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

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Bunuel
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The figure above represents a semicircular archway over a flat street. [#permalink] New post   22 Jun 2018, 00:30
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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

(PS05957)

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EgmatQuantExpert
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Re: The figure above represents a semicircular archway over a flat street. [#permalink] New post   22 Jun 2018, 05:24
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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.

Answer: D
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BrentGMATPrepNow
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Re: The figure above represents a semicircular archway over a flat street. [#permalink] New post   Updated on: 15 Mar 2020, 05:36
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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)


Show Spoiler
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

Answer: D

Cheers,
Brent

Originally posted by BrentGMATPrepNow on 23 Jun 2018, 09:52.
Last edited by BrentGMATPrepNow on 15 Mar 2020, 05:36, edited 2 times in total.
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Re: The figure above represents a semicircular archway over a flat street. [#permalink] New post   25 Jun 2018, 12:06
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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

Answer: D
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Re: The figure above represents a semicircular archway over a flat street. [#permalink] New post   26 Jan 2020, 11:19
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)


Show Spoiler
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

Answer: 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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BrentGMATPrepNow
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Re: The figure above represents a semicircular archway over a flat street. [#permalink] New post   26 Jan 2020, 14:44
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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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Re: The figure above represents a semicircular archway over a flat street. [#permalink] New post   09 May 2021, 06:43
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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



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Re: The figure above represents a semicircular archway over a flat street. [#permalink] New post   16 Mar 2023, 08:43
why is not possible to solve this using 1:sqr3:2

if we know that the triangles are right triangles
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Re: The figure above represents a semicircular archway over a flat street. [#permalink] New post   16 Mar 2023, 08:59
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Maria240895chile wrote:
why is not possible to solve this using 1:sqr3:2

if we know that the triangles are right triangles


The sides are in 1:√3:2 ratio only in a 30°-60°-90° triangle, which is not the case here.
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Re: The figure above represents a semicircular archway over a flat street. [#permalink] New post   19 Jun 2023, 22:31
Bunuel can we do approximation here

We know height can't be 6, it will be something less than 6
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Re: The figure above represents a semicircular archway over a flat street. [#permalink]
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