Q&A - PSLE Math
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Hi MathIzzzFun,
I got it now. Thank you for your clear picture.
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Hi…I have a doubt in an Angle sum but can’t understand if i just write here without a picture. Anybody pls. highlight any chance i can attach an image here.? If so pls. explain how to attach an image so that draw the angle & post here.Thanks.
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Mary Joy:
Hi..I have a doubt in an Angle sum but can't understand if i just write here without a picture. Anybody pls. highlight any chance i can attach an image here.? If so pls. explain how to attach an image so that draw the angle & post here.Thanks.
Click on \"Post Reply\" at the end of the last post to go into the edit/post mode.
scroll down the screen and you will see the \"tinypic\" plugin.. attach your picture file (in image format - jpg, png, bmp), resize image size to \"message board\" -- click upload and then add to post.
cheers. -
MathIzzzFun:
Mary Joy:
Hi..I have a doubt in an Angle sum but can't understand if i just write here without a picture. Anybody pls. highlight any chance i can attach an image here.? If so pls. explain how to attach an image so that draw the angle & post here.Thanks.
Click on \"Post Reply\" at the end of the last post to go into the edit/post mode.
scroll down the screen and you will see the \"tinypic\" plugin.. attach your picture file (in image format - jpg, png, bmp), resize image size to \"message board\" -- click upload and then add to post.
cheers.
Hi Mathizzfun...Thanks for the feedback. -
Hi…Pls. help me to find out the answer.
The figure below is not drawn to scale.CDEF is a trapezium and AB=BC.
<EFC is twice of <FCA. BCD is a right angled triangle.
a) Find <FCD
b) Find <DBA -
Mary Joy:
This was discussed on pg 862 ... http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&start=8610![](\"http://i47.tinypic.com/2n8xah4.jpg\")
http://i47.tinypic.com/2n8xah4.jpg\"> Hi..Pls. help me to find out the answer.
The figure below is not drawn to scale.CDEF is a trapezium and AB=BC.
<EFC is twice of <FCA. BCD is a right angled triangle.
a) Find <FCD
b) Find <DBA
cheers. -
quote=\"SAHMwith2boys\"]Can someone help with the below question:
![](\"http://i46.tinypic.com/14qyd3.jpg\")
It's taken from 2012 CHIJ CA1 paper.[/quote]![](\"http://i48.tinypic.com/hvqj42.png\")
cheers.[/quote]
This method seems a bit tough for Primary level, was wondering is there any other way to do this?
[/quote]
The solution is not beyond P6 students.
2 key information are provided - triangle ABC is isosceles and angle AGE = 75 deg. So, one has to made use of these to work out the solution.
another approach is to work out the angle CAG in terms of the angles EFC and FCA and then use the interior angles of quadrilateral ACFG to work out angle FCA (1u)
angle CAG = 180 -(90-1u) = 1u + 90
in quadrilateral ACFG
angle EFC + angle FCA + angle CAG + angle AGE = 360
2u + 1u + 1u + 90 + 75 = 360
4u = 195, 1u = 48.75![](\"http://i45.tinypic.com/qstohc.png\")
cheers.[/quote]
Hi..,
Thanks for the answer. Actually this is from CHIJ question paper and the answer given is 6 degree. But your's is 7.5 degree. -
quote=\"SAHMwith2boys\"]Can someone help with the below question:
It's taken from 2012 CHIJ CA1 paper.[/quote]
cheers.[/quote]
This method seems a bit tough for Primary level, was wondering is there any other way to do this?[/quote]
The solution is not beyond P6 students.
2 key information are provided - triangle ABC is isosceles and angle AGE = 75 deg. So, one has to made use of these to work out the solution.
another approach is to work out the angle CAG in terms of the angles EFC and FCA and then use the interior angles of quadrilateral ACFG to work out angle FCA (1u)
angle CAG = 180 -(90-1u) = 1u + 90
in quadrilateral ACFG
angle EFC + angle FCA + angle CAG + angle AGE = 360
2u + 1u + 1u + 90 + 75 = 360
4u = 195, 1u = 48.75
cheers.[/quote]
Hi..,
Thanks for the answer. Actually this is from CHIJ question paper and the answer given is 6 degree. But your's is 7.5 degree. -
Mary Joy:
quote=\"SAHMwith2boys\"]Can someone help with the below question:
It's taken from 2012 CHIJ CA1 paper.
cheers.[/quote]
This method seems a bit tough for Primary level, was wondering is there any other way to do this?[/quote]
The solution is not beyond P6 students.
2 key information are provided - triangle ABC is isosceles and angle AGE = 75 deg. So, one has to made use of these to work out the solution.
another approach is to work out the angle CAG in terms of the angles EFC and FCA and then use the interior angles of quadrilateral ACFG to work out angle FCA (1u)
angle CAG = 180 -(90-1u) = 1u + 90
in quadrilateral ACFG
angle EFC + angle FCA + angle CAG + angle AGE = 360
2u + 1u + 1u + 90 + 75 = 360
4u = 195, 1u = 48.75
cheers.[/quote]
Hi..,
Thanks for the answer. Actually this is from CHIJ question paper and the answer given is 6 degree. But your's is 7.5 degree.[/quote]
use the given answer and work backwards to see whether it is correct.
The answer key is not always correct
cheers -
MathIzzzFun:
This method seems a bit tough for Primary level, was wondering is there any other way to do this?Mary Joy:
quote=\"SAHMwith2boys\"]Can someone help with the below question:
It's taken from 2012 CHIJ CA1 paper.
cheers.[/quote]
The solution is not beyond P6 students.
2 key information are provided - triangle ABC is isosceles and angle AGE = 75 deg. So, one has to made use of these to work out the solution.
another approach is to work out the angle CAG in terms of the angles EFC and FCA and then use the interior angles of quadrilateral ACFG to work out angle FCA (1u)
angle CAG = 180 -(90-1u) = 1u + 90
in quadrilateral ACFG
angle EFC + angle FCA + angle CAG + angle AGE = 360
2u + 1u + 1u + 90 + 75 = 360
4u = 195, 1u = 48.75
cheers.[/quote]
Hi..,
Thanks for the answer. Actually this is from CHIJ question paper and the answer given is 6 degree. But your's is 7.5 degree.[/quote]
use the given answer and work backwards to see whether it is correct.
The answer key is not always correct
cheers[/quote]
Hi.. Thanks.. Yeah we have checked and your answer is correct.
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