Pond Problem
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Source: Beat The GMAT — Problem Solving |
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kevincanspain
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Do you see that we are dealing with a right-isosceles triangle?
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Stuart@KaplanGMAT
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Since it's a Kaplan problem, let's apply the Kaplan method!
Step 1 of the Kaplan Method for PS: analyze the problem
We see that AB is the diameter of the circle. We see that each bridge is equal, creating an isosceles triangle with the diameter. We know that any triangle that has the diameter at one side and the opposite vertex on the side of the circle will be right (i.e. the angle opposite the diameter is 90 degrees).
So, we have a 45/45/90 triangle, with sides of x, x and x(root2).
Step 2 of the Kaplan Method for PS: state the task
Here, we need to find the ratio of the sum of the bridges to the distance from A to B along the outside of the pond, i.e. 1/2 the circumference of the circle (or 1/2(pi)d).
So, the answer will be:
(AC + BC)/(1/2)(pi)d
Step 3 of the Kaplan Method for PS: approach strategically
We could work with "x"s, but we prefer numbers; so, to make life simple, let's set the diameter of the triangle equal to root2, giving us bridge lengths of 1 and 1.
Accordingly, the distance via bridge is 1+1=2.
d=root2, so circumference = (pi)d = (pi)root2
Plugging into (AC + BC)/(1/2)(pi)d:
(1+1)/(1/2)(pi)(root2)
2 /(1/2)(pi)(root2)
2 * 2/(pi)(root2)
4 / (pi)(root2)
On the GMAT, it's very unlikely you'll see a root in the denominator of a fraction; accordingly, to match up to a choice, we need to "rationalize the denominator", which we accomplish by multiplying the top and bottom by the root in question:
4 / (pi)(root2) * root2/root2
4(root2)/2(pi)
2(root2)/pi
choose (A).
Step 1 of the Kaplan Method for PS: analyze the problem
We see that AB is the diameter of the circle. We see that each bridge is equal, creating an isosceles triangle with the diameter. We know that any triangle that has the diameter at one side and the opposite vertex on the side of the circle will be right (i.e. the angle opposite the diameter is 90 degrees).
So, we have a 45/45/90 triangle, with sides of x, x and x(root2).
Step 2 of the Kaplan Method for PS: state the task
Here, we need to find the ratio of the sum of the bridges to the distance from A to B along the outside of the pond, i.e. 1/2 the circumference of the circle (or 1/2(pi)d).
So, the answer will be:
(AC + BC)/(1/2)(pi)d
Step 3 of the Kaplan Method for PS: approach strategically
We could work with "x"s, but we prefer numbers; so, to make life simple, let's set the diameter of the triangle equal to root2, giving us bridge lengths of 1 and 1.
Accordingly, the distance via bridge is 1+1=2.
d=root2, so circumference = (pi)d = (pi)root2
Plugging into (AC + BC)/(1/2)(pi)d:
(1+1)/(1/2)(pi)(root2)
2 /(1/2)(pi)(root2)
2 * 2/(pi)(root2)
4 / (pi)(root2)
On the GMAT, it's very unlikely you'll see a root in the denominator of a fraction; accordingly, to match up to a choice, we need to "rationalize the denominator", which we accomplish by multiplying the top and bottom by the root in question:
4 / (pi)(root2) * root2/root2
4(root2)/2(pi)
2(root2)/pi
choose (A).
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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kstv
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<ACB will always be 90°
i.e double of the < AdB cos <AdB is a st line = 180°
AC²+BC² = d² given that AC = BC
2AC² = d² AC = d/√2
Semi circle ACB = Pi d/2
Ratio of 2AC / (Pi D/2)
(2*d/√2)/(Pi d/2)
2*2/(Pi √2)
(2√2)/Pi
i.e double of the < AdB cos <AdB is a st line = 180°
AC²+BC² = d² given that AC = BC
2AC² = d² AC = d/√2
Semi circle ACB = Pi d/2
Ratio of 2AC / (Pi D/2)
(2*d/√2)/(Pi d/2)
2*2/(Pi √2)
(2√2)/Pi
