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 By: parent603 (offline)  Sunday, July 01 2012 @ 10:53 PM CDT (Read 3156 times)
parent603

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Registered: 11/07/11
Posts: 14

 By: jo sarah (offline)  Wednesday, July 04 2012 @ 08:55 AM CDT
jo sarah

Hi,

Label the square starting from top left corner clockwise as ABCD, and centre of square is centre of circle P.

Area of a leaf between points A and P = area of semicircle radius 6 cm - area of square base 12 cm and height 6 cm
= 0.5 x (pi) x 6 cm x 6 cm - 0.5 x 12 cm x 6 cm
= 20.5486 cm^2

therefore,
Area of shaded region = area of triangle ABC - area of leaf
= 0.5 x 12 cm x 12 cm - 20.5486 cm^2
= 51.4513 cm^2
= 51.45 cm^2 (to 2 d.p.)

Hope this helps.
jo sarah

Regular Member

Registered: 03/20/12
Posts: 111

 By: jo sarah (offline)  Wednesday, July 04 2012 @ 08:56 AM CDT
jo sarah

Hi,

Label the square starting from top left corner clockwise as ABCD, and centre of square is centre of circle P.

Area of a leaf between points A and P = area of semicircle radius 6 cm - area of square base 12 cm and height 6 cm
= 0.5 x (pi) x 6 cm x 6 cm - 0.5 x 12 cm x 6 cm
= 20.5486 cm^2

therefore,
Area of shaded region = area of triangle ABC - area of leaf
= 0.5 x 12 cm x 12 cm - 20.5486 cm^2
= 51.4513 cm^2
= 51.45 cm^2 (to 2 d.p.)

Hope this helps.
jo sarah

Regular Member

Registered: 03/20/12
Posts: 111

 By: jo sarah (offline)  Wednesday, July 04 2012 @ 08:59 AM CDT
jo sarah

Hi,

Let's label the square starting from top left corner clockwise as ABCD, and centre of square is centre of circle P.

Area of a leaf between points A and P = area of semicircle radius 6 cm - area of square base 12 cm and height 6 cm
= 0.5 x (pi) x 6 cm x 6 cm - 0.5 x 12 cm x 6 cm
= 20.5486 cm^2

therefore,
Area of shaded region = area of triangle ABC - area of leaf
= 0.5 x 12 cm x 12 cm - 20.5486 cm^2
= 51.4513 cm^2
= 51.45 cm^2 (to 2 d.p.)

Hope this helps.
jo sarah

PS: is isn't clear if there is to be a small portion to be added on the left of the figure. In my solution, i have treated it as not to be there and ignored it.

Regular Member

Registered: 03/20/12
Posts: 111

 By: Jane Lim (offline)  Tuesday, July 10 2012 @ 12:34 AM CDT
Jane Lim

Hi,
Thank you.

Newbie

Registered: 10/01/10
Posts: 11

 By: jo sarah (offline)  Tuesday, July 10 2012 @ 09:27 AM CDT
jo sarah

hi,

someone has pointed out that i seem to have missed even a small portion (on the left), and that could mean that the answer might even be larger than mine.

It is a little uncertain if the "little portion" on the left of the figure is included in the original question (for we know that many of these papers are cleaned out copies.

Please also note that we might not necessarily go by the so-called given answers all the time because i have encountered many an incorrect answer before though i am not sure about this one yet.

many thanks.

Regular Member

Registered: 03/20/12
Posts: 111

 By: echeewh (offline)  Monday, July 16 2012 @ 01:38 AM CDT
echeewh

Dear All,

I have already done some research on the question itself, esp on the diagram. And came up with a 're-engineered' one of what I believe could be the actual one. Appreciate that you take a look at that in a new post created by me entitled "Henry Park Prelim 2011 Paper2 Qn 10 - Updates".

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 625

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