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Three parallel lines r, s and t pass, respectively, through three cons

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Three parallel lines r, s and t pass, respectively, through three cons  [#permalink]

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New post 26 Sep 2018, 14:37
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Three parallel lines r, s and t pass, respectively, through three consecutive vertices A, B and C of square ABCD. If the distance between r and s is 5 and the distance between s and t is 7, the numerical value of the area of ABCD is:

(A) prime
(B) divisible by 7
(C) divisible by 11
(D) divisible by 23
(E) divisible by 37

Source: http://www.GMATH.net

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Re: Three parallel lines r, s and t pass, respectively, through three cons  [#permalink]

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New post 26 Sep 2018, 19:51
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fskilnik wrote:
Three parallel lines r, s and t pass, respectively, through three consecutive vertices A, B and C of square ABCD. If the distance between r and s is 5 and the distance between s and t is 7, the numerical value of the area of ABCD is:

(A) prime
(B) divisible by 7
(C) divisible by 11
(D) divisible by 23
(E) divisible by 37

Source: http://www.GMATH.net



Draw the figure as attached...

Now since the three lines are parallel, ∆AEB and ∆BFC are similar and congruent triangles as they have similar angles X,90-x and 90 . Also the hypotenuse is same - 'a'
So CF =EB=5.. therefore side of square =a=√(7^2+5^2)=√74

Area =a^2=(√74)^2=74
So area is divisible by 37

E
Attachments

Screenshot_2018-09-27-08-17-20-898_com.viettran.INKredible.png
Screenshot_2018-09-27-08-17-20-898_com.viettran.INKredible.png [ 102.31 KiB | Viewed 547 times ]


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Three parallel lines r, s and t pass, respectively, through three cons  [#permalink]

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New post 27 Sep 2018, 13:27
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fskilnik wrote:
Three parallel lines r, s and t pass, respectively, through three consecutive vertices A, B and C of square ABCD. If the distance between r and s is 5 and the distance between s and t is 7, the numerical value of the area of ABCD is:

(A) prime
(B) divisible by 7
(C) divisible by 11
(D) divisible by 23
(E) divisible by 37

Source: http://www.GMATH.net

Perfect, chetan2u. Thanks for the contribution!

There are two possible configurations. As we will see, any one of them will go to the same answer! (Otherwise the question would be ill posed.)

\(?\,\,\,:\,\,\,{L^{\,2}}\,\,{\text{divisib}}{\text{.}}\,\,{\text{property}}\)

From the images below, please note that (in both cases):

Image

1. Right triangles AEB and BFC are similar. (All corresponding internal angles are equal.)
2. The ratio of similarity is 1:1 (AB and BC are homologous sides and AB=BC)

Conclusion: Right triangles AEB and BFC are congruent ("equal"), hence (in both cases):

3. Each triangle has hypotenuse (length) L, and legs (with lengths) 5 and 7.

Finally:

\(?\,\,\,:\,\,\,\,{L^{\,2}} = {5^{\,2}} + {7^{\,2}}\, = 74\,\,\left( { = 2 \cdot 37} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( E \right)\,\,\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Three parallel lines r, s and t pass, respectively, through three cons   [#permalink] 27 Sep 2018, 13:27
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