HKDSE數學必修部分模擬試題卷二題解 Solutions to HKDSE Maths Compulsory Part Paper 2 (Sample Paper) - Part 3

以下是HKDSE數學試卷必修部分卷二（英文版下載/ 中文版下載）第11～20題的答案。

（注意：若你正在尋找考評局2012年1月出版的練習試題答案，請參閱此頁。）

如欲查閱第1～10題答案，請參閱題解Part 1。

如欲查閱第11～20題答案，請參閱題解Part 2。

如欲查閱第31～45題答案，請參閱題解Part 4。

請注意：此試題題解只供英文版本，歡迎轉載，但請提供來源（網址： http://hkdsefighter.blogspot.com/2011/06/hkdse-solutions-to-hkdse-maths_19.html）

21. The answer is B.

Join BD.

∠ADB=90° (∠ in semi-circle)

In triangle BCD,

∠BCA=90° (∠ in semi-circle)

∠BCD=90°+28°=118°

∠CBD=∠CDB (base ∠s, isos triangle)

∠BCD+∠CBD+∠CDB=180° (∠sum of triangle)

∠BCD+2∠CDB=180°

118°+2∠CDB=180°

∠CDB=31°

∠ADC=∠ADC+∠CDB=90°+31°=121°

22. The answer is C.

Join DA.

∠ABE=∠ADE=30° (∠s in the same segment)

∠ADC=105°-30°=75°

∠AOC=2∠ADC=75°×2=150° (∠ at centre= 2∠ at circumference)

23. The answer is B.

TriangleAFB~TriangleDFE~TriangleCBE (AAA)

CD:DE=AF:FD=2:1

AD=BC (opp. sides, //gram)

AF:BC=AF:AD=AF:(AF+FD)=2:(1+2)=2:3

24. The answer is D.

∠ADB=20°=∠DAB (ext. ∠ of triangle)

Therefore, AB=BD=10 cm (base ∠s, isos. triangle)

As BD=CD (given),

∠DBC=∠DCB=40° (base ∠s, isos. triangle)

∠CDB+∠DBC+∠DCB=180° (∠ sum of triangle)

∠CDB=180°-40°-40°=100°

CB÷sin100°=BD÷sin40° (sine law)

CB=10÷sin40°×sin100°

CB=15cm

25. The answer is A.

The figure on the right can be obtained by rotating the figure on the left by 90 degrees clockwise.

26. The answer is D.

As the point is rotated by 180 degrees, the MAGNITUDE of the x-coordinate and the y-coordinate DO NOT CHANGE.

Point (-4,3) is in Quadrant II. After rotating about the point of origin by 180 degrees, the image will be locate in Quadrant IV. So the image is at (4, -3).

27. The answer is D.

The lower quartile= 50 marks (The left end of the box)

So 25% of the students got less than 50 marks.

Passing rate=1-25%=75%

28. The answer is B.

There are 23 staff members.

As (23+1)÷2=12, the height of the 12th shortest staff member is the median.

The median= 165cm.

29.

Mean of the first set= [(a-7)+(a-1)+a+(a+2)+(a+4)+(a+8)]÷6=a+1

Mean of the second set= a+1/6

So I is wrong.

Median of the first set= [a+(a+2)]÷2=a+1

Median of the second set= [(a-1)+(a+3)]÷2= a+1

So II is wrong.

Standard deviation of the first set= 5.05…

Standard deviation of the second set= 7.06…

So III is wrong.

30.

I is correct: Only 2 people out of 950 are selected.

II is correct: Only the chairpersons of the Student Union are selected.

III is correct: Other students have no opportunity to be selected.

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