show that x = 2,(iv) Calculate the ratio. Area of MCD: area of MCD. B

(QSQ6 ABCD is a quadrilateral. AB is parallel to. CD. F is the point on AD such that 3DF = DA. The

lines CD and BF produced meet at E.(a) Explain why MBF and I1DEF are similar.(b) Find the area of I1DEF given that~~f MBFis 8 cm2

.j\-

(.\) .i\6C.D r- beAD

(jr)

Cb')

LAFB -=- L Df=E (veA. Off- Ls)

LABF ==LDEr (a\·L is)

LFAB =L fDE (a.l~, Ls)

r 6Ae,P (V ADEF CAAA) ~

A(ea ~ .6.DEF (-DF\ 2-

Area a?AM3F = AP-)

Areo ~ bDEf :;: ( ~ ~L 'f- %"1...

- \%uY\)J(

7

DA

A

c

B

QbQ1 In the diagram, C lies on AR s ch that AC : CR = 2 : 3 and PQ is parallel to BC such that PQ :

BC=2: 3.(i) Find CQ : QR.* (ii) If MBC has an area 0 18 crrr', find the area of

~QR. 3

~: QR ::::0 ,; 2 ~ B· R

P 3 ,2'cii) ABCar.d..6 GaQ. hQIl~ Ar-eo.cr-? ~i2>a<. = :L x. 2-

mmct) rweh+ = 21 emA eo. <YfQ.-A,~ = 2:-. AI3CIZ IV A f'QR ::L 2 21

~ ~ B::.fl."3 " _ A-I'e\::ur-~APQ.R = (3'"') x = 12cm2

D = 2 em, DE = 4 em, BC = 10 ern and EC = 2 em. If AE = x em, ~

CQ'-tQK. _ 3Q~ - 2:'

:2 c..Q ~ :lQR. = 3 QR2c..Q = Q~

~ In the figure, LAED = LABC,calculate(a) the value of x,(b) the ratio

(i) area of MED: area of MBC,(ii) area of DECB : area of MBC.

30\") 6.AED tv tdA'BG b i ) A 6AED _ (~) 1.

X 4- 2 (to cr.f .QABC - 10

BDt2 ::. 10 -=:. X+2 = -±-25 ~

In the diagram, ABD, AMN, A(fE, BMC and DNE are straight lines. AD II NC, BC IIDE and DE= 3NE. Find(a) the ratio area of MND: area of MNE,(b) the ratio area of MCN : area of MAD,(c) the area of trapezium A'DNC if the area of MCN = 17 cm2

.

0) 4AN1) CAt1d ~ANE ho...\JQ. ~m YlD('\ \-'~h

Afro.. 4~N -= 2Area ~,ANE -,- ¥f

(i) .6f6cR.. A" ~ PQR.

CR \3C~R -:: PQ

- - Lf.::;c + 8 ~ :LO

+,x. -=- 12

:;,c :::: 3cm *

b) ~ 'EcN /'./ ~ E AD

-- Ar~ d AEC!JN(I.Jl ~ A '(;A D

-J...- 2..CQ

Q(Z

A

B

D

rea. 0-# DECB

reel of. A p·JjG

;It

DE 4-

25-'+25

=~25 ~.

A

E

B f----+--"<------,,~

D

c) Area. <>f 6EA1) =. q)<. If2-

15~cm

C==.·,S.3-112-

13GcrYl ~

8