Page 47 - Mathematics_Form_Two

Similarity

P Example 3�4

In the following figure, find the
7.5 cm 6 cm 10 cm constant of proportionality needed

to obtain a pair of similar triangles, if
37 0 CE : DE = 1.2:1. Name this pair of
PM FOR ONLINE READING ONLY
L N
M similar triangles.
4.5 cm 8 cm
A
Solution 12 cm D
Consider PMN and LPN  18 cm E Mathematics for Secondary Schools
ˆ
ˆ
MNP = LNP (common) 10 cm 15 cm
ˆ
ˆ
NMP = LPN = 90°(given). C B
ˆ
ˆ
MPN = PLN (third angles of the
triangles) Solution
( ) 1
Therefore, PMN ~ LPN .........................................   (1) The ratio of lengths of corresponding
sides are given by;
Consider ∆ ÄLMP and ÄLPN:ÄLMP and ÄLPN:
and ∆
ˆ
ˆ
MLP = PLN (common) CE : DE = CE = 1.2 = 12 cm = 6
5
10 cm
1
ˆ
ˆ
LMP = NPL = 90° (given) DE
ˆ
ˆ
LPM = LNP (third angles of the AE : BE = AE = 12 cm = 6
5
BE
10 cm
triangles)
AC 18 cm 6
ΔLMP~ΔLNP .............................................. 2 AC = =
( ) : BD =

(2)

Relating (1) and (2), it can be observed BD 15 cm 5
that CE AE AC 6
Thus, = = = .
DE BE BD 5
Since LPN ~ PMN,  then the constant Therefore, ACED ! D BDE
of proportionality is given by (corresponding sides are proportional)
6
LP PN NL and is the constant of proportionality.
= = . 5
MN NP
7.5 cm 10 cm 12.5 cm 5
Thus, = = =. Exercise 3�1
6 cm 8 cm 10 cm 4
1. (a) Given that PQR ~ TSM, 
Therefore, the constant of proportionality identify the corresponding
needed to show their similarity is 5:4 or angles and the corresponding
5 . sides.
4

41
Student's Book Form Two

11/10/2024 20:11:30
MATHEMATIC F2 v5.indd 41 11/10/2024 20:11:30
MATHEMATIC F2 v5.indd 41