# Mathematics 2010 | Study Mode

Question 61
A
$$\frac{6}{35}$$
B
$$\frac{10}{21}$$
C
$$\frac{21}{10}$$
D
$$\frac{35}{6}$$
##### Explanation
W $$\alpha$$ U

W = ku

u = $$\frac{w}{k}$$; $$\frac{2}{7}$$ x $$\frac{3}{5}$$

= $$\frac{6}{35}$$

Question 62
A
x < -1 or x > 1
B
-1 < x < 1
C
x > 0
D
x < -1
x(x - 1) > 0

x < -1 or x > 1

Question 63
A
75o
B
65o
C
55o
D
50o
##### Explanation
In the diagram above, < STU = < TRS = 25o

(< between tangent to circle and a circle and a chord through the point of contact = < in the alternate segment)

< RTS = 90o

(< in a semicircle = 90o)
xo + < RTS + < STU = 180o

xo + 90o + 25o = 180o

xo + 115o = 180o

xo = 180o - 115o = 65o

Question 64
A
20o
B
23o
C
24o
D
26o
##### Explanation
In the diagram above 88o + (3x + 20)o = 180o

(oppsite < s of circlic quad. are supplementary)

88o + 20o + 3x = 180o

108o + 3x = 180o

3x = 180o - 108o

= 72o

Question 65
A
7cm
B
8cm
C
5cm
D
6cm
##### Explanation
Let A denote the area of $$\bigtriangleup$$PQR, then A = $$\frac{1}{2} bh$$

Using Sin 60o = $$\frac{h}{q}$$

h = q sin 60o

So A = $$\frac{1}{2}b(q \sin 60^o)$$

12$$\sqrt{3} = \frac{1}{2} \times 8 \times q \times \frac{\sqrt{3}}{3}$$

12$$\sqrt{3}$$ - 2q$$\sqrt{3}$$

q = $$\frac{12}{2} = 6$$cm

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