Page 152 - Mathematics_Form_3

Circles

to ‾ XY and ‾ ON is perpendicular to ‾ PQ . But, MN = OM ON+
Draw ‾ OX and ‾ OP . It implies that, = 3cm 4.58cm+

OX = OP 5cm= (radius). = 7.58 cm

1 Therefore, the distance between the
FOR ONLINE READING ONLY
But, XM = XY ( since XY ⊥ MN )
2 chords is 7.58 cm.
Thus,
1
XM = × 8cm
2 Exercise 5.10
= 4 cm .
In questions 1 to 4, O is the centre
Similarly, of each circle. Find the value of x
1 (all measurements are in centimetres).
PN = PQ ( since PQ MN⊥ )
2 1.
1
= × 4cm
2

= 2 cm O
In triangle OXM, apply the Pythagoras'
7
theorem as follows. x
( OM ) ( ) ( XM ) 2 A 5 P B
2
2
=
OX −
2
2
2
( ‾ OM ) = (5 cm) − (4 cm)
2.
2
= 25 cm – 16 cm R
2
= 9 cm 8
2
OM = 9 cm 2 O
= 3 cm 3
x
In triangle OPN, apply the Pythagoras'
A P B
theorem as follows.
2 2 2 Mathematics for Secondary Schools
( ‾ ON ) = ( ‾ OP ) − ( ‾ PN ) 3. Q
2
2
2
( ‾ ON ) = (5 cm) − (2 cm) 3.6 B
= 25 cm − 4 cm
2
2
O
= 21 cm 5
2
⇒ ON = 21cm 2
A x P
= 4 . 58 cm

Student\s Book Form Three 145

18/09/2025 09:59:48
MATHEMATIC F3 SB.indd 145 18/09/2025 09:59:48
MATHEMATIC F3 SB.indd 145