Point A and B are at opposite ends of a circular pond with diameter d.

Question

https://gmatclub.com/forum/download/file.php?id=55639 Point 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. 1//π B. (√2)/π C. (2)/π D. (2√2)/π E. (3√2)/π download.png

rajat96 · Answer

The point "c" subtends a 90 degree angle, hence making triangle ABC and isosceles right angles triangle. So if AC = BC =x, therefore d(square) = x(square)+x(square) => x = d/root(2). Distance AC+BC = Root(2)*d. Also distance along circumference = pie*r = pie*d/2.
ratio is root(2)d/(pie*d/2) = 2 root(2)/pie .
Option "D".

nick1816 · Answer

Since AC = AB and AB is a diameter, ACB is an isosceles right angle triangle.

AC^2+BC^2 =d^2

2AC^2 = d^2

AC= \frac{d}{√2}

2AC = √2d

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

Ratio =  √2d/(\frac{\pi*d}{2})  =   \frac{(2√2)}{π}

NitishJain · Answer

IMO D.

AB = Diameter =d
Radius =  \frac{d}{2}

Angle C =90
Since AC = BC. => angle CAB = angle CBA = 45

If O being the center; In triangle AOC, AO = OC = radius & AC =  \sqrt{2}  radius (Using Pythagoras theorem)
BC = AC =  \sqrt{2}  radius =

Distance from A to B when traveling along the two bridges = AC + BC = 2 AC = 2 *  \sqrt{2}  radius = 2 *  \sqrt{2}  *  \frac{d}{2} 
 =  \sqrt{2}  *d

Distance when traveling along the edge of the pond = Semi perimeter of the circle =  \frac{πd}{2}

Required Ratio = ( \sqrt{2}  *d) / ( \frac{πd}{2} )
 = (2√2)/π

MARVEL13 · Answer

One way is to simply let the diameter be  \sqrt{2} 
As we know that the two bridges are of equal lengths and

any triangle with diameter as one side and all vertices on circle will be Right angle triangle with diameter being the Hypotenuse.

The two sides come out to be 1 & 1

and radius be  \sqrt{2}/2  --->  1/\sqrt{2}

Therefore ratio is (1+1)/ Pi*(radius) ---> 2/Pi*(1/ \sqrt{2})  -----> 2 \sqrt{2} /Pi

Which is Option D

Answer is Option D

vipulshahi · Answer

Given : d = diameter

Required: AC+BC / edge of the pond = ?
Solution: AB = d
AC = BC thus Tri. ABC is right angle triangle with right angle at C.
Angle ratio = 45:45:90
Side ratio = x : x : sqrt2 x
sqrt(2) x = d
x = d / sqrt(2)
AC+BC = d/sqrt(2) +d/sqrt2 = sqrt(2)*d ----equation 1

distance travelling edge of the pond= 3.14* d / 2
 
Thus, AC+BC / edge of the pond = sqrt(2) * d *2 / 3.14*d
= 2*sqrt(2)/3.14

Answer : D

rajatchopra1994 · Answer

Explanation:
d = AB , r = d/2
AC = BC, C when C points connects with AB on center, it will be radius in length.
By Pythagoras, we can find the length of one bridge: i.e. d/√2., for 2 bridges: 2xd/√2 = d√2.

Distance traveling along the edge(semicircle)= πr = π(d/2)

Ratio = (d√2) /(πd/2)
= (2√2)/π

IMO-D

bM22 · Answer

The distance when travelling along the pond = πd/2.
The distance when travelling along the bridges = AC + CB.
Now in Triangle ACB,
Let AC = BC = x
=> d^2= 2x^2
=> x = d/\sqrt{2}.

AC + BC = \sqrt{2}d.

Ratio = \sqrt{2}d / πd/2.

Answer D.

Point A and B are at opposite ends of a circular pond with diameter d.

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Point A and B are at opposite ends of a circular pond with diameter d.

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