C. & M.A.

Sun Kei Secondary School S3 Mathematics

Chapter 7 Supplementary Notes: Ratio of Areas of Triangles

Name: Class: ( )

(A) Four Basic Diagrams

Look for the following basic diagrams. When they occur, ratio of areas of triangles and ratio of
lengths of sides are related.

(1) Similar Triangles (2) Similar Triangles (3) Equal Height (4) Equal Height
Ratio and (Ratio)2 Ratio and (Ratio)2 Equal Ratio Equal Ratio

(B) Similar Triangles

In the figure, ΔABC ~ ΔPQR . P
A

B C Q R
We have AB : PQ = BC : QR = CA : RP = Ratio of Corresponding Sides,
and Area of ΔABC : Area of ΔPQR = (Ratio of Corresponding Sides)2 .

Exercise
1. In the figure, ABCD is a parallelogram. AEC, AFD and BEF are straight lines.
If BE : EF = 7 : 4, find AF : FD.

(If the question is only about ratio, it is not required to perform

any deductive proof. You only need to show key steps.)
△AEF ~ △CEB (AAA)
BC : AF = BE : EF
=7:4
Since AD = BC,
AD : AF = BC : AF = 7 : 4
AF : FD = 4 : (7 – 4) = 4 : 3

Ratio of Areas of Triangle P.1/11

2. In the figure, ABCD is a parallelogram. CDE, AFD and BFE are straight lines.
If CD : DE = 5 : 3, find AF : BC.
E

△EDF ~ △ECB (AAA) or Since AB = DC, A F

D
FD : BC = ED : EC AB : DE = CD : DE
= 3 : (5 + 3) =5:3
=3:8 △EDF ~ △BAF (AAA) B C
Since AD = BC, AF : FD = AB : DE
AF : BC = (AD – FD) : BC =5:3
= (8 – 3) : 8 Since AD = BC,
=5:8 AF : BC = AF : (AF + FD)
= 5 : (5 + 3)
=5:8

3. In the figure, ABCD is a parallelogram. E is a point on DC. AD and BE are produced to meet at
F. It is given that AD : DF = 3 : 2 and the area of ΔABF is 450 cm2.
(a) Write down all similar triangles in the figure.
(b) Find the area of ΔDEF.
(c) Find the area of ΔBCE and area of parallelogram ABCD.
(Reference: Example 7.23 and Quick Practice 7.23 on textbook)
(a) ΔFDE, ΔFAB and ΔBCE
(Remark: if you are asked to “write down”, proof is not required.)
(b) Note that ΔFDE ~ ΔFAB
and FD : FA = 2 : (3 + 2)
= 2 : 5,
Area of DEF  2 
2

 
Area of ABF  5 
4
Area of DEF   450 cm 2
25
 72 cm 2
(c) Note that ΔFDE ~ ΔBCE
and BC : DF = 3 : 2,
Area of BCE  3 
2

 
Area of DEF  2 
9
Area of BCE   72 cm 2
4
 162 cm 2
Area of parallelogram ABCD = 162 + (450 – 72)
= 540 cm2

Ratio of Areas of Triangle P.2/11

(C) Triangles with an Equal Height
In the figure, ABC is a straight line. If we take AB and BC as the bases of ΔDAB and ΔDBC
respectively, then the two triangles have an equal height, say h. D
1
AB  h
Area of ABD 2 AB h
Since   ,
Area of BCD 1 BC  h BC
2 A B C
the ratio of their areas is equal to ratio of the lengths of their bases. That is,
Area of ΔABD : Area of ΔBCD = AB : BC .

C
Exercise
4. In the figure, M is a point on AB such that AM : MB = 3 : 4. It is given
the area of ΔBCM is 56 cm2. Find the area of ΔACM.
Area of ACM AM

Area of BCM MB A M B
3
Area of ACM   56 cm 2
4
 42 cm 2

C
5. In the figure, M and N are points on AB and AC respectively such that
N
AM : MB = 3 : 4 and AN : NC = 5 : 3. It is given the area of ΔAMN is
120 cm2. Find the area of BCNM. (Hint: Join BN.)

Join BN. A M B
Area of ABN AB
 Ratios of Areas of triangles with an equal height
Area of AMN AM
3 4

3
7
Area of ABN   120 cm 2
3
 280 cm 2

Area of ABC AC
 Ratios of Areas of triangles with an equal height
Area of ABN AN
53

5
8
Area of ABC   280 cm 2
5
 448 cm 2
Area of BCNM = 448 – 120
= 328 cm2

Ratio of Areas of Triangle P.3/11

(D) Trapezium
The figure shows a trapezium ABCD with AB//CD. If we take AB and A B
CD as the bases of ΔABC and ΔACD respectively, then the two
triangles and the trapezium have an equal height.
D C
By considering the formulas of their areas, we have
Area of ABCD : Area of ΔABC : Area of ΔACD = (AB + CD) : AB : CD .

(E) Parallelogram A D
The figure shows a paralleogram ABCD. The diagonal cuts the
parallelogram into two equal halves. That is,
Area of ΔABD = Area of ΔBCD . B C

Exercise
6. In the figure, ABCD is a parallelogram. M is a point on AD
A M D
such that AM : MD = 3 : 1 and the area of ΔABM is 48 cm2.
Find the area of ΔBCM.
B C

BC : AM = (3 + 1) : 3
=4:3
Area of BCM BC
 Trapezium
Area of ABM AM
4
Area of BCM   48 cm 2
3
 64 cm 2

7. [HKCEE 2004 Paper II Q17] In the figure, ABCD is a parallelogram and E is a point on AD
such that AE : ED  1 : 3 . If the area of ABE is 3 cm2 , then the area of the shaded region is
A. 9 cm2 . A E D
B. 15 cm2 .
C. 21 cm2 .
D. 24 cm2 .

B C
Area of shaded region DE  BC
 Trapezium
Area of ABE AE
3 4
Area of shaded region   3 cm 2
1
 21 cm 2

Ratio of Areas of Triangle P.4/11

8. [HKCEE 2004 Paper II Q18] In the figure, AD and BC meet at E . If CE : EB  3 : 1 , then
area of ABD : area of CDE =
A. 1 : 1 . A B
B. 1 : 3 .
C. 2 : 3 . E
D. 4 : 9 .
C D
Note that ΔBAE ~ ΔCDE
area of BAE : area of CDE = EB2 : CE2 Ratios of Areas of similar triangles
=1:9
area of BDE : area of CDE = EB : CE Ratios of Areas of triangles with an equal height
=1:3
area of BAE : area of BDE : area of CDE = 1 : 3 : 9
area of ABD : area of CDE = (1 + 3) : 9
=4:9

Alternative:
Note that ΔABE ~ ΔDCE
and DE : EA = CE : EB = 3 : 1.
Let area of ABE = a,
Area of DCE  3 
2

  Ratios of Areas of similar triangles

Area of ABE  1 
9
Area of DCE   a
1
 9a

Area of BDE 3
 Ratios of Areas of triangles with an equal height
Area of ABE 1
Area of BDE  3a

area of ABD : area of CDE = (a + 3a) : 9a

= 4a : 9a
=4:9

Ratio of Areas of Triangle P.5/11

9. [HKCEE 2003 Paper II Q17] In the figure, ABDF and ACEG are straight lines. If the area of
ABC is 16 cm2 and the area of quadrilateral BDEC is 20 cm2, then the area of quadrilateral
DFGE is A
2
A. 24 cm .
B. 28 cm2 .
C. 36 cm2 . B C
D. 44 cm2 . D E

F G
Note that ΔABC ~ ΔADE ~ ΔAFG,
Area of ABC  AC 
2

  Ratios of Areas of similar triangles

Area of ADE  AE 
2
16 cm 2  AC 
 
16 cm  20 cm  AE 
2 2

2
 AC  4
  
 AE  9
AC 2

AE 3

Since CE = EG,
AC : AE : AG = 2 : 3 : (3 + 1).

Area of AFG  AG 
2

  Ratios of Areas of similar triangles

Area of ABC  AC 
2
4
Area of AFG     16 cm 2
2
 64 cm 2

Area of quadrilateral DFGE = 64 – (16 + 20)

= 28 cm2

Ratio of Areas of Triangle P.6/11

10. [HKDSE 2012 Paper II Q17] In the figure, ABCD is a parallelogram. E and F are points lying
on AB and CD respectively. AD produced and EF produced meet at G . It is given that
DF : FC  3 : 4 and AD : DG  1 : 1 . If the area of DFG is 3 cm2 , then the area of the
parallelogram ABCD is
A. 12 cm2 .
B. 14 cm2 . G
C. 18 cm2 .
D. 21 cm2 .

F
Note that ΔDFG ~ ΔAEG, D C
GD : GA = 1 : 2

Area of AEG  2 
2 Ratios of Areas of
  similar triangles
Area of DFG  1 
A E B
Area of AEG  4  3 cm 2
 12 cm 2
Area of ADFE = area of ΔAEG – area of ΔDFG
= 12 – 3
= 9 cm2

DF : AE = GD : GA
=1:2
(DF + AE) : (AB + CD) = (3 + 6) : (7 + 7)
= 9 : 14
Area of ABCD 14
 Trapezium
Area of ADFE 9
14
Area of ABCD   9 cm 2
9
 14 cm 2

Ratio of Areas of Triangle P.7/11

11. [HKDSE 2016 Paper II Q20] In the figure, ABCD, CDEF and EFGH are squares. AG cuts CD
and EF at P and Q respectively. Find the ratio of the area of quadrilateral DEQP to the area of
quadrilateral ABCP .
A. 1 : 2
B. 2 : 3
C. 3 : 5
D. 4 : 9

Note that ΔADP ~ ΔAEQ ~ ΔAHG,

AD : AE : AH = 1 : 2 : 3
Area of ΔADP : Area of ΔAEQ : Area of ΔAHG = 12 : 22 : 32 Ratios of Areas of similar triangles
=1:4:9
Area of ΔADP : Area of DEQP : Area of EHGQ = 1 : (4 – 1) : (9 – 4)
=1:3:5
Note that area of ABCP = Area of EHGQ,
Area of DEQP : Area of ABCP = 3 : 5

Ratio of Areas of Triangle P.8/11

12. [HKDSE 2017 Paper II Q16] In the figure, ABCD and BEDF are parallelograms. E is a point
lying on BC such that BE : EC = 2 : 3. AC cuts BF and DE at G and H respectively. If the area
of ΔABG is 135 cm2, then the area of the quadrilateral DFGH is
A. 60 cm2 D F A
B. 81 cm2 G
C. 90 cm2 H
D. 144 cm2

C E B
Note that ΔAFG ~ ΔCBG,
FG : GB = AF : CB
= 3 : (2 + 3)
=3:5
Area of AGF 3
 Ratios of Areas of triangles with an equal height
Area of ABG 5
3
Area of AGF  135 cm 2
5
 81 cm 2

Note that ΔAGF ~ ΔAHD,

AF : AD = 3 : 5
Area of AHD  5 
2

  Ratios of Areas of similar triangles

Area of AGF  3 
25
Area of AHD   81 cm 2
9
 225 cm 2

Area of DFGH = area of ΔAHD – area of ΔAGF

= 225 – 81
= 144 cm2

Ratio of Areas of Triangle P.9/11

13. In the figure, ABCD is a parallelogram. E is a point on AD such that AE : ED = 2 : 1. AC and
BE intersect at F. If the area of ΔABF is 84 cm2, find E
A D
(a) the area of ΔAEF,
(b) the area of ABCD,
F
(c) the area of CDEF.
B C
(a) Note that ΔAEF ~ ΔCBF,
BF : FE = BC : AE
= (2 + 1) : 2
=3:2
Area of AEF 2
 Ratios of Areas of triangles with an equal height
Area of ABF 3
2
Area of AEF   84 cm 2
3
 56 cm 2

(b) (AD + BC) : AE = 6 : 2 = 3 : 1

Area of ABCD 3
 Trapezium
Area of ABE 1
Area of ABCD  3  (84  56) cm 2
 420 cm 2

(c) Area of CDEF = area of ΔACD – area of ΔAEF

1
Area of CDEF   420  56
2
 154 cm 2

Ratio of Areas of Triangle P.10/11

14. In the figure, ABD and ACE are straight lines where AB : BD = 3 : 1 and C is the mid-point of
AE. If the area of the quadrilateral BCDE is 25 cm2, find the area of ΔABC.
E
Join CD.
Area of ΔABC : Area of ΔADC = AB : AD Ratios of Areas of triangles C
= 3 : (3 + 1) with an equal height
=3:4
A B D

Area of ΔADC : Area of ΔADE = AC : AE Ratios of Areas of triangles

= 1 : (1 + 1) with an equal height
=1:2
Area of ΔABC : Area of ΔADE = 3 : 8

Area of ΔABC : Area of BCDE = 3 : (8 – 3)

=3:5
3
Area of ABC   25 cm 2
5
 15 cm 2

Alternative:
Let the area of ΔABC be a cm2.
Area of ADC 4
 Ratios of Areas of triangles with an equal height
Area of ABC 3
4a
Area of ADC 
3

Area of ADE 2
 Ratios of Areas of triangles with an equal height
Area of ADC 1
4a
Area of ADE  2 
3
8a

3
Area of BCDE = Area of ΔADE – Area of ΔABC
8a
25  a
3
5a
25 
3
a  15
Thus, Area of ABC  15 cm2 .

Ratio of Areas of Triangle P.11/11