## 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

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( 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 – 26

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