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Surds Questions
1. Given that and is an acute angle, find without using tables or calculators, Sin (90 – ), leaving your answer in simplified surd form. (2mks)
2. Given that , express in surd form, rationalize the denominator and then find the value of the expression below to 5 significant figures without using the calculator. (3mks)
3. Simplify and hence evaluate to 3 significant figures given that . (3mks)
4. Without using mathematical tables or calculators, find the volume of a container whose base is a regular hexagon of side cm and height cm (4 mks)
5. Simplify; 3
1 leaving the answer in the form a + b c , where a, b and c are
rational numbers 7 –2 7
6. Given that:-
Find the values of a and b where a and b are rational numbers
7. If:- 14 – 14 = a 7 + b 2
7 – 12 7 + 2 Find the values of a and b, where a and b are rational numbers *
8. Rationalize the denominator 2- 2 and express your answer in the form of a + c 2
( 2 – 1)3
9. The figure below is a right pyramid with a rectangular base ABCD and vertex V.
O is the centre of the base and M is a point on OV such that OM = 1/3 OV, AB = 8 cm, BC = 6 cm
and VA = VB=VD = VC = 15 cm. Find
i) The height OV of the pyramid.
ii) The angle between the plane BMC and base ABCD.
10. Find the value of y which satisfies the equation
Log 10 5 – 2 + log10 (2y + 10) = log10 (y – 4)
11. Simplify the expression √ 3 – √ 2 giving your answer in the for of a + b √ c.
√3 + √2
Surds Answers
1. |
Sin (90 – ) |
B1
B1
B1 | |
02 |
1.. 3 + 1 = 3 7 + 2 + 1 7
7 – 2 7 7 – 4 7
3 + 1 = 3 7 + 7 -2)
7 – 2 7 7 -2 7
3 7 + (7-2)
7 – 2 7
= 3 7 + 7 -2 7 + 2 7
7 – 2 7 7 + 2 7
= 49 – 28
= (3 7 + 7-2) (7 + 2 7
21
= (4 7 – 2) 7 + 2 7
21
2.
4 + 4 5 + 5 – (6-3 5 + 2 5-5)
4-5
8 + 5 5
-1
a = -8 b = -5
3. 14 ( 7 + 2 ) – 14 ( 7 – 12 )
7 – 12
14 . 7 + 14. 12 – 14 . 7 + 14 . 12
-5
4.
( 2-1 )2= 2 – 2 2 + 1 = 3 2 2
( 2 – 1)3 = 2 – 1(3-2 2
= 5 2 – 7
2 – 2 x 5 2 + 7 = 2 2 + 7) – 2 2 +2)
5 2-7 5 2+7 1
= 17 2-6 = -6+17 2
5. (2 – 3 ) (3 + 2)
3(2)2 – 2)2
3×2 – 3 + 2 – 2
9×2 – 4×3
6 – 3 + 2 – 6
18 – 12 = 6
6. i) Or = 162 – 52
= 256 – 25
= 15.198 cm
ii) tan = 5.066 = 1.2665
4
∴
51.710
7. Log 105 – log 10 102 + log 10 (2y + 10) = log 10 (y – 4)
Log 10
5 (2y + 10) = log 10 (y-4)
102
10y + 50 = 100 y – 400
90 y = 450
y = 5
8. 3 – 2 3 – 2
3 + 2 3 – 2
= 3 – 6 – 6 + 2
3 – 6 + 6 -2
= 5 – 2 6
3 – 2
= 5 – 26