Welcome to Test-paper.info
Sunday, February 25 2018 @ 03:29 AM CST
 Select a Forum » General Chat » Primary 1 Matters » Primary 2 Matters » Primary 3 Matters » Primary 4 Matters » Primary 5 Matters » Primary 6 Matters » Question, Feedback and Comments » Education Classified
 Forum Index >  Test Paper Related >  Primary 6 Matters Maths Question
 | Printable Version
 By: sundarshoba (offline)  Friday, June 22 2012 @ 05:43 PM CDT (Read 1816 times)
sundarshoba

Please assist to work out the steps for the attached questions. Thks

Chatty

Registered: 12/31/06
Posts: 53

 By: GIM (offline)  Sunday, June 24 2012 @ 09:58 AM CDT
GIM

oQ15)Here is my solution

Since AE=6cm and BC=12cm
Therefore, ratio of AE:BC is 1:2 and ratio of AF:FC is also 1:2
Length AB is divided into 3 equal parts of 4+4+4 due to the above ratio

Area BCF = 1/2 x 12 x 8cm sq = 48 cm sq

Area ABE = 1/2 x 12x 6 = 36cm sq

Shaded Area = (12 x12) - 48 -36 = 60 cm sq (answer)

Regular Member

Registered: 11/07/10
Posts: 70

 By: echeewh (offline)  Sunday, June 24 2012 @ 09:50 PM CDT
echeewh

Hey sundarshoba

Following pls find worked solution for Q14 (as per attached).

Pls refer to attached diagram in question.

Label diagram as shown.
(*** - buffer up for alignment purpose)

Join diagonal BD. 2 Big (B ) Quadrants (CBD, centre C) and (ABD, centre A); 4 Small (S) Quadrants (FOD, centre F), (GDO, centre G), (HBO, centre H), (EOB, centre E)

A***********E***********B
**|-----------------------------|
**|***********|************|
**| **********|************|
**|*********O|************|
G|-----------------------------|H
**|***********|************|
**|***********|************|
**|***********|************|
**|-----------------------------|
D***********F***********C

in Triangle ABD, label the larger unshaded part, X. Label the 2 unshaded half-leaves, Y (enclosed by S [(FOD, centre F)]) and Z (enclosed by S [(HBO, centre H)]) respectively.

Unshaded X = Big Square (ABCD) - B (CBD, centre C)
= (20 x 20) - [(pi x 20 x 20)/4]
= 400 - 100 pi

in Triangle ABD, there are 2 unshaded half-leaves. Label one of them Y.

Y = S [(FOD, centre F)] - Triangle FOD
= [(pi x 10 x 10)/4] - [(10 x 10) / 2]
= 25 pi - 50

1 Shaded area (in Triangle ABD)
= Triangle ABD - 2Y - X
= [(20 x 20) / 2) - 2 (25 pi - 50) - (400 - 100 pi)]
= 200 - 50 pi + 100 - 400 + 100 pi
= 50 pi - 100

in Big Square ABCD, there are 2 of these shaded areas.
Total Shaded area = 2 x (50 pi - 100)
= 100 pi - 200
= 114.16 cm2 (nearest 2 dec places)

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

Trust this helps.

Do let me know again if this does not agree with the answerkey.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 By: sundarshoba (offline)  Monday, June 25 2012 @ 12:00 AM CDT
sundarshoba

Thank you very much for the elaborative answer, Thanks

Chatty

Registered: 12/31/06
Posts: 53

 By: echeewh (offline)  Tuesday, June 26 2012 @ 05:48 PM CDT
echeewh

Hey Sundar,

An alternative solution to Q15 (as provided by GIM)

Q15

Join the line DF.

Triangle FDC + Triangle FAB = 1/2 Square ABCD = Triangle FDC + Triangle FAD

In other words, Triangle FAB = Triangle FAD

Given that AE = ED, Triangle FAE = Triangle FED. This implies that Triangle FAD --> 2u area. So is Triangle FAB.

Hence, Triangle BAE --> 3u area.

Area of Triangle BAE = (6 x 12) / 2 = 36 --> 3u
Triangle FAE = 1u --> 12
= [(12 x 12) / 2] - 12
= 72 - 12 = 60 cm2

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

Trust this helps.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 All times are CST. The time is now 03:29 am.
 Normal Topic Locked Topic Sticky Topic
 New Post Sticky Topic w/ New Post Locked Topic w/ New Post
 View Anonymous Posts Able to Post HTML Allowed Censored Content