BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Sphere Problem

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 460
Joined: Sun Mar 25, 2007 7:42 am
Thanked: 27 times
by samirpandeyit62 » Mon Nov 26, 2007 12:37 am
The answer choices are not very legible to be, but I would solve it as follows

Now min lenght from a vertice to a pt on the sphere will be when moving diagonally from a vertice & the diameter of sphere is 10

diagonal of base of square is rt(2) *10

so diagonal from side is rt(10^2 + 10*rt(2) ^2)

= 10* rt(3)

so min lenght is (10*rt(3) - 10)/2 = 5(rt(3) -1)
Regards
Samir
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Legendary Member
Posts: 645
Joined: Wed Sep 05, 2007 4:37 amFollowed by: 5 members
Location: India
Thanked: 34 times

by camitava » Mon Nov 26, 2007 1:31 am
Sorry Samir! Probably I am missing something. I am getting some different answer at all.
Refer the figure below -

Image
We get OA = 5sqrt(2)
Now OD = 5
So AD = OA - OD = 5(sqrt(2) - 1)

I am not getting why you are talking about 5(sqrt(3) - 1)
Correct me If I am wrong


Regards,

Amitava
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 645
Joined: Wed Sep 05, 2007 4:37 amFollowed by: 5 members
Location: India
Thanked: 34 times

by camitava » Mon Nov 26, 2007 1:32 am
Sorry Samir! Probably I am missing something. I am getting some different answer at all.
Refer the figure below -

We get OA = 5sqrt(2)
Now OD = 5
So AD = OA - OD = 5(sqrt(2) - 1)

I am not getting why you are talking about 5(sqrt(3) - 1)

Image
Correct me If I am wrong


Regards,

Amitava
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 460
Joined: Sun Mar 25, 2007 7:42 am
Thanked: 27 times

by samirpandeyit62 » Mon Nov 26, 2007 1:46 am
Hey Amitava the figures mentioned in the Q are 3D figures, ie cube, sphere, & diagonal of a cube is rt(3)*s (lenght of side)
Regards
Samir
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 645
Joined: Wed Sep 05, 2007 4:37 amFollowed by: 5 members
Location: India
Thanked: 34 times

by camitava » Mon Nov 26, 2007 1:51 am
Shi...! A miss again ... God pls help meeeeeeeeeeeeeee :cry: :oops:
Correct me If I am wrong


Regards,

Amitava
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 195
Joined: Sun Oct 21, 2007 4:33 am
Thanked: 10 times

by sankruth » Mon Nov 26, 2007 3:07 am
The distance between 2 opposite vertices of the cube =
SqRt(10^2 + 10^2 + 10^2) = 10*SqRt(3)

Since the Sphere is inscribed in the cude, radius = 5 and the centre will be the midpoint of the line joining the two opposite vertices i.e. [10SqRt(3)/2] = 5SqRt(3)

So, Shortest distance between the surface of Sphere and vertex of square = 5SqRt(3) - 5 = 5[SqRt(3)-1]

Hope that helps!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1