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

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.

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.