
By: sundarshoba (offline) Friday, June 22 2012 @ 05:43 PM CDT (Read 1645 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************
GH
*************************
*************************
*************************
**
D***********F***********C
in Triangle ABD, label the larger unshaded part, X. Label the 2 unshaded halfleaves, 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 halfleaves. 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: 625





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
Hence, Shaded area = Triangle ADC  12
= [(12 x 12) / 2]  12
= 72  12 = 60 cm2
======================================
Trust this helps.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 625



