P6 Maths Question

• baiyun

  12 Sep 08, 09:12

  Thank you for your help.

   

  

• Mospeada

  12 Sep 08, 09:15

  where got 4 suqares?

• Girl_Next_Door

  12 Sep 08, 09:25

  Let AD be X and AB be Y

   

  Solve this:

   

  2 X (square) + 2 Y (square) = 80 -----------------(1)

  2 X + 2 Y = 20 ---------------------------------------(2)

   

  When you find out X and Y, multipy it and you will get the area of your shaded area.

   

   

• baiyun

  12 Sep 08, 09:30

  Originally posted by Mospeada:

  where got 4 suqares?

  
  Two big squares and 2 small squares.Hope you can see them as the picture is not clear.

• baiyun

  12 Sep 08, 11:36

  Originally posted by Girl_Next_Door:

  Let AD be X and AB be Y

   

  Solve this:

   

  2 X (square) + 2 Y (square) = 80 -----------------(1)

  2 X + 2 Y = 20 ---------------------------------------(2)

   

  When you find out X and Y, multipy it and you will get the area of your shaded area.

   

   

  Hi

  Could you help to elaborate on

  2 X (square) + 2 Y (square) = 80 -----------------(1)

   

• weewee

  12 Sep 08, 11:40

  2 big square + 2 small square = 80cm2 area

  means 1 big square + 1 small square = 40cm2 area

  which means you take 36 + 4= 40. Lengthof big square is 6 cm. Length of small square is 2 cm.

  Area of shaded region = 6 x 2 = 12cm2

   

• baiyun

  12 Sep 08, 11:52

  Originally posted by weewee:

  2 big square + 2 small square = 80cm2 area

  means 1 big square + 1 small square = 40cm2 area

  which means you take 36 + 4= 40. Lengthof big square is 6 cm. Length of small square is 4 cm.

  Area of shaded region = 6 x 4 = 24cm2

   

  
  Hi

  Could you help to elaborate on

  which means you take 36 + 4= 40.

  If length of small square is 4, then it's area is 16sqcm, leaving area of big square as 24sqcm

• maurizio13

  12 Sep 08, 11:56

   

  6 + 2 = 8

  then perimeter only 16 not 20.

   

  i wonder if there is a problem with the question.

   

• Ahm97sic

  12 Sep 08, 15:44

  Let AD be x cm and AB be y cm

  Perimeter of the rectangle ABCD = 2x + 2y = 20 cm

                                                        x +   y = 10 

                                                        x         = 10 - y    ------------ (1)

  Total areas of the 4 squares = 2x^2 + 2y^2 = 80 cm^2

                                                 x^2  +  y^2  = 40        ------------- (2)

  Substitute x = 10 - y into equation (2) ie

                                                 (10 - y)^2 + y^2 = 40

                                                 100 - 20y + y^2 + y^2 = 40

                                                  2y^2 - 20 y + 60 = 0

                                                    y^2 - 10 y + 30 = 0

                                                    no real solution for y

  Hence, the question is WRONG !

  Alternatively, we can use the guess and check method

  x   y   Perimeter of the rectangle ABCD = 20 cm   Areas of 4 squares = 80 cm^2 

  1  9    1+9+1+9 = 20                                              2(1)^2 + 2(9)^2 = 163

  2  8    2+8+2+8 = 20                                              2(2)^2 + 2(8)^2 = 136

  3  7    3+7+3+7 = 20                                              2(3)^2 + 2(7)^2 = 116

  4  6    4+6+4+6 = 20                                              2(4)^2 + 2(6)^2 = 104

  5  5    5+5+5+5 = 20                                              2(5)^2 + 2(5)^2 = 100

  6  4    6+4+6+4 = 20                                              2(6)^2 + 2(4)^2 = 104

  So, it is impossible for the areas of the 4squares to be 80 cm^2. Hence, the question is Wrong ! 

   

   

                                                 

                                          

• weewee

  13 Sep 08, 00:00

  Originally posted by baiyun:

  
  Hi

  Could you help to elaborate on

  which means you take 36 + 4= 40.

  If length of small square is 4, then it's area is 16sqcm, leaving area of big square as 24sqcm

  
  paiseh made an error. question indeed got error and i was led into believing the sides were 6 cm and 4cm respectively.

  Question cannot be solved .

  x^2 + y2 = 40
  2x + 2y = 20
  x + y = 10
  x = 10 -y

  (10-y)^2 + y^2 = 40
  100 - 20y +2y^2 = 40
  2y^2 - 20y + 60 = 0
  which cannot be solved. discriminant b^2 - 4ac < 0.

• latte kid

  13 Sep 08, 00:12

  wah now p6 qns like amath siol. :X

• maurizio13

  13 Sep 08, 00:14

  very soon primary school kids will be learning C maths and F maths.

   

• skythewood

  13 Sep 08, 01:17

  how come got wrong question one...
• latte kid

  13 Sep 08, 11:15

  Originally posted by skythewood:

  how come got wrong question one…

  Teacher fail Amath also

• sinicker

  13 Sep 08, 11:18

  wow soon P6 can learn partial fractions already. differentiation/integration all!