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Math Expert V
Joined: 02 Sep 2009
Posts: 60688
The figure above represents a semicircular archway over a flat street.  [#permalink]

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25 00:00

Difficulty:   5% (low)

Question Stats: 91% (01:34) correct 9% (01:57) wrong based on 868 sessions

### HideShow timer Statistics 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 [ 34.62 KiB | Viewed 11269 times ]

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e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3230
Re: The figure above represents a semicircular archway over a flat street.  [#permalink]

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4
1

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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##### General Discussion
Senior Manager  P
Joined: 13 Feb 2018
Posts: 494
GMAT 1: 640 Q48 V28
The figure above represents a semicircular archway over a flat street.  [#permalink]

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Let's disturb Pythagoras once again:

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

In My Opinion
Ans: D
GMAT Club Legend  V
Joined: 12 Sep 2015
Posts: 4234
Re: The figure above represents a semicircular archway over a flat street.  [#permalink]

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4
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

RELATED VIDEO (simplifying roots)

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Target Test Prep Representative V
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Posts: 9144
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Re: The figure above represents a semicircular archway over a flat street.  [#permalink]

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2
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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Senior Manager  P
Joined: 10 Apr 2018
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Location: United States (NC)
Re: The figure above represents a semicircular archway over a flat street.  [#permalink]

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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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Joined: 21 Aug 2019
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The figure above represents a semicircular archway over a flat street.  [#permalink]

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GMATPrepNow wrote:
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

RELATED VIDEO (simplifying roots)

Can you please tell me, why can't we do like this?

H^2 = P^ + B^2
H^2=36 + 4
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I don't believe in giving up!
GMAT Club Legend  V
Joined: 12 Sep 2015
Posts: 4234
Re: The figure above represents a semicircular archway over a flat street.  [#permalink]

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Top Contributor
Farina wrote:
Can you please tell me, why can't we do like this?

H^2 = P^ + B^2
H^2=36 + 4

Your solution suggests that 6 is the length a leg and h is the hypotenuse.
However, 6 is the hypotenuse since it's the side opposite the 90-degree angle.
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Joined: 21 Aug 2019
Posts: 21
Location: Pakistan
GPA: 3.5
Re: The figure above represents a semicircular archway over a flat street.  [#permalink]

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GMATPrepNow wrote:
Farina wrote:
Can you please tell me, why can't we do like this?

H^2 = P^ + B^2
H^2=36 + 4

Your solution suggests that 6 is the length a leg and h is the hypotenuse.
However, 6 is the hypotenuse since it's the side opposite the 90-degree angle.

Yes i know the concept of hypotenuse, i got confused with "h" written in diagram, i thought it was for hypotenuse. Height is perpendicular and it is unknown. Thanks a lot for your help

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I don't believe in giving up! Re: The figure above represents a semicircular archway over a flat street.   [#permalink] 26 Jan 2020, 14:52
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