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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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21 Jun 2018, 23:30
14
00:00

Difficulty:

5% (low)

Question Stats:

89% (01:37) correct 11% (01:40) wrong based on 396 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)

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

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22 Jun 2018, 00:41
Let's disturb Pythagoras once again:

$$h^2+2^2=6^2$$
h=$$\sqrt{32}$$

In My Opinion
Ans: D
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Posts: 2269
Re: The figure above represents a semicircular archway over a flat street.  [#permalink]

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22 Jun 2018, 04:24

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

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.

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

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23 Jun 2018, 08:52
Top Contributor
Bunuel wrote:

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:

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

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

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25 Jun 2018, 11:06
Bunuel wrote:

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

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

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02 Aug 2018, 12:30
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
Re: The figure above represents a semicircular archway over a flat street. &nbs [#permalink] 02 Aug 2018, 12:30
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