Page 53 - Mathematics_Form_Two

Similarity

that is: Therefore, MY 3 cm.= 1
3
DB = BA = AD
DA AC CD (b) From 6 cm = 9 cm ,
MN 12cm
DB BA

Therefore, = . 6 cm 12 cm
FOR ONLINE READING ONLY
DA AC MN = = 8 cm.
9 cm
Therefore, MN 8 cm.=
Example 3�7

In the following figure, calculate: Example 3�8 Mathematics for Secondary Schools

(a) MY (b) MN
In the following figure, ABC ~ PQR. 
If AC = 4.8 cm, AB = 4 cm , and
PQ = 9 cm, find the value of PR.

Solution
(a) Since ZY / /MN, it follows that,

ˆ
ˆ
XYZ = XMN (corresponding angles)
ˆ
ˆ
XZY = XNM (corresponding angles)
Hence, XYZ ~ XMN  (by AA – Solution
similarity theorem)
Given ABC ~ PQR,  it follows that,
XY XZ YZ
Thus, = = , this means AB AC
XM XN MN =
that, PQ PR
10 cm = 9 cm = 6 cm 4 cm 4.8 cm


XM 12cm MN That is, 9 cm = PR  4cm PR = 9 cm 4.8 cm

From 10 cm = 9 cm , it implies that 4.8 cm 4 cm = 9 cm 4.8 cm = 10.8 cm.
4 cm
PR
4.8 cm



XM 12cm That is, 9 cm = That is, 4cm PR = = 9 cm 4.8 cm 9 cm 4.8 cm
 4cm
4cm PR =


PR
10 cm 12 cm 9 cm PR
XM = = 13 cm. 9 cm 4.8 cm 9 cm 4.8 cm

1
10.8 cm.
=
9 cm 3 PR = 4cm PR = = 10.8 cm.
But, 4cm
MY XM XY 13 cm 10 cm 3 cm= − = 1 3 − = 1 3 Therefore, PR 10.8 cm.=
47
Student's Book Form Two
MATHEMATIC F2 v5.indd 47
MATHEMATIC F2 v5.indd 47 11/10/2024 20:11:34
11/10/2024 20:11:34