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Section - 7 Problem - 24

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Section - 7 Problem - 24

by camitava » Mon Oct 29, 2007 3:57 am
Guys help me on this Qs -

24. The figure above shows the dimensions of a square picture frame that was constructed using four pieces of frame as shown. If w is the width of each piece of the frame, what is the area of each piece?
(1) w = 3 inches
(2) PQ = inches

[spoiler]IMO : C but OA is D[/spoiler]
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by samirpandeyit62 » Mon Oct 29, 2007 5:21 am
each piece is a trapezium whose area is 1/2 * sum of Perallel sides *height

stmt 1: w =3 = height of trapezium & as well as the side of the inner square as 3 feet - 2*3 inches SUFF

stmt2: PQ = ? inches

if we know PQ we can use pythgoras theorem to find w & then find area using above method

SUFF

D
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by camitava » Mon Oct 29, 2007 8:34 pm
Sorry Samir, I somehow missed the value of PQ. It is sqrt(18 ) inch. Now coming back to ur approach -
Refer the figure below -
By A, we can get the value of w i.e RQ. But what will be the QT? Are u taking it will be QT = 3ft - 2w?
Now by B, we know the value of PQ but not PR. So how can we use pythgoras theorem? Again are you taking PR = w?
If so, then I now understand D is suff but otherwise it should be C! What do u say Samir?
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by samirpandeyit62 » Mon Oct 29, 2007 9:34 pm
Q says all pieces have same width so we can take QT = 3ft - 2w

& in B if we have sqrt(18) =3sqrt(2) we have w =2

coz w is same for all pieces hence if we join RQ then triangle PRQ will be iscoceles right angled traingle i.e sides are in ratio 1:1:rt(2) so if PQ = 3rt(2) then BQ =w = 3

Hope this clarifies
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by camitava » Mon Oct 29, 2007 10:00 pm
Ohhhhhhhhh! Got it Samir... Sorry its my mistake and thanks a lot man for pointing out my fault... Thanks once again.
Correct me If I am wrong


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