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805+ Level|   Geometry|      
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mikemcgarry
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1) In the figure above, triangle ABC is inscribed in the circle with c 15 Sep 2016, 15:03
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

43% (03:17) correct 57% (02:54) wrong based on 92 sessions

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Not Attempted Yet
This is the first of two problems inspired by this flawed problem:
in-the-figure-attached-triangle-abc-is-inscribed-215771.html#p1735468
The second is here:
2-in-the-figure-above-triangle-abc-is-inscribed-in-the-circle-with-c-225609.html
Attachment:
inscribed triangle.JPG
inscribed triangle.JPG [ 18.92 KiB | Viewed 6080 times ]

In the figure above, triangle ABC is inscribed in the circle with center O, such that CD is perpendicular to AB. If the length of side AC is 5 and the radius equals \(r = \frac{25}{8}\), find the perimeter of the triangle.

(A) 14
(B) 15
(C) 16
(D) 20
(E) none of the above

MorningRunner, this might be of interest to you.
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C
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0ld
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1) In the figure above, triangle ABC is inscribed in the circle with c 18 Sep 2016, 07:28
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Lets take AD = x , OD = y , radius of circle = r = \(\frac{25}{8}\)

ADO, makes a right angle triangle. AO = radius of circle = r.

Using Pythagoras theorem,\(OA^2 = OD^2 + AD^2\)==> \(r^2= X^2 + y^2\) --Equation 1

In Triangle ADC, CD = OC + OD = r + y. ( OC -- radius of triangle)

\(AC^2 = CD^2 + AD^2\)
==> \(5^2 = (r+y)^2 + x^2\)
==> 25 = \(r^2 + 2ry + (y^2 + x^ 2 )\)
==> 25 = \(r^2 + r^2\)+2ry ( substituting values from Eq1)
==> 25 = 2 (\(r^2\)) + 2 ry
==> 25 = 2 (\(\frac{25}{8}\) ) ^ 2 + 2 * (\(\frac{25}{8}\) ) * y
==> 1 = \(\frac{50}{64}\) + \(\frac{1}{4}\) * y
==> 64 = 50 + 16y
==> y = \(\frac{7}{8}\)


Putting value of y in equation 1.

\(x^2\) = \(\frac{625}{64}\) - \(\frac{49}{64}\) = \(\frac{576}{64}\) = 9

==> x = 3

In triangle BDC, BD = x and CD = r + y = \(\frac{25}{8}\) + \(\frac{7}{8}\) = 4

\(BC^2 = CD^2 + BC^2\)= \(4^2 + 3^2\) = 16+9 = 25

BC = 5

Perimeter = AB + BC + CA = 6 + 5 + 5 = 16 (Ans. C)

Edit : Adding formatting for equations.
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1) In the figure above, triangle ABC is inscribed in the circle with c 02 Mar 2017, 20:10
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mikemcgarry
This is the first of two problems inspired by this flawed problem:
https://gmatclub.com/forum/in-the-figure ... l#p1735468
The second is here:
https://gmatclub.com/forum/2-in-the-figu ... 25609.html
Attachment:
inscribed triangle.JPG

In the figure above, triangle ABC is inscribed in the circle with center O, such that CD is perpendicular to AB. If the length of side AC is 5 and the radius equals \(r = \frac{25}{8}\), find the perimeter of the triangle.

(A) 14
(B) 15
(C) 16
(D) 20
(E) none of the above

MorningRunner, this might be of interest to you.

Hi mikemcgarry,

I took somewhere around 5 minutes to work out the solution! Had to use Pythagoras thrice and a bit of substitution. Is there a shorter, more elegant way to solve this? Would such questions appear on the actual test? :shock:

Thanks in advance :)
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AbhiJung
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1) In the figure above, triangle ABC is inscribed in the circle with c 03 Mar 2017, 00:49
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I solved this problem rather quickly using the rule of Pythagoras i.e 3:4:5.
But since the diagram is not drawn to scale, is this method flawed?
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mikemcgarry
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1) In the figure above, triangle ABC is inscribed in the circle with c 03 Mar 2017, 15:58
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Shruti0805
Hi mikemcgarry,

I took somewhere around 5 minutes to work out the solution! Had to use Pythagoras thrice and a bit of substitution. Is there a shorter, more elegant way to solve this? Would such questions appear on the actual test? :shock:

Thanks in advance :)
AbhiJung
I solved this problem rather quickly using the rule of Pythagoras i.e 3:4:5.
But since the diagram is not drawn to scale, is this method flawed?
Dear Shruti0805 & AbhiJung,

I'm happy to respond. :-)

I came up with this problem on the fly, without much planning. I was inspired by another flawed problem here on GMAT Club. AbhiJung, essentially you guessed in the dark, and the flaw of this problem was that an uninformed guess could lead to the right answer. In general, if all you know is that a right triangle has a hypotenuse of length 5, the GMAT will punish you for blindly assuming that the triangle has to be a 3-4-5 triangle. Insofar as this problem did not punish this poor approach, this is not really a good GMAT question.

Nevertheless, Shruti0805, for the harder questions on the Quant section, the GMAT absolutely could give you a question that involves a holiday festival of algebra. Many harder Quant questions have an elegant solution, but one of the many things the Quant section assesses, at the upper level, is having the stomach to handled an ocean of algebra--and having the skill to wade through it efficiently and purposefully. Any question that separates those with what it takes to solve a problem from those who don't is a useful question, in terms of formal statistically question design. If what this question takes is doing a lot of algebra, then the right question is not about how to avoid that, but instead how to do it all efficiently and strategically. The respond of Old in this thread was a particularly skillful use of algebra. If you had to use the Pythagorean Theorem three times, this suggests that you did not pursue the most strategic route to the solution.

Does all this make sense?
Mike :-)
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1) In the figure above, triangle ABC is inscribed in the circle with c 26 May 2017, 08:55
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0ld
Lets take AD = x , OD = y , radius of circle = r = \(\frac{25}{8}\)

ADO, makes a right angle triangle. AO = radius of circle = r.

Using Pythagoras theorem,\(OA^2 = OD^2 + AD^2\)==> \(r^2= X^2 + y^2\) --Equation 1

In Triangle ADC, CD = OC + OD = r + y. ( OC -- radius of triangle)

\(AC^2 = CD^2 + AD^2\)
==> \(5^2 = (r+y)^2 + x^2\)
==> 25 = \(r^2 + 2ry + (y^2 + x^ 2 )\)
==> 25 = \(r^2 + r^2\)+2ry ( substituting values from Eq1)
==> 25 = 2 (\(r^2\)) + 2 ry
==> 25 = 2 (\(\frac{25}{8}\) ) ^ 2 + 2 * (\(\frac{25}{8}\) ) * y
==> 1 = \(\frac{50}{64}\) + \(\frac{1}{4}\) * y
==> 64 = 50 + 16y
==> y = \(\frac{7}{8}\)


Putting value of y in equation 1.

\(x^2\) = \(\frac{625}{64}\) - \(\frac{49}{64}\) = \(\frac{576}{64}\) = 9

==> x = 3

In triangle BDC, BD = x and CD = r + y = \(\frac{25}{8}\) + \(\frac{7}{8}\) = 4

\(BC^2 = CD^2 + BC^2\)= \(4^2 + 3^2\) = 16+9 = 25

BC = 5

Perimeter = AB + BC + CA = 6 + 5 + 5 = 16 (Ans. C)

Edit : Adding formatting for equations.

Great approach! However, I assume a couple of shortcuts:
1) Because CD is perpendicular and the center of the circle is on it - the triangle should be isosceles (BC=AC=5)
2) Instead of working with 25/8 you can work with 3 (25/8=3,125). The same with 7/8=1. Just simplifies the math.

Anyway - great approach
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