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Re: If the perimeter of a right triangle is 20 units, what is the approxim [#permalink]

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06 Sep 2014, 12:43

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The best strategy to solve this problem is to just dive in.

Largest circle inscribed within a right triangle means that this is a isosceles right triangle. For an isosceles right triangle there are two equal sides a. Using the Pythagoream theorem you can determine that the hypotenuse side:

\(a^2\) + \(a^2\) = \(c^2\) Thus, c = \(\sqrt{2}a\)

With that in mind, the perimeter of the triangle is 20 units.

\(a + a + \sqrt{2}a\) = 20 a ~ 6

Now, look at the diagram shown below. For a right triangle with sides a, b, and hypotenuse c, the area is \(A = \frac{1}{2}ab\)

The inradius can be found by equating the area of the triangle ABC with sum of the areas of the triangles ACI, BCI and ABI.

The altitudes for these triangles serve as the inradii for the circle. Let's call this radius (shown in red), r. This gives us

In this case we already established above that \(a = b\) and \(c = \sqrt{2}a\) and that \(a = 6\) Thus, using these values we establish that the radius of the circle, \(r = 1.8\)

Plugging this into the formula for area of a circle, \(A = pi (1.8)^2\) This gives us approx. \(4pi\)

If the perimeter of a right triangle is 20 units, what is the approximate area of the largest circle that can be inscribed in this triangle?

(A) pi (B) 6.3 pi (C) 4 pi (D) pi/2 (E) 3 pi/2

Solving this question using basic geometry:

First think, what kind of right triangle will have the largest circle inscribed in it? How will you split the 20 units of the perimeter among the three sides?

You can vary the length of the two legs and the hypotenuse will be already defined. If you keep the length of the two legs very different from each other, the circle you can inscribe in the triangle will be very small. As you make the lengths equal, the circle keeps getting bigger. So this should help you realize that you are looking for an isosceles right triangle.

Attachment:

Ques3.jpg [ 17.88 KiB | Viewed 2041 times ]

\(S + S + \sqrt{2}*S = 20\) \(S = 20/(2+\sqrt{2})\) \(S = 10(2 - \sqrt{2}) = 6\) approximately (actually it is smaller than 6 but taking \(\sqrt{2} = 1.4\) helps approximate. Ideally, they should have mentioned this value of root 2 in the question since the calculations are tedious otherwise)

Attachment:

Ques4.jpg [ 15.37 KiB | Viewed 2034 times ]

Now notice that we can calculate area of the triangle in two ways: (1/2)*leg1*leg2 = (1/2)*Altitude*Hypotenuse

\(6 * 6 = Altitude * \sqrt{2} * 6\)

\(Altitude = 3\sqrt{2} = r + \sqrt{2}r\)

\(r = 3*\sqrt{2}*(\sqrt{2} - 1)\)

\(r = 3*1.4*.4 = 1.7\)

Area of circle \(= \pi*r^2 = \pi*1.7^2 = 3.89\pi\) approximately
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Re: If the perimeter of a right triangle is 20 units, what is the approxim [#permalink]

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12 Apr 2016, 16:19

VeritasPrepKarishma wrote:

shreyas wrote:

If the perimeter of a right triangle is 20 units, what is the approximate area of the largest circle that can be inscribed in this triangle?

(A) pi (B) 6.3 pi (C) 4 pi (D) pi/2 (E) 3 pi/2

Solving this question using basic geometry:

First think, what kind of right triangle will have the largest circle inscribed in it? How will you split the 20 units of the perimeter among the three sides?

You can vary the length of the two legs and the hypotenuse will be already defined. If you keep the length of the two legs very different from each other, the circle you can inscribe in the triangle will be very small. As you make the lengths equal, the circle keeps getting bigger. So this should help you realize that you are looking for an isosceles right triangle.

Attachment:

Ques3.jpg

\(S + S + \sqrt{2}*S = 20\) \(S = 20/(2+\sqrt{2})\) \(S = 10(2 - \sqrt{2}) = 6\) approximately (actually it is smaller than 6 but taking \(\sqrt{2} = 1.4\) helps approximate. Ideally, they should have mentioned this value of root 2 in the question since the calculations are tedious otherwise)

Attachment:

Ques4.jpg

Now notice that we can calculate area of the triangle in two ways: (1/2)*leg1*leg2 = (1/2)*Altitude*Hypotenuse

\(6 * 6 = Altitude * \sqrt{2} * 6\)

\(Altitude = 3\sqrt{2} = r + \sqrt{2}r\)

\(r = 3*\sqrt{2}*(\sqrt{2} - 1)\)

\(r = 3*1.4*.4 = 1.7\)

Area of circle \(= \pi*r^2 = \pi*1.7^2 = 3.89\pi\) approximately

Hi VeritasPrepKarishma This is quite tough problem. Do you think is possible on test day to face such problem. I think even in 3 minutes is quite hard to get to the solution .

Anyway I would like to point something that I found in your solution and of course correct me if I'm wrong.

So you wrote \(1.7^2 = 3.89\) but \(1.7^2 = 2.89\) now can we round up 2.89 to 4, other than that we can see that the answers are quite far.

If the perimeter of a right triangle is 20 units, what is the approximate area of the largest circle that can be inscribed in this triangle?

(A) pi (B) 6.3 pi (C) 4 pi (D) pi/2 (E) 3 pi/2

Solving this question using basic geometry:

First think, what kind of right triangle will have the largest circle inscribed in it? How will you split the 20 units of the perimeter among the three sides?

You can vary the length of the two legs and the hypotenuse will be already defined. If you keep the length of the two legs very different from each other, the circle you can inscribe in the triangle will be very small. As you make the lengths equal, the circle keeps getting bigger. So this should help you realize that you are looking for an isosceles right triangle.

Attachment:

Ques3.jpg

\(S + S + \sqrt{2}*S = 20\) \(S = 20/(2+\sqrt{2})\) \(S = 10(2 - \sqrt{2}) = 6\) approximately (actually it is smaller than 6 but taking \(\sqrt{2} = 1.4\) helps approximate. Ideally, they should have mentioned this value of root 2 in the question since the calculations are tedious otherwise)

Attachment:

Ques4.jpg

Now notice that we can calculate area of the triangle in two ways: (1/2)*leg1*leg2 = (1/2)*Altitude*Hypotenuse

\(6 * 6 = Altitude * \sqrt{2} * 6\)

\(Altitude = 3\sqrt{2} = r + \sqrt{2}r\)

\(r = 3*\sqrt{2}*(\sqrt{2} - 1)\)

\(r = 3*1.4*.4 = 1.7\)

Area of circle \(= \pi*r^2 = \pi*1.7^2 = 3.89\pi\) approximately

Hi VeritasPrepKarishma This is quite tough problem. Do you think is possible on test day to face such problem. I think even in 3 minutes is quite hard to get to the solution .

Anyway I would like to point something that I found in your solution and of course correct me if I'm wrong.

So you wrote \(1.7^2 = 3.89\) but \(1.7^2 = 2.89\) now can we round up 2.89 to 4, other than that we can see that the answers are quite far.

Thanks a lot

You are right. A much closer value is about \(2.94 * \pi\) and that would mean about \(3*\pi\). \(4*\pi\) is too much of a stretch but there are no options closer to this value.

The likelihood of such a question in actual GMAT is not very high though it is based on very basic concepts. Also, this formula, Leg1 * Leg2 = Altitude*Hypotenuse comes in very handy. It is something we discuss in our book and it makes complicated questions simple.

The "isosceles right triangle" part is logical deduction, getting the altitude is mechanical but splitting the altitude into r and sqrt(2)r is again logic.

Overall, a high level question. Such a question is more likely in DS format.
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Re: If the perimeter of a right triangle is 20 units, what is the approxim [#permalink]

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02 Oct 2017, 11:05

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