Puzz1
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| sree |
 Aug 20 2006, 07:48 PM
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#1
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 CORPORAL  Group: Members Posts: 80 Joined: 20-August 06 From: Hyderabad Member No.: 77  |
ABCD is a square, E is the mid-point of BC and F is the mid-point of CD. The ratio of the area of triangle AEF to the area of the square ABCD is _____
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| ymhatre |
Aug 20 2006, 10:55 PM
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#2
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SERGEANT  Group: Moderators Posts: 160 Joined: 10-June 06 Member No.: 36 |
1:8...i guess
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| sree |
Aug 22 2006, 09:16 AM
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#3
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CORPORAL Group: Members Posts: 80 Joined: 20-August 06 From: Hyderabad Member No.: 77 |
its 3:8
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| ymhatre |
Aug 22 2006, 10:21 PM
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#4
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SERGEANT Group: Moderators Posts: 160 Joined: 10-June 06 Member No.: 36 |
Ohh..i said na..my maths is damn...weak..so can u put more light on this solution
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| sree |
Aug 22 2006, 11:07 PM
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#5
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CORPORAL Group: Members Posts: 80 Joined: 20-August 06 From: Hyderabad Member No.: 77 |
QUOTE(ymhatre @ Aug 22 2006, 10:21 PM) Ohh..i said na..my maths is damn...weak..so can u put more light on this solution  ym an engg shldnt say tat he/she s wk in maths .... its all practice yar .... jst thinking ltl bit logically tats it .... lets improve all our skills by solvng such puzzles daily ... lets wt 4 sometime 4 others 2 give a try 4 this .... then i ll post the soln logic .... |
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| bakuryu |
Aug 23 2006, 05:39 PM
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#6
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SPECIALIST Group: Members Posts: 32 Joined: 14-June 06 From: India Member No.: 51 |
yes .. easy .. 3:8
for simplicity consider each side of the square to be 2a Drawing the figure we find 4 triangles. ADF, ECF, ABE, and AEF. Area of triangle = 0.5 * base * height Area of ADF = 0.5 * 2a * a = a^2 Area of ABE = 0.5 * 2a * a = a^2 Area of CEF = 0.5 * a * a = (a^2)/2 Area of Square ABCD = 4a^2 So, area of AEF = 4a^2 - (a^2 + a^2 + (a^2)/2) == (3a^2)/2 Ratio = [(3a^2)/2)]/(4a^2) = 3/8 or 3:8 |
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| sree |
Aug 23 2006, 06:11 PM
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#7
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CORPORAL Group: Members Posts: 80 Joined: 20-August 06 From: Hyderabad Member No.: 77 |
wow ym i tld u naa .... finally baku came out with wonderfull xplanation .... thnx 4 tat baku ....
lets c more such in future too ... |
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| christo |
Aug 23 2006, 06:55 PM
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#8
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PRIVATE FIRST CLASS Group: Members Posts: 18 Joined: 23-August 06 Member No.: 89 |
2:6.3352
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| ymhatre |
Aug 23 2006, 08:42 PM
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#9
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SERGEANT Group: Moderators Posts: 160 Joined: 10-June 06 Member No.: 36 |
Wow.....so a conclusion am a BAD Engineer......LOL....a bit more practice for me..sree..now im desparate to improve my maths...
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| sree |
Aug 23 2006, 08:47 PM
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#10
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CORPORAL Group: Members Posts: 80 Joined: 20-August 06 From: Hyderabad Member No.: 77 |
yea ym v all r here 2 improve our skills while njoyng .... talkng ... fighting ... correcting etc etc ... none r genius r good r bad ...
lets make this place as a cool chit-chat cum learning pt ... |
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