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Section - 34 Problem - 15

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Section - 34 Problem - 15

by camitava » Sun Nov 04, 2007 11:47 pm
Guys need help -
15. The shaded region in the figure above represents a rectangular frame with length 18 inches and width 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture, in inches?
A. 9*sqrt(2)
B. 3/2
C. 9/sqrt(2)
D. 15(1 - 1/sqrt(2))
E. 9/2

Will tell u OA after some discussion!
Image
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by samirpandeyit62 » Mon Nov 05, 2007 2:23 am
let l = lenght of picture
w = width of picture

so area =lw = area of frame

now 2lw = 18*15
so lw = 9 *15 so w =9*15/l

also given that l/w = 18/15

so l^2/9*15 = 18/15

or l^2 = 9 *18

so l = 9 *sqrt(2)

A
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by camitava » Mon Nov 05, 2007 2:50 am
Samir, thanks for the sol! But I am still having some doubts! I think the Qs is missing some more explanation or so. Because, we generally take that area of the frame means the area bounded by the checks and how u took are of the photo + area of the frame = 18 * 15? In the Qs, it is saying that area of the frame is 18*15. Am I missing something? If possible, pls explain me in detail ...
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by samirpandeyit62 » Mon Nov 05, 2007 3:02 am
The frame is the outer part which is shaded (between picture & the outer rectangle), the picture is the inner rectangle & 18*15 is the combined area of both of them.

also logically if area of the frame is 18*15 then how can area of frame = area of picture.
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by camitava » Mon Nov 05, 2007 3:14 am
Ohhhhhhhhhhhhhhhhhhhhhhhh! Now got it. Sorry my mistake to understand the Qs.
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by camitava » Mon Nov 05, 2007 3:16 am
Ohhhh! Sorry I forgot to mention the OA. It is A, obviously.
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