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

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Wed Sep 26, 2018 1:30 pm

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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

Answer: [spoiler]__(E)_____ [/spoiler]
Difficulty Level: 650 - 700
Source: www.GMATH.net

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
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Source: Beat The GMAT — Problem Solving | Wed Sep 26, 2018 1:30 pm

deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

by deloitte247 » Thu Sep 27, 2018 10:37 am

The 3 parallel lines are certain to form similar and congruent triangle AEB and $$\triangle$$ BFC when they respectively pass through vertices A, B, and C of a square A B C D.
$$hey\ also\ have\ similar\ angles\ x^0,\ \left(90\ -x\right)^0\ and\ 90$$
Let the hypotenuse = a
CF = EB =5 <distance between r and s
$$Side\ of\ Square\ ABCD\ =\ a=\sqrt{7^2\ +\ 5^2}$$
$$\ a=\sqrt{49\ +\ 25}$$
$$\ a=\sqrt{74}$$

Area of square A B C D = side of square $$\ =\ a^2$$
$$\ =\ \left(\sqrt{74}\right)^2$$
$$\ =\ 74^{\frac{1}{2}\cdot\ \frac{2}{1}}$$
= 74
And 74 is divisible by 37.
Hence, option E is CORRECT.

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

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

by fskilnik@GMATH » Thu Sep 27, 2018 12:33 pm

fskilnik@GMATH 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: www.GMATH.net

Perfect, deloitte247. 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):

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.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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