NEW HERE? LEARN MORE
closeclose
Last visit was: 24 Apr 2024, 19:08 It is currently 24 Apr 2024, 19:08
Decision Tracker
My Rewards
New posts
New comers' posts

Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

What is the circumference of the semicircle in the figure shown?

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92900
Own Kudos [?]: 618823 [0]
Given Kudos: 81588
Send PM
CAT Tests Forum Quiz Mode
What is the circumference of the semicircle in the figure shown? [#permalink] New post   17 Nov 2020, 01:15
Expert Reply
00:00
A
B
C
D
E

Difficulty:

45% (medium)

Question Stats:

64% (01:57) correct 36% (02:15) wrong based on 98 sessions
Hide Show timer Statistics

What is the circumference of the semicircle in the figure shown?


A. \(14\pi\sqrt{6}\)

B. \(7\pi\sqrt{6}\)

C. \(\frac{7\pi\sqrt{6}}{2}\)

D. \(\frac{7\pi\sqrt{6}}{3}\)

E. \(\frac{7\pi\sqrt{6}}{4}\)


Project PS Butler


Subscribe to get Daily Email - Click Here | Subscribe via RSS - RSS



Show Spoiler
Attachment:
2020-10-27_14-08-58.png
2020-10-27_14-08-58.png [ 30.13 KiB | Viewed 2567 times ]
ShowHide Answer
Official Answer
C
User avatar
G
Pankaj0901
Senior Manager
Senior Manager
Joined: 18 Dec 2018
Posts: 425
Own Kudos [?]: 43 [1]
Given Kudos: 738
Location: India
WE:Account Management (Hospitality and Tourism)
Send PM
Re: What is the circumference of the semicircle in the figure shown? [#permalink] New post   17 Nov 2020, 02:19
1
Kudos
Length of side of an isosceles triangle with angles 45 deg = 14 cos30 = 14 \sqrt{3} /2 = 7 \sqrt{3} = the other equal side of an isosceles triangle
The hypotenuse of this isosceles triangle = Diameter of the semi-circle = \sqrt{2} x 7 \sqrt{3} = 7 \sqrt{6}

Radius of semi circle = 7 \sqrt{6} / 2
Circumference of semi-circle = pi x 7 \sqrt{6} / 2 = \(\frac{7\pi\sqrt{6}}{2}\)

ANSWER C IMO

Bunuel wrote:

What is the circumference of the semicircle in the figure shown?


A. \(14\pi\sqrt{6}\)

B. \(7\pi\sqrt{6}\)

C. \(\frac{7\pi\sqrt{6}}{2}\)

D. \(\frac{7\pi\sqrt{6}}{3}\)

E. \(\frac{7\pi\sqrt{6}}{4}\)


Project PS Butler


Subscribe to get Daily Email - Click Here | Subscribe via RSS - RSS



Show Spoiler
Attachment:
2020-10-27_14-08-58.png
User avatar
P
mdsaddamforgmat
Manager
Manager
Joined: 05 Dec 2019
Posts: 114
Own Kudos [?]: 154 [1]
Given Kudos: 155
Location: Nepal
Schools: Tuck '23
GMAT 1: 420 Q33 V15
GMAT 2: 650 Q48 V31
GMAT 3: 640 Q47 V31
Send PM
Re: What is the circumference of the semicircle in the figure shown? [#permalink] New post   17 Nov 2020, 03:31
1
Kudos
IMO: C

Top Triangle's is of 30,60,90 deg property hence, Side will be in the ration of 1:Sqrt3:2
therefore, Base = 14 x (sqrt3)/2 = 7 x (sqrt3)
Therefore hypotenuse of below triangle in diagram = 7 x (sqrt6) (note: the lower triangle is isosceles right-angle triangle hence, we can apply Pythagoras theorem)

Therefore Dia of Circle= 7 x (sqrt6)
Hence, circumference of semi cirlce = (pi x Dia)/2 = Pi x 7 x (sqrt6) /2
ANSWER: C
User avatar
L
akadiyan
Retired Moderator
Joined: 31 May 2017
Posts: 749
Own Kudos [?]: 670 [2]
Given Kudos: 53
Concentration: Technology, Strategy
Schools: IIMA PGPX'23 Bayes Said '23
Send PM
GMAT ToolKit User Reviews Badge
Re: What is the circumference of the semicircle in the figure shown? [#permalink] New post   17 Nov 2020, 19:39
1
Kudos
1
Bookmarks
Triangle 1:
The triangle on top is 30-60-90 triangle
So, we know the ratio of sides of the triangle is \(x:x\sqrt{3}:2x\)

From the triangle 1 on top, we know the side 2x=14, so x=7

Base of triangle 1 = \(7\sqrt{3}\)

Triangle 2:
The triangle in the bottom is 45-45-90 triangle
So, we know the ratio of sides fof the triangle is x:x:\(x\sqrt{2}\)
The hypotenuse of the triangle with side \(x\sqrt{2}\) = \(7\sqrt{3}\)*\(\sqrt{2}\)

The hypotenuse of the triangle 2 = \(7 \sqrt{6}\)

Circle:
The hypotenuse of the triangle is the diameter of the circle.
Diameter = \(7 \sqrt{6}\)
Circumference = d\(\pi\) = \(7 \sqrt{6}*\pi\)

Circumference of semicircle = \( (7 \sqrt{6}*\pi) / 2\)

Ans: C
User avatar
B
Abir77
Intern
Intern
Joined: 06 Nov 2019
Posts: 24
Own Kudos [?]: 12 [0]
Given Kudos: 79
Location: Bangladesh
Send PM
Re: What is the circumference of the semicircle in the figure shown? [#permalink] New post   20 Nov 2020, 10:37
Bunuel wrote:

What is the circumference of the semicircle in the figure shown?


A. \(14\pi\sqrt{6}\)

B. \(7\pi\sqrt{6}\)

C. \(\frac{7\pi\sqrt{6}}{2}\)

D. \(\frac{7\pi\sqrt{6}}{3}\)

E. \(\frac{7\pi\sqrt{6}}{4}\)


Project PS Butler


Subscribe to get Daily Email - Click Here | Subscribe via RSS - RSS



Show Spoiler
Attachment:
2020-10-27_14-08-58.png


Bunuel Please help me out here

Isn't circumference for semi circle = pi*r+2r ???
User avatar
L
ScottTargetTestPrep
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 18756
Own Kudos [?]: 22049 [0]
Given Kudos: 283
Location: United States (CA)
Send PM
Re: What is the circumference of the semicircle in the figure shown? [#permalink] New post   28 Nov 2020, 13:33
Expert Reply
Bunuel wrote:

What is the circumference of the semicircle in the figure shown?


A. \(14\pi\sqrt{6}\)

B. \(7\pi\sqrt{6}\)

C. \(\frac{7\pi\sqrt{6}}{2}\)

D. \(\frac{7\pi\sqrt{6}}{3}\)

E. \(\frac{7\pi\sqrt{6}}{4}\)



Solution:

Since the top triangle is a 30-60-90 right triangle and its hypotenuse is 14, its long leg (the one that is opposite 60-degree angle) is 14/2 * √3 = 7√3. Since that leg is also a leg of the 45-45-90 right triangle at the bottom, the hypotenuse of the 45-45-90 right triangle is 7√3 * √2 = 7√6. Since that hypotenuse is also the diameter of the semicircle, the length of its semicircular arc is ½ * 7√6 * π = 7π√6/2.

Answer: C
GMAT Club Bot
gmatclubot
Re: What is the circumference of the semicircle in the figure shown? [#permalink]
28 Nov 2020, 13:33
Moderators:
Bunuel
Math Expert
92900 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions