Page 156 - Mathematics_Form_3

Circles

Solution From the figure, it follows that
(a) From the figure, it follows that AXBX× = CXDX× (intersecting
AEBE× = CEDE× (intersecting chords)
chords). Thus, Thus,
FOR ONLINE READING ONLY
8cm x×= 6cm 4cm×
24cm 2 4cm 3cm× = 2cm DX×
x
⇒= ⇒ 12cm = 2 2cm DX×
8cm
= 3cm. ⇒ DX = 12cm 2
Therefore, x = 3 cm. 2cm
(b) From the figure, it follows that = 6cm

UW XW× = VW YW× In BXD∆ , apply Pythagoras theorem
(intersecting secants). Thus, 2 2 2

8cm 6cm 12cm x× = × BD = BX + DX
2
BD =
2
48cm = 12cm x× Thus, 3 + 2 6 = 2 2 45 2
2
2
2
3 5 cm
45 =
BD =
x = 48cm BD = (3cm) + (6cm)
12cm = 45cm 2
4cm=
Therefore, x = 4cm. BD = 45 cm
= 3 5 cm
Example 5.21
In AXC∆ , apply Pythagoras theorem
Two chords AB and CD intersect 2 2 2

at right angles at X. If CX = 2 cm, AC = AX + CX
2
AC =
AX = 4 cm, and BX = 3 cm, find the Thus, 4 + 2 2 = 2 20
2 2 2
2
AC AC = (4cm) + (2cm)
BD =
2 5 cm
20 =
value of .
BD = 20cm 2
Solution
Consider the following figure. AC = 20 cm Mathematics for Secondary Schools
C
= 2 5 cm
AC 2 5 2
Hence, = = .
2 cm
A 4 cm X 3 cm B BD 35 3
2 5
AC 2 AC 5 2 2
= .
Therefore, = = = .
BD 35 3 3
35
BD
D

Student\s Book Form Three 149

18/09/2025 09:59:51
MATHEMATIC F3 SB.indd 149
MATHEMATIC F3 SB.indd 149 18/09/2025 09:59:51