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GMAT Official Guide 2019 The figure above represents a

by BTGmoderatorDC » Fri Jun 29, 2018 10:14 pm

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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
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by Brent@GMATPrepNow » Sat Jun 30, 2018 4:36 am
BTGmoderatorDC 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
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

Cheers,
Brent
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by ceilidh.erickson » Mon Jul 02, 2018 10:40 am
Brent provided a nice solution, but there's also an easy way to GUESS on this problem:

Since h is not the radius but is perpendicular to the diameter, we can infer that it's shorter than the radius:
Image

Therefore, we can eliminate E.

We can also infer that h must be greater than 4, since the angle that's across from must be greater than 45 degrees. Compare these two:

Image

Image

We don't have to calculate the angle of x to see that it's opening wider than 45*. Thus angle y must be less than 45 degrees. Since larger sides are across from larger angles, h must be larger than 4.

Eliminate any answer choices less than 4: A, B, and C.

The only thing left must be the answer: D.
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by ceilidh.erickson » Mon Jul 02, 2018 10:50 am
For another exact approach, you could use SIMILAR TRIANGLES:

We can create 2 similar right triangles inside the semicircle using h as a side length for each:

Image

Since side lengths in similar triangles are proportional, we can infer:
(4/h) = (h/8)
h² = 32
h = √32
h = 4√2
Ceilidh Erickson
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Harvard Graduate School of Education
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by hallock » Mon Jul 02, 2018 5:33 pm
Another way to solve without doing any math:

-You know h < 6; the height above the center of the circle is the highest point
-Draw a circle with radius 4 from the bottom of line h. It is a smaller arc and won't reach the top of the arc where "h" is, so you know h > 4
-Only option for 4 < h < 6 is D
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by Shahrukh_mbabreakspace » Tue Jul 03, 2018 1:25 am
Join the centre and the point on circumference, its length is 6 feet(radius)
Now look at the triangle formed, it is a right angled triangle, with hypotenues being 6, base 2.
Now hieght will be sqrt(36-4)= 4*sqrt(2)
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by Scott@TargetTestPrep » Wed Jul 04, 2018 6:29 pm
BTGmoderatorDC 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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