1991-02-18 — Page 26

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報日僑華

一期星

日八十月二 { 一九九一)年十八國民華中 (26

As shown in the figure,

Tet r =radius x = OD

AD = JAL BD

108

= 6 cm

X. r. 6

In rt.

Figure 5

002 + 002

(r-6)2 + 82

12r+36 +64

12r = 100

r = 8.33 cm

Mathematics (20)

Exercise, 20 Mensmation (1).

Section A

H. K. Lo

1.As shown 'in figure 1, AB=AC=10 cm, BC= 16 cm. Find the radius of the

circle.

Section

6. As shown in figure ba right circular cone is divided into 3 proportions A, Bànd C by planes The heights parallel to the base.

of the portions 1, 2 and 3 units respectively. Find

(a):the ratio of the volume of Ato.

that of B.

(b) the ratio of the volume of B to

that of C.

[{Ans.)

= area of semi-circle AEC

HACH

= area of semi-circle BDC

(a) Let r

radius of the circle.

2 x 15°

= (BC)

=.30°

area of pt. a

Area:of shaded ran (AC)(BC)

(Az+A3)={Aj-Ag)

BCZY

19.1 cm

(Ans.)

AC+

(b) Area of shaded sector

{AC2+BC2-AB?} +

AC •BC

AC-BC

2x (19.1) x

.95.5 cm2

(Ains.)

Section. B

16cm

Figure.

2. As shown in figure 2, 0 is the centre of the circle, A=0B=10 cm and AOB 120°,

the shaded region.

Find the area.of.

Figure 6

7. In figure 7, ABC is a right-angled:

triangle with right angled at C. AFC, CD and ACB are semi-circles with diameters AC, CB and AB.

Area of CATAOB

(OA)(QB)ṣinZAOB

+(10)(10)sin120°

43.3 cm2

Area of sector ADB

120°

30 XX (10)?

104.7 cm2:

Area of the shaded region

104.7 --43.3

= 61.4 cm2

Let VVolume of

120

Figure 2.

3. In figure 3, ABCD is a parallelogram with area:72cm2. HK are the mid-

points of AB and CD find the area

of the shaded region.

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Figure 3

Infigure 4, ABC is an equilateral

triangle of side 6 cm. A circle,

centre at-0, passes through A,B and

C. Find 'the radius of the circle.

Area of rt. & ACB

the required ratio is 1:1.. (Ans.)

(a) tan/BAD

Figure 7

Find the ratio of the area of

shaded :region.

triangle ABC.

8. As shown in figurex

with centres A, B d

Area of paralieldgram KHBC

x 72 cm2

circles

"Touch each

4-cm and

= 36 cm2

Area of shaded region

(a)

of the shaded:

(h) the area of the shaded region.

other. Of the radit of the circles ares respectively. 5. cm,

régie

9 ст

8.4 cm

4 cm)

HONG

圖

(b)

Volume of 8

Volume of C

= Curved surface area of A

= Curved surface area of B

= Curved surface area of C

41:27

Area of a EOH + Area of quad. TOFB

= Area of 4KOF+ Area of quad, HOFH

= Area of HRA

* (Area of HBCK)

3x36cm2

18 cm2

#1: (27-1) : (216-27)

#1: 26: 189

the required. ratiois

(Sa+Sg) (S

12(1+2)2

(Ans.):

(Ans.)

189.

IBRARIES

(Ans.)

袋

LBAD ¥ 23.96 0.

LBAC='2. X23.96°

47.929

Length of arc (EF

360 X (2x)(5)

= 4.182.cm

LABC = 2 × (180°)

=66040

Length of arc pf

Length of arc.DE.

66.04 360

4.61 cm

(2)(4)

47.92")

Perimeter of the shaded region

4.19 2.x 4.61

13.4 cm.

(b) Area of sector AFE.

47.92 x (π){5}2

360

= 10.45 cm2

Area of sector OCE

Area of sector BDF

66.04

{8}{4}2

{Ans,}

igure 8

Solutions to Exercise 20

Section A

Let r = 'radius of the circle:

KONG PUBLIC LIB

10 cm

9 cm 0 8 em

BC.4 2rco530°

BC 200530

2cos30

3.464 cm

(Ans)

佳麗安

area of semi-circle ACB

(AB)

CARRY ON TRAVEL

5264549-0

360°

9.22 cm2

Area of triangle. ABC

子(AB)(AC)sinLBAC.

+7 (9}(9)sin47.920.

=.30.06 cm2.

Area of the shaded region

= 30.06cm2 - (9,22cm2+9,22cm2+10,45cm?)

= 1.17 cm2

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Figure 4

5. In figure 5. 0. is the centre of the

circle. Length. of-arc APB = 10. cl

and LACB = 15°, find

(a) radius of the circle.

(b) area of the shaded sector.

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1991-02-18 — Page 26

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