GMAT Official Guide 2019 The figure above represents a

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

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. âˆš2

B. 2

C. 3

D. 4âˆš2

E. 6

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Source: Beat The GMAT — Problem Solving | Fri Jun 29, 2018 10:14 pm

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

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

BTGmoderatorDC 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. âˆš2

B. 2

C. 3

D. 4âˆš2

E. 6

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: hÂ² + 2Â² = 6Â²
Simplify: hÂ² + 4 = 36
So, we get: hÂ² = 32
This means h = âˆš32
Check the answer choices. . . âˆš32 is not among them.
Looks like we need to simplify âˆš32

We'll use the fact that âˆš(xy) = (âˆšx)(âˆšy)
So, âˆš32 = âˆš[(16)(2)] = (âˆš16)(âˆš2) = (4)(âˆš2) = 4âˆš2
Check the answer choices. . . D

Answer: D

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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ceilidh.erickson: GMAT Instructor; Posts: 2095; Joined: Tue Dec 04, 2012 3:22 pmFollowed by: 247 members; Thanked: 1443 times

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:

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:

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.

Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education

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ceilidh.erickson: GMAT Instructor; Posts: 2095; Joined: Tue Dec 04, 2012 3:22 pmFollowed by: 247 members; Thanked: 1443 times

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:

Since side lengths in similar triangles are proportional, we can infer:
(4/h) = (h/8)
hÂ² = 32
h = âˆš32
h = 4âˆš2

Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education

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hallock: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Mon Jul 02, 2018 5:17 pm

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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Shahrukh_mbabreakspace: Senior | Next Rank: 100 Posts; Posts: 32; Joined: Mon Jul 02, 2018 1:56 amFollowed by: 2 members

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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Scott@TargetTestPrep: GMAT Instructor; Posts: 8138; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Wed Jul 04, 2018 6:29 pm

BTGmoderatorDC 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. âˆš2

B. 2

C. 3

D. 4âˆš2

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 = âˆš16 x âˆš2 = 4âˆš2

Answer: D

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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