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Can somebody please explain a step-by-step solution please.

Thanks!!!

The diagonal of the rectangle is diameter in the circle and therefore its length is 4. The right triangle formed by a diagonal and two adjacent sides of the rectangle is a 30-60-90 rectangle (because the short leg is half of the hypotenuse). The other side of the rectangle is therefore \(2\sqrt{3}.\)

The shaded area can be obtained by subtracting from the circular sector defined by two radii (half diagonals) with an angle of \(120^o\) between them the area of the triangle formed by the same two radii and the long side of the rectangle. The are of the circular sector is \(\pi\cdot2^2\cdot\frac{120}{360}=\frac{4\pi}{3}\) and the area of the triangle is 1/4 of the area of the rectangle, which is \(2\cdot2\sqrt{3}=4\sqrt{3}.\) In any parallelogram, so also in a rectangle, the two diagonals divide the paralellogram into four triangles with equal areas. Two radii with the short side of the rectangle form an equilateral triangle, while two radii with the long side of the rectangle form an isosceles triangle with angles 30-30-120.

The requested area is \(\frac{4\pi}{3}-\sqrt{3}.\) _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Apply Pythagoras starting with the triangle with sides 1 and 1. The hypotenuse is \(\sqrt{2}.\) In the next right triangle, the hypotenuse is \(\sqrt{3}\) ... the longest hypotenuse is \(\sqrt{12}=\sqrt{2^2\cdot3}=2\sqrt{3}.\)

EvaJager wrote:

NYC5648 wrote:

In the figure to the right what is the length of the longest side, if all of the triangles are right triangles?

Apply Pythagoras starting with the triangle with sides 1 and 1. The hypotenuse is \(\sqrt{2}.\) In the next right triangle, the hypotenuse is \(\sqrt{3}\) ... the longest hypotenuse is \(\sqrt{12}=\sqrt{2^2\cdot3}=2\sqrt{3}.\)

Thanks for the answer. I got the idea behind it! Thanks. But still I dont get that the longest hypotenuse is 2 srt 3. When I apply Pythagoras theorem I get 4 as an answer. Could you please explain. Thanks!

Can somebody please explain a step-by-step solution please.

Thanks!!!

The diagonal of the rectangle is diameter in the circle and therefore its length is 4. The right triangle formed by a diagonal and two adjacent sides of the rectangle is a 30-60-90 rectangle (because the short leg is half of the hypotenuse). The other side of the rectangle is therefore \(2\sqrt{3}.\)

The shaded area can be obtained by subtracting from the circular sector defined by two radii (half diagonals) with an angle of \(120^o\) between them the area of the triangle formed by the same two radii and the long side of the rectangle. The are of the circular sector is \(\pi\cdot2^2\cdot\frac{120}{360}=\frac{4\pi}{3}\) and the area of the triangle is 1/4 of the area of the rectangle, which is \(2\cdot2\sqrt{3}=4\sqrt{3}.\) In any parallelogram, so also in a rectangle, the two diagonals divide the paralellogram into four triangles with equal areas. Two radii with the short side of the rectangle form an equilateral triangle, while two radii with the long side of the rectangle form an isosceles triangle with angles 30-30-120.

The requested area is \(\frac{4\pi}{3}-\sqrt{3}.\)

Hi EVA, thanks a lot for you detailed explanation. But I still dont get it. Where I have problems to follow is the following:

The shaded area can be obtained by subtracting from the circular sector defined by two radii (half diagonals) with an angle of \(120^o\) between them the area of the triangle formed by the same two radii and the long side of the rectangle.

Two radii with the short side of the rectangle form an equilateral triangle, while two radii with the long side of the rectangle form an isosceles triangle with angles 30-30-120.

The two circles are inscribed in the rectangle as shown [#permalink]

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29 Aug 2012, 13:06

The two circles are inscribed in the rectangle as shown above. If line segment AB, length 0.4, is perpendicular to line segment CD and line segment CD is tangent to the circle, what is the approximate area of the shaded region of the rectangle?

Apply Pythagoras starting with the triangle with sides 1 and 1. The hypotenuse is \(\sqrt{2}.\) In the next right triangle, the hypotenuse is \(\sqrt{3}\) ... the longest hypotenuse is \(\sqrt{12}=\sqrt{2^2\cdot3}=2\sqrt{3}.\)

EvaJager wrote:

NYC5648 wrote:

In the figure to the right what is the length of the longest side, if all of the triangles are right triangles?

Apply Pythagoras starting with the triangle with sides 1 and 1. The hypotenuse is \(\sqrt{2}.\) In the next right triangle, the hypotenuse is \(\sqrt{3}\) ... the longest hypotenuse is \(\sqrt{12}=\sqrt{2^2\cdot3}=2\sqrt{3}.\)

Thanks for the answer. I got the idea behind it! Thanks. But still I dont get that the longest hypotenuse is 2 srt 3. When I apply Pythagoras theorem I get 4 as an answer. Could you please explain. Thanks!

The first hypotenuse squared is \(1^2+1^2=2\). This hypotenuse is one of the legs in the next right triangle. We are going to apply Pythagoras's again, so no need to take the square root of it. The second hypotenuse squared is \(2+1^2=3\). The third hypotenuse squared is \(3+1^2=4\). ...

Finally, the longest hypotenuse is \(\sqrt{12}=\sqrt{2^2\cdot3}=2\sqrt{3}.\)

Make a drawing and mark where you have the right angles. It will help you when applying Pythagoras's. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Can somebody please explain a step-by-step solution please.

Thanks!!!

The diagonal of the rectangle is diameter in the circle and therefore its length is 4. The right triangle formed by a diagonal and two adjacent sides of the rectangle is a 30-60-90 rectangle (because the short leg is half of the hypotenuse). The other side of the rectangle is therefore \(2\sqrt{3}.\)

The shaded area can be obtained by subtracting from the circular sector defined by two radii (half diagonals) with an angle of \(120^o\) between them the area of the triangle formed by the same two radii and the long side of the rectangle. The are of the circular sector is \(\pi\cdot2^2\cdot\frac{120}{360}=\frac{4\pi}{3}\) and the area of the triangle is 1/4 of the area of the rectangle, which is \(2\cdot2\sqrt{3}=4\sqrt{3}.\) In any parallelogram, so also in a rectangle, the two diagonals divide the paralellogram into four triangles with equal areas. Two radii with the short side of the rectangle form an equilateral triangle, while two radii with the long side of the rectangle form an isosceles triangle with angles 30-30-120.

The requested area is \(\frac{4\pi}{3}-\sqrt{3}.\)

Hi EVA, thanks a lot for you detailed explanation. But I still dont get it. Where I have problems to follow is the following:

The shaded area can be obtained by subtracting from the circular sector defined by two radii (half diagonals) with an angle of \(120^o\) between them the area of the triangle formed by the same two radii and the long side of the rectangle.

Two radii with the short side of the rectangle form an equilateral triangle, while two radii with the long side of the rectangle form an isosceles triangle with angles 30-30-120.

Maybe I am blind but I dont see it.

Thanks!!!

Copy the drawing from the question and add the two diagonals. Mark the angles and don't forget that the two diagonals are equal and bisect each other. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: A train passes a man standing on a platform in 9 sec. and pa [#permalink]

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23 Oct 2013, 01:14

"Two men plan to run in opposite directions around a park 1 km long. The first man can run 15 meters in a minute and the second can run 25 meters per minute. How much further will the second runner have run at the point they cross paths? "

What does "How much further will the second runner have run at the point they cross paths?" mean? Is "point they cross paths" mean the meeting place? Should the 1000 m be considered a circle or a straight line? "further will the second runner.." - could not understand this term too.

Can you please help.

Thanks

gmatclubot

Re: A train passes a man standing on a platform in 9 sec. and pa
[#permalink]
23 Oct 2013, 01:14

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