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length Questions

1.  Two coils which are made by winding copper wire of different gauges and length have the same mass. The first coil is made by winding 270 metres of wire with cross sectional diameter 2.8mm while the second coil is made by winding a certain length of wire with cross-sectional diameter 2.1mm. Find the length of wire in the second coil . (4 marks)

2.  The figure below represents a model of a hut with HG = GF = 10cm and FB = 6cm. The four slanting edges of the roof are each 12cm long.

Image From EcoleBooks.com

 

 

 

 

 

 

 

 

 

 

 

 

ecolebooks.com

 

Calculate

  1. Length DF. (2 mks)
  2. Angle VHF (2 mks)
  3. The length of the projection of line VH on the plane EFGH. (1 mk)
  4. The height of the model hut. (2 mks)
  5. The length VH. (1 mk)
  6. The angle DF makes with the plane ABCD. (2 mks)

3.  A square floor is fitted with rectangular tiles of periemeter 220 cm. each row (tile length wise) carries 20 less tiles than each column (tiles breadth wise). If the length of the floor is 9.6 m.

Calculate:

a.  The dimensions of the tiles (6 marks)

b.  The number of tiles needed  (2 marks)

c.  The cost of fitting the tiles, if tiles are sold in dozens at sh. 1500 per dozen and the labour cost is sh. 3000 (2 marks)

4.  Simplify; by factorization:

 15x2 + xy – 6y2

5x2 – 8xy + 3y2

 

5.  Given the matrices M = 3 0 , R = -1 2 and N = 2/3 1 . Find the value of value

of 3n + ½ (R-M) -1 4 0 0   2 4  

Length Answers

 

1.

Mass = Density x volume

But Density is constant.

x y

 

Vol (270000 x 2.8): x2.1

 

= 270000 x 2.8 = 360m

2.1

M1

 

M1

 

M1

 

A1

 
  

4

 

2.

a.)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b.)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C.

 

 

 

 

 

 

 

d.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

e.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

f.

Image From EcoleBooks.com

 

 

 

 

 

 

 

 

 

 

 

DF1 = √62 + (√200)2

 

= 15.362291

 

 

 

Image From EcoleBooks.com

 

 

 

 

 

 

 

 

 

 

 

½ 200

 

 


2

144 – ½ 200

 

= 9.6953597

 

6

+ 15. 6953597

 

Tan Ө = 15.6953597

7.0710678

 

= 2.219659

 

Ө = 65.747499o

 

½ HF

 

½ x √200

 

= 7.0710678

 

V

 

 

 

 

h 12 7.071

 

 

 

D B

√200

 

√144 – 50 = 9.6953597

 

Height = 9.6953597 + 6

 

= 15.69536 V

 

25.13234

15.219544

 

 

 


H

14.142136

 

 

 

 

14.1421362 + 15.2195442

 

= 25.13234

 

BDF

 

D 14.142136 B

Ө

 

15.362291 6 cm

 

F


Ө = 90

6 15.362291

 

Ө = 90 X 6

15.362291

 

= 35.150

 

 

 

 

 

 

 

 

 

 

 

M1

 

 

A1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M1

 

 

 

 

A1

 

 

 

 

 

B1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M1

B1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M1

 

 

 

 

 

A1

10

 

3

 Image From EcoleBooks.com

Dimensions 30 by 80 cm

 Image From EcoleBooks.com

No of tiles =

 Image From EcoleBooks.com

Cost =

 Image From EcoleBooks.com

 

 

  
  

10

 

 

4.   3x + 2y) (5x – 3y)

 (5x – 3y) (x-y)

= 3x + 2y

  x -y

 

5.  3N + ½ (R-M)

Image From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.com

= 3 + ½  –

Image From EcoleBooks.com

 

=   + ½

Image From EcoleBooks.comImage From EcoleBooks.com

Image From EcoleBooks.com =   +   =

 

 


 




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EcoleBooks | Length Questions and Answers

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