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The figure above represents a semicircular archway over a flat street.

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The figure above represents a semicircular archway over a flat street.  [#permalink]

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New post 22 Jun 2018, 00:30
22
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A
B
C
D
E

Difficulty:

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Question Stats:

91% (01:33) correct 9% (01:53) wrong based on 761 sessions

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The figure above represents a semicircular archway over a flat street. The semicircle has a center at O and a radius of 6 feet. What is the height h, in feet, of the archway 2 feet from its center?


A. √2

B. 2

C. 3

D. 4√2

E. 6



NEW question from GMAT® Official Guide 2019


(PS05957)


Attachment:
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The figure above represents a semicircular archway over a flat street.  [#permalink]

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New post 22 Jun 2018, 01:41
Let's disturb Pythagoras once again:

\(h^2+2^2=6^2\)
h=\(\sqrt{32}\)

In My Opinion
Ans: D
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Re: The figure above represents a semicircular archway over a flat street.  [#permalink]

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New post 22 Jun 2018, 05:24
3

Solution



Given:
    • The figure depicts a semi-circular archway over a flat street
    • The semi-circle has a center at O and a radius of 6 feet

To find:
    • The length of the height h, in feet, of the archway 2 feet from its center

Approach and Working:

If we add the point O and A, we get a right-angled triangle OAB, right-angled at point B and hypotenuse AO
Image

As the radius is 6 feet, we can say
    • AO = 6 feet

Applying Pythagoras Theorem in triangle OAB, we can write:
    • \(AO^2 = AB^2 + BO^2\)
    Or, \(6^2 = h^2 + 2^2\)
    Or, \(h^2 = 36 – 4 = 32\)
    Or, \(h = 4\sqrt{2}\)

Hence, the correct answer is option D.

Answer: D
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Re: The figure above represents a semicircular archway over a flat street.  [#permalink]

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New post 23 Jun 2018, 09:52
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Top Contributor
Bunuel wrote:
Image
The figure above represents a semicircular archway over a flat street. The semicircle has a center at O and a radius of 6 feet. What is the height h, in feet, of the archway 2 feet from its center?


A. √2

B. 2

C. 3

D. 4√2

E. 6



NEW question from GMAT® Official Guide 2019


(PS05957)


Attachment:
PS05957_f009.jpg


Let's add the radius of 6 feet to the diagram to get:
Image

From here, we can see the right triangle hiding within the diagram, which means we can apply the Pythagorean Theorem.
Se can write: h² + 2² = 6²
Simplify: h² + 4 = 36
So, we get: h² = 32
This means h = √32
Check the answer choices. . . √32 is not among them.
Looks like we need to simplify √32

We'll use the fact that √(xy) = (√x)(√y)
So, √32 = √[(16)(2)] = (√16)(√2) = (4)(√2) = 4√2
Check the answer choices. . . D

Answer: D

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Re: The figure above represents a semicircular archway over a flat street.  [#permalink]

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New post 25 Jun 2018, 12:06
2
Bunuel wrote:
Image
The figure above represents a semicircular archway over a flat street. The semicircle has a center at O and a radius of 6 feet. What is the height h, in feet, of the archway 2 feet from its center?


A. √2

B. 2

C. 3

D. 4√2

E. 6


Since the radius is 6 feet, we can create a right triangle with hypotenuse of 6, leg of 2, and leg of h; thus, we have:

2^2 + h^2 = 6^2

4 + h^2 = 36

h^2 = 32

h = √16 x √2 = 4√2

Answer: D
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Re: The figure above represents a semicircular archway over a flat street.  [#permalink]

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New post 02 Aug 2018, 13:30
if the radius is 6 max height of the arch can be 6 from the center of semi-circle. So height asked in question which is two feet away from center is definitely less than 6 . Also point 2 feet away from center to end of semicircle the dist is 4 so height is little more than 4 . So the only answer choice that matches it is D
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Re: The figure above represents a semicircular archway over a flat street.  [#permalink]

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New post 23 Sep 2019, 05:19
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Re: The figure above represents a semicircular archway over a flat street.   [#permalink] 23 Sep 2019, 05:19
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