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# Planifying the lateral surface area of a cone, we get the circular sec

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GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Planifying the lateral surface area of a cone, we get the circular sec  [#permalink]

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28 Feb 2019, 08:04
00:00

Difficulty:

55% (hard)

Question Stats:

53% (02:18) correct 47% (02:30) wrong based on 15 sessions

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GMATH practice exercise (Quant Class 11)

Planifying the lateral surface area of a cone, we get the circular sector with center O and radius 18 shown in the figure. Which of the following is closest to the length of the height of this cone?

(A) 12
(B) 14
(C) 16
(D) 18
(E) 20

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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
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Joined: 25 Feb 2019
Posts: 336
Re: Planifying the lateral surface area of a cone, we get the circular sec  [#permalink]

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28 Feb 2019, 09:21
IMO C .

join the two end points of the circular area. You get a chord here .

Draw a perpendicular (h)on the chord from center O.
now we get two congruent triangles in which one angle will be 80.

So , by trigonometry ,

we get sin80=h/18

h = sin80*18
Now we know that Sin90 =1

So Sin80 will be little less than 1 which makes it appear less than 18 but more than 14 .
So answer should be 16 .
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: Planifying the lateral surface area of a cone, we get the circular sec  [#permalink]

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28 Feb 2019, 12:47
fskilnik wrote:
GMATH practice exercise (Quant Class 11)

Planifying the lateral surface area of a cone, we get the circular sector with center O and radius 18 shown in the figure. Which of the following is closest to the length of the height of this cone?

(A) 12
(B) 14
(C) 16
(D) 18
(E) 20

$$?\,\, \cong \,\,h$$

$${\rm{circular}}\,\,{\rm{base}}\,{\rm{of}}\,\,{\rm{cone}}\,\,:\,\,\,\,2\pi r\,\, = \,\,{{{{160}^ \circ }} \over {{{360}^ \circ }}}\left( {2\pi \cdot 18} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,r = {{16 \cdot 18} \over {36}} = 8$$

$$\left( {{\text{Pythagoras}}} \right)\,\,\,\,{h^2} = {18^2} - {8^2} = \underleftrightarrow {\left( {18 - 8} \right)\left( {18 + 8} \right)} = 260$$

$${h^2} \cong 256 = {16^2}\,\,\,\,\,\mathop \Rightarrow \limits^{h\, > \,\,0} \,\,\,\,\,? = 16$$

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Re: Planifying the lateral surface area of a cone, we get the circular sec   [#permalink] 28 Feb 2019, 12:47
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