Points A and B are at opposite ends of a circular pond with

Question

130914 Kaplan Geometry.jpg Points A and B are at opposite ends of a circular pond with diameter d. A bridge connects point A with point C, and another bridge connects point C with point B. The two bridges are of equal length. What is the ratio of the distance from A to B when traveling along the two bridges to the distance when traveling along the edge of the pond? A) \frac{(2\sqrt{2})}{\pi} B) \frac{(d\sqrt{2})}{\pi} C) \frac{2}{\pi} D) \frac{\sqrt{2}}{2\pi} E) \frac{2\sqrt{2}}{d\pi}

Bunuel · Answer

https://gmatclub.com/forum/download/file.php?id=21052 
Points A and B are at opposite ends of a circular pond with diameter d. A bridge connects point A with point C, and another bridge connects point C with point B. The two bridges are of equal length. What is the ratio of the distance from A to B when traveling along the two bridges to the distance when traveling along the edge of the pond?

A)  \frac{(2\sqrt{2})}{\pi}

B)  \frac{(d\sqrt{2})}{\pi}

C)  \frac{2}{\pi}

D)  \frac{\sqrt{2}}{2\pi}

E)  \frac{2\sqrt{2}}{d\pi}

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.

According to the above angle C must 90 degrees. Next, since AC=BC, then triangle ABC is 45-45-90 triangle, so, its sides are in the ratio  1:1:\sqrt{2} . Thus,  AC=BC=\frac{d}{\sqrt{2}} .

The distance from A to B when traveling along the two bridges =  AC+BC=\frac{d}{\sqrt{2}}+\frac{d}{\sqrt{2}}=\frac{2d}{\sqrt{2}} .

The distance when traveling along the edge of the pond = half of the circumference =  \frac{\pi{d}}{2} .

The ratio =  \frac{(\frac{2d}{\sqrt{2}})}{(\frac{\pi{d}}{2})}=\frac{2*2}{\sqrt{2}\pi}=\frac{2*2*\sqrt{2}}{\sqrt{2}*\sqrt{2}\pi}=\frac{2\sqrt{2}}{\pi} .

Answer: A.

mvictor · Answer

I assigned values for R, so that it would be easier to work with..
R=1, then d=2.

now, we can draw another line from O to C, and get another line, of the length 1. thus, AC must be, according to the 45-45-90 right angle properties, sqrt(2). The CB is as well sqrt(2).
to travel from A to B across the bridges, we need to walk a distance of 2 * sqrt(2)

now, the edge:
C=2*pi*R
C=2*pi.
since AB is a semicircle, the length of the arc AB thus must be pi.

since we are asked for the ratio, we then have:
2 sqrt(2) / pi.

A

KrishnakumarKA1 · Answer

let the radius be r

since the distance of two bridges are equal
triangle BAC is isosceles right triangle
therefore the length of each bridge will be √2r and the total length via bridge = 2r√2

length of the semi perimeter = pi*r

hence the ratio = 2r√2/ pi*r = 2√2/pi

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