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studybuddy:

aceurmaths:

[quote=\"clblinym\"]Need help with a question... many thanks. Amy

http://i44.tinypic.com/2lc8snl.jpg\">

You still need any help ?

Q12 part a.
height of water = 3/8 * 48cm = 18cm
surface area = base area + rectangular area + 2 trapezium areas + slant area
= 90 *56 + 56 *18 + 2* (.5*(90+75)*18) + 56* 23.4
= 10330 = 10300 sqcm (to 3sf)



the slant height is 23.4cm using pythagoras theorem for bottom right triangle with other lengths 18cm and (90-75) = 15cm.

part b.
volume of water in part a = 75 * 56 * 18 + 0.5 *15*56*18
= 83160 cm3
25% increase = vol of 9 cylinders = 83160*0.25 = 20790 cm3
vol of 1 cylinder = 2310 cm3
= pi*r^2 * h
r^2 = 2310 / (3.142 * 16) = 45.950
r = 6.7786
diameter = 2r = 13.557 = 13.6cm (to 3 sf)[/quote]
thanks.. I understand it now

Need help with a question... many thanks. Amy

http://i44.tinypic.com/2lc8snl.jpg\">

http://i49.tinypic.com/2mfnion.png\">

Another one is from Rosyth 2001 SA1 Q8.

The answer given is 2. Can anybody explain why? Many thanks. Amy

http://i49.tinypic.com/14shc9h.jpg\">

This is from RGS 2012 Sa1 Q25
Can anybod here explain why the answer given is 3. TIA. Amy

kitty2:

Simply Mum:

Any body has good secondary 1 maths tutor to recommend? Tks in advance!

Do you mind tuition centre?

My son is also weak in maths. Can you tell me the tuition centre? TIA.

tianzhu:

clblinym:

From the solution, I do not understand why the area of AOB + COD = that of AOC + BOD. Can somebody help to explain? TIA. Amy

Hi

The bases of AOB + COD are the same as the length of the rectangle (AB=CD).The sum of the heights of these two triangles is equal to the breadth of the rectangle (AC and BD).Hence, their combined areas is equal to half the area of the rectangle.

As an illustration, let the height of triangle AOB be h1 and the height of triangle COD be h2.

Area of triangle AOB ------ ½*AB*h1
Area of triangle COD ------ ½ *CD*h2

Both triangles have similar bases as AB=CD.
AB=CD=length of rectangle ABCD

Their sum of areas is equal to ½ *length of rectangle*(h1+h2), where (hi+h2) is equal to the breadth (AC and BD) of the rectangle.

Therefore, their areas is equal to ½ *CD*BD which is equal to half the area of the rectangle.

The same logic is applicable to the triangles AOC and BOD.

AOB + COD = half the area of rectangle

AOC + BOD = half the area of rectangle

Hence,the area of AOB + COD = that of AOC + BOD.

Best wishes

Tianzhu,

Now I and my son understand. I really appreciate your prompt and detailed explanation. Thank you very much.

Pls help with the following question with thanks. It is from Henry Park, P5 2011, SA1, Paper2, Qn15.



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and the solution is as follows:
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From the solution, I do not understand why the area of AOB + COD = that of AOC + BOD. Can somebody help to explain? TIA. Amy

My son had some problem in solving the question below. Can anyone help? TIA http://i44.tinypic.com/33w43yr.png\">

Dear MathIzzzFun,


Thank you very much for your help. My son and I can understand. Really appreciate it.