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

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

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

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Re: The figure above represents a semicircular archway over a flat street. [#permalink]
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]
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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]
Bunuel can we do approximation here

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