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
Three parallel lines r, s and t pass, respectively, through
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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.
$$hey\ also\ have\ similar\ angles\ x^0,\ \left(90\ -x\right)^0\ and\ 90$$
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
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Perfect, deloitte247. Thanks for the contribution!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
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)
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