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In the preceding figure, what is the perimeter of ∆ABC inscribed withi

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In the preceding figure, what is the perimeter of ∆ABC inscribed withi  [#permalink]

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New post 17 Jan 2016, 10:53
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A
B
C
D
E

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In the preceding figure, what is the perimeter of ∆ABC inscribed within the semicircle with center O?

A. 9
B. \(6 + 3 \sqrt{2}\)
C. \(9 + 3 \sqrt{3}\)
D. \(6 + 6 \sqrt{2}\)
E. \(12 \sqrt{2}\)

Attachment:
2016-01-17_2250.png
2016-01-17_2250.png [ 5.35 KiB | Viewed 1393 times ]

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Re: In the preceding figure, what is the perimeter of ∆ABC inscribed withi  [#permalink]

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New post 19 Jan 2016, 10:04
Because the triangle is inscribed in a semicircle, we can see that the radius of the circle is 3 and the diameter is 6. Points A and C will have coordinates (0,3) and (0,-3) respectively. Therefore the long side of the triangle is 6.

The two short sides of the triangle each form a smaller right isosceles triangle. \(\triangle\)AOB and \(\triangle\)COB. The ratio of lengths of a right isosceles triangle are \(1:1:\sqrt{2}\). So the lengths of our triangles are \(3:3:3\sqrt{2}\)

So the total perimeter of the large triangle is \(6+ 2*3\sqrt{2} = 6+6\sqrt {2}\)

Answer D

Alternatively, we could consider the large triangle, which is also an isosceles right triangle with short sides = x and hypotenuse = 6. Then with the same \(1:1:\sqrt{2}\) ratio, our triangle has side lengths x:x:6. \(x\sqrt{2} = 6\)

\(x=\frac{6}{\sqrt{2}} = 3\sqrt{2}\)

Again the perimeter is \(6+ 2*3\sqrt{2} = 6+6\sqrt {2}\)
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In the preceding figure, what is the perimeter of ∆ABC inscribed withi  [#permalink]

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New post 19 Jan 2016, 10:27
Bunuel wrote:
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In the preceding figure, what is the perimeter of ∆ABC inscribed within the semicircle with center O?

A. 9
B. \(6 + 3 \sqrt{2}\)
C. \(9 + 3 \sqrt{3}\)
D. \(6 + 6 \sqrt{2}\)
E. \(12 \sqrt{2}\)

Attachment:
2016-01-17_2250.png


This question is tailormade for approximations and not employing any particular formulae.

The triangle in inscribed inside the semicircle. Thus the perimeter of the triangle will be just less than (half circumference of the circle + 2r)

----> Perimeter of triangle < \(3\pi+2r\) ---> P< 9+6 ---> P < 15 (assuming \(\pi \approx\)3)

Analyse the options keeping in mind \(\sqrt{2}=1.4\) and \(\sqrt{3} =1.7\)

A. 9 . Too small. Eliminate.
B. \(6 + 3 \sqrt{2}\). 10. Too small. Eliminate.
C. \(9 + 3 \sqrt{3}\). 14. Relatively small. Eliminate.
D. \(6 + 6 \sqrt{2}\). \(\approx\) 14.4. This should be it.
E. \(12 \sqrt{2}\). 18. Too large. Eliminate.

In order to eliminate 1 out of C and D, I went with which one will you get out of \(\sqrt{2}\)or \(\sqrt{3}\). You can get \(\sqrt{2}\) as one of the multipliers as triangle ABC is an iscosceles traingle while for getting \(\sqrt{3}\), you need to have the 30-60-90 triangle which is not the case in this question.

D is thus the correct answer.

Hope this helps.
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Re: In the preceding figure, what is the perimeter of ∆ABC inscribed withi  [#permalink]

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New post 20 Jan 2016, 09:42
AO=OB=radious = 3 units
So from from Right angle triangle , height and base are equal in length.
So ABO=45 Degree
Similarly , Angle CBO=45 Dgree
So, the angle ABC is a right angle and triangle ABC is a right angle triangle with its base and height =3\sqrt{2}
so paremeter = 2x3\sqrt{2}+6

Hence Answer is D
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Re: In the preceding figure, what is the perimeter of ∆ABC inscribed withi  [#permalink]

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New post 20 Jan 2016, 11:53
AC = Diameter = 6
AO = OB = AC = 3
AB and AC = 3sqrt2 (using Pythagoras theorem)

Perimeter = 6 + 2*3sqrt2 = 6 + 6sqrt2
Option D is the correct answer.
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Re: In the preceding figure, what is the perimeter of ∆ABC inscribed withi  [#permalink]

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Re: In the preceding figure, what is the perimeter of ∆ABC inscribed withi   [#permalink] 29 Jan 2019, 16:30
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In the preceding figure, what is the perimeter of ∆ABC inscribed withi

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