Page 92 - MATHEMATICS COURSE FOR SECONDARY SCHOOLS BOOK 2
P. 92
4.9 cm B
= BD/OD
= 4.9 cm 5.4 cm
= 0.907
= arc tan 0.907
= 42.2° (correct to 1 d.p.) the minor sector angle AOB = 42.2° 2
D 5.4 cm
(e) The area of the minor segment, A = A1 - A2
(b) New tan BOD So BODO
= (39.1 – 26.5) cm2 = 12.6 cm2
EXERCISE 4.6
A chord, AB, of the length 6.8 cm is 4.5 cm away from the centre, O, of a circle.
Calculate:
(a) the radius,OB, of the circle
(b) the minor sector angle AOB
(c) the area of the minor sector AOB (d) the area of triangle AOB
(e) the area of the minor segment.
A chord, PQ, of length 5.4 cm is 3.5 cm away from the centre, O, of a circle. Evaluate:
(a) the radius OP, of the circle
(b) the minor sector angle POQ
(c) the area of the minor sector POQ (d) the area of triangle POQ
(e) the area of the minor segment.
A chord, AB, of length 9.6 cm is 6.5 cm away from the centre, O, of a circle. Determine:
(a) the radius OA, of the circle (b) the minor sector angle AOB
(c) A
= 84.4°
A=39.1 cm 4.9 cm
1.
minor sector AOB, A1, = π r2 0 360
2.
A
A
A=12.6 cm2 B
86
= 3.14 7.292 84.4 A A=26.5 cm2 360cmB2
(d) The area of triangle AOB, A2, = 2 bh
= 39.1 cm2 (correct to 3 s.f.)
h=5.4 cm
A=39.1 cm
O
b=9.8 cm
A=26.5 cm2
A=12.6 cm2
h=5.4 cm B
= 2 9.8 5.4 cm2
= 26.5 cm2 (correct to 3 s.f.)
B
The area of the
5.4 cm
O
O
b=9.8 cm
D
B
3.
42.2o
r=7.29 cm
42.2o
r=7.29 cm
r=7.29 cm
r=7.29 cm
0=84.40o
0=84.40o