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 By: JessicaTan (offline)  Thursday, October 18 2012 @ 07:26 AM CDT (Read 853 times)
JessicaTan

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Registered: 10/17/09
Posts: 82

 By: echeewh (offline)  Friday, October 19 2012 @ 05:02 AM CDT
echeewh

Hey Jessica,

(Henry Park 2011 SA2 Paper 2 Q13).

Kindly refer to figure in question. Such kind of question is very typical in most schools' papers.

Following pls find my worked solution:

Apply <Restating the problem> method,

T : R
1 : 3

T (Triangle) consists 2 parts:
(a) ST (Shaded Triangle) --> 2u [ given that (2/5) of T is shaded ]
(b) UT (Unshaded Triangle) --> 3u [ (3/5) of T is unshaded ]

Since T is made up of 5u, multiply the ratio, T : R , 5 times.

T : R
(1 : 3) × 5

(5 : 15)

R (Rectangle) consists of 15u, which makes up these 2 parts:

(a) SR (Shaded Rectangle) --> 2u
(b) UR (Unshaded Rectangle) --> 13u

Hence , whole figure = UT + S(T/R) + UR
= 3u + 2u + 13u
= 18u

Unshaded part of figure = UT + UR = 3u + 13u = 16u

Fraction of Whole figure that is Unshaded:
16u/18u = 8/9

==========================

Trust this helps.

Do let me know again if there's further clarification.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 By: JessicaTan (offline)  Friday, October 19 2012 @ 10:57 PM CDT
JessicaTan

Thanks for clear working . I have seen quite similar this kind of questions, sometimes I feel confuse also.

Rdgs
Jessica

Regular Member

Registered: 10/17/09
Posts: 82

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