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Surds Questions

1.  Given that Image From EcoleBooks.comand Image From EcoleBooks.comis an acute angle, find without using tables or calculators, Sin (90 – Image From EcoleBooks.com), leaving your answer in simplified surd form. (2mks)

2.  Given that Image From EcoleBooks.com, 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)

Image From EcoleBooks.com

3.  Simplify Image From EcoleBooks.comand hence evaluate Image From EcoleBooks.com to 3 significant figures given that Image From EcoleBooks.com. (3mks)

4.  Without using mathematical tables or calculators, find the volume of a container whose base is a regular hexagon of side Image From EcoleBooks.comcm and height Image From EcoleBooks.comcm (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  

ecolebooks.com

( 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.

Image From EcoleBooks.com

 

Image From EcoleBooks.com

Sin (90 – Image From EcoleBooks.com) Image From EcoleBooks.com

 

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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