Last visit was: 14 Jun 2024, 02:02 It is currently 14 Jun 2024, 02:02
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.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Show Tags
Hide Tags
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 10147
Own Kudos [?]: 16830 [17]
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
Send PM
Current Student
Joined: 25 May 2013
Posts: 22
Own Kudos [?]: 25 [4]
Given Kudos: 78
Location: India
Concentration: Finance, Strategy
GMAT 1: 710 Q49 V38
GPA: 2.8
WE:Account Management (Consulting)
Send PM
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 10147
Own Kudos [?]: 16830 [1]
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
Send PM
Manager
Manager
Joined: 03 Jun 2013
Posts: 72
Own Kudos [?]: 9 [0]
Given Kudos: 1
Send PM
Re: As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink]
Where are the assumption provided that FC=EC=BC–BE=BC–BD, is it a rule .
Intern
Intern
Joined: 18 Dec 2018
Posts: 13
Own Kudos [?]: 1 [0]
Given Kudos: 26
Send PM
As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink]
Hello Moderators,

Please confirm if my assumption and understanding to the solution provided by Math Revolution are correct.

Concept - Length of the two tangents, only two possible, drawn from the same point are equal.

Points from which the tangents are drawn are A,B,C

Tangents from A : - AD and AF ; AD = AF = 3

Tangents from B :- BD and BE ; BD = BE = 5

Tangents from C :- CE and CF; CE = CF = 4 (BC = 9 ; BE = 5; EC = FC = (9-5 = 4)

Total length of AC = AF+FC = 3+4 = 7


Thanks

Originally posted by adizephyr on 15 Feb 2020, 08:54.
Last edited by adizephyr on 15 Feb 2020, 09:18, edited 1 time in total.
Intern
Intern
Joined: 20 Jul 2015
Posts: 24
Own Kudos [?]: 18 [0]
Given Kudos: 7
Send PM
Re: As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink]
DUtkarsh wrote:
Considering the radius of Incircle as r. Know that DI, EI and FI are perpendiculars on their respective sides.
In triangles BDI and BEI,
BI2 + DI2 = BI2 = BE2 + EI2
BI2 + DI2 = BE2 + EI2
BI2 = BE2 (Since, DI = EI = r)
Hence, BI = BE = 5
Therefore, EC = 9 – 5 = 4

Similarly, using triangles ADI and AFI
AD = AF = 3
And, using triangles CEI and CFI
EC = CF = 4

Solution = AF + CF = 3 + 4 = 7 i.e. Option E.


DUtkarsh

can u please elaborate above highlighted portion....
Current Student
Joined: 25 May 2013
Posts: 22
Own Kudos [?]: 25 [1]
Given Kudos: 78
Location: India
Concentration: Finance, Strategy
GMAT 1: 710 Q49 V38
GPA: 2.8
WE:Account Management (Consulting)
Send PM
Re: As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink]
1
Kudos
Rohit2015 wrote:
DUtkarsh wrote:
Considering the radius of Incircle as r. Know that DI, EI and FI are perpendiculars on their respective sides.
In triangles BDI and BEI,
BI2 + DI2 = BI2 = BE2 + EI2
BI2 + DI2 = BE2 + EI2
BI2 = BE2 (Since, DI = EI = r)
Hence, BI = BE = 5
Therefore, EC = 9 – 5 = 4

Similarly, using triangles ADI and AFI
AD = AF = 3
And, using triangles CEI and CFI
EC = CF = 4

Solution = AF + CF = 3 + 4 = 7 i.e. Option E.


DUtkarsh

can u please elaborate above highlighted portion....


Rohit2015 - Thanks for pointing out the typo, it'd be BD and not BI in the first half of the equation. I hope it is clear now, and I have also updated the aforementioned post.
Manager
Manager
Joined: 16 Jul 2018
Posts: 213
Own Kudos [?]: 68 [0]
Given Kudos: 261
Send PM
Re: As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink]
MathRevolution wrote:
=>

Attachment:
2.3ps(a).png


Since \(AD = AF,\) we have \(AF = 3.\)

Since \(BC = 9\) and \(BD = 5,\) we have \(FC = EC = BC – BE = BC – BD = 9 – 5 = 4.\)

Then we have \(AC = AF + FC = 3 + 4 = 7.\)

Therefore, E is the answer.
Answer: E


Hello MathRevolution,
could you please explain why AD=AF and FC=BC?
Intern
Intern
Joined: 04 Feb 2023
Posts: 12
Own Kudos [?]: [0]
Given Kudos: 33
Send PM
Re: As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink]
adizephyr

The tangents drawn from an external point to a circle are equal.
GMAT Club Bot
Re: As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink]
Moderator:
Math Expert
93707 posts